{"id":1704,"date":"2015-12-26T03:46:45","date_gmt":"2015-12-26T03:46:45","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1704"},"modified":"2024-08-20T10:25:34","modified_gmt":"2024-08-20T10:25:34","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-campo-eletrico","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletrostatica\/campo-eletrico\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-campo-eletrico\/","title":{"rendered":"Campo el\u00e9trico – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/b><\/span><\/span><\/span><\/p>\n

Campo el\u00e9trico<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>I. Certo\u00a0 —\u00a0 s\u00f3 surge for\u00e7a sobre a carga se existir campo el\u00e9trico<\/b><\/span><\/span><\/p>\n

II. Certo\u00a0 —\u00a0 se n\u00e3o tiver carga, pode existir o campo el\u00e9trico sem existir for\u00e7a el\u00e9trica.<\/b><\/span><\/span><\/p>\n

III. Errado —\u00a0 n\u00e3o surgir\u00e1 for\u00e7a sobre a carga se n\u00e3o houver campo el\u00e9trico<\/b><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>II e IV s\u00e3o falsas\u00a0 —\u00a0 os dois vetores devem ter suas origens no ponto P, onde est\u00e1 q\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) E=F\/q\u00a0 —\u00a0 5.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>=F\/1,5.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F=7,5.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

b) Se move na dire\u00e7\u00e3o e sentido da for\u00e7a el\u00e9trica que, no caso, \u00e9 horizontal e para a direita, pois q \u00e9 positiva.<\/b><\/span><\/span><\/p>\n

c) Nesse caso, o campo el\u00e9trico\"\"\u00a0e a for\u00e7a el\u00e9trica\u00a0\"\"\u00a0ter\u00e3o sentidos contr\u00e1rios e q se desloca no sentido da for\u00e7a, ou seja,para a esquerda.<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Colocando nessa regi\u00e3o uma pequena carga el\u00e9trica de prova e verificar que se sobre ela surgir uma for\u00e7a el\u00e9trica, nessa regi\u00e3o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

existe um campo el\u00e9trico, caso contr\u00e1rio, n\u00e3o existe.<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>E=F\/q\u00a0 —\u00a0 E=10\/2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>E=5.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>N\/C\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 como q>0,\u00a0\"\"\u00a0e\u00a0\"\"tem mesmo sentido\u00a0 —\u00a0 sentido para o<\/b><\/span><\/span>\u00a0norte.<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>E=F\/q=1,2\/4.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E=3.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/C\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe na express\u00e3o E=KQ\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0que E \u00e9 inversamente proporcional ao quadrado da dist\u00e2ncia\u00a0 —\u00a0 se a dist\u00e2ncia \u00e9 dobrada, E fica 2<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=4vezes menor\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Veja esquema abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- A<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0E=F\/q\u00a0 —\u00a0 2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=18.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\/q\u00a0 —\u00a0 q=9,0.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span> R- E<\/b><\/span><\/span>\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Se q>0,\u00a0\"\"\u00a0e\u00a0\"\"\u00a0tem a mesma dire\u00e7\u00e3o e o mesmo sentido\u00a0 —\u00a0 E=F\/q\u00a0 —\u00a0 5=F\/3.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 intensidade\u00a0<\/b><\/span><\/span>F=1,5.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span>\u00a0 —\u00a0 dire\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0—\u00a0<\/b><\/span><\/span>horizontal\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 sentido\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0para a esquerda<\/b><\/span><\/span><\/p>\n

b) q<0,\u00a0\"\"\u00a0e\u00a0\"\"\u00a0tem a mesma dire\u00e7\u00e3o e sentidos opostos —\u00a0 E=F\/q\u00a0 —\u00a0 5=F\/3.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 intensidade\u00a0<\/b><\/span><\/span>F=1,5.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span>\u00a0 —\u00a0 dire\u00e7\u00e3o\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0horizontal\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 sentido\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0para a direita\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Veja esquema abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- D<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Veja esquema abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- E<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Cargas negativas criam campos de aproxima\u00e7\u00e3o e positivas, de afastamento.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- D<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>As duas cargas positivas originam no ponto P campos el\u00e9tricos de afastamento de intensidades\u00a0 —\u00a0 E<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=KQ<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/(2.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0\u00a0—<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.4.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\/4.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=9.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/C\u00a0 —\u00a0 E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\/(1,0)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=9.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/C\u00a0 —\u00a0 E<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=9.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0\u2013 9.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=0\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>O ponto P onde o campo el\u00e9trico resultante deve se anular deve estar \u00e0 direita de Q (observe figura)\u00a0 —\u00a0 E<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=K4Q\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=KQ\/(d \u2013 3)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 4KQ\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=KQ\/(d \u2013 3)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 (d \u2013 3)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=1\/4\u00a0 —\u00a0 (d \u2013 3)\/d=1\/2\u00a0 —\u00a0 d=6 unidades\u00a0 —\u00a0 \u00e9 nulo no ponto 10\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Cargas positivas criam campos de afastamento e negativas de aproxima\u00e7\u00e3o.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- C<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Cargas negativas originam campos de aproxima\u00e7\u00e3o e positivas, de afastamento. Veja figuras abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- B<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Prolongando os vetores\u00a0\"\"\u00a0e\u00a0\"\", a carga Q estar\u00e1 na intersec\u00e7\u00e3o de seus prolongamentos e ser\u00e1 positiva, pois os vetores s\u00e3o de afastamento.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) Observe na escala da figura que o vetor\u00a0\"\"\u00a0tem duas unidades e sua intensidade \u00e9 E<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=2.2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N\/C\u00a0 —\u00a0 a dist\u00e2ncia entre Q e\u00a0\"\"\u00a0\u00e9 de 4 unidades e de Q a M, onde est\u00e1 q \u00e9 de duas unidades, assim, E<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=2<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=4 vezes maior que E<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 E<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=4.4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=16.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N\/C, horizontal<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

e para a direita, pois a carga Q (positiva) cria em M um campo\u00a0 campo\u00a0\"\"\u00a0de afastamento\u00a0 —\u00a0 em M, onde est\u00e1 q=3.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 E<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=F\/q\u00a0 — 16.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=F\/3.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0F=4,8.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>N, horizontal e para a direita.<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- E<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Veja esquema abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- C<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Veja esquema abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- B
\n<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>O campo el\u00e9trico deve ter dire\u00e7\u00e3o vertical e sentido para cima, pois a carga \u00e9 positiva e a for\u00e7a deve tamb\u00e9m ser para cima<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0para anular a for\u00e7a peso que \u00e9 vertical e para baixo\u00a0 —\u00a0 F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 q.E=m.g\u00a0 —\u00a0 10.1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>.E=4.10<\/b><\/span><\/span>-10<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.10\u00a0 —\u00a0 E=4.10<\/b><\/span><\/span>-12<\/b><\/span><\/span><\/sup>\/16.10<\/b><\/span><\/span>-9<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E=0,25.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0N\/C\u00a0 — \u00a0E=2,5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0N\/C, dire\u00e7\u00e3o vertical e sentido para cima.<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe na figura abaixo, que o vetor campo el\u00e9trico resultante no centro da circunfer\u00eancia \u00e9<\/b><\/span><\/span>\u00a0nulo,\u00a0<\/b><\/span><\/span>pois o \u00e2ngulo entre cada<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0um deles \u00e9 de 120<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>. .<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe na express\u00e3o E=KQ\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0que o campo \u00e9 diretamente proporcional ao m\u00f3dulo de cada carga\u00a0 —\u00a0 como uma \u00e9 o dobro da outra, o campo por ela originado ser\u00e1 o dobro do campo gerado pela outra\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Veja esquema abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- B<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>01. Falsa\u00a0 —\u00a0 E=F\/q<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0 observe nessa express\u00e3o q<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=E\/F que E e F s\u00e3o diretamente proporcionais.<\/b><\/span><\/span><\/p>\n

02. Verdadeira\u00a0 —\u00a0 veja teoria.<\/b><\/span><\/span><\/p>\n

04. Verdadeira\u00a0 —\u00a0 E=KQ\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

08. Falsa\u00a0 —\u00a0 ter\u00e1 sentido contr\u00e1rio ao da for\u00e7a el\u00e9trica.<\/b><\/span><\/span><\/p>\n

16. Falsa\u00a0 — veja teoria<\/b><\/span><\/span><\/p>\n

R- (2 + 4) = 6<\/b><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Sendo a dist\u00e2ncia do ponto A a cada uma das cargas iguais e q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0voc\u00ea observa que E<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 E<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=Kq<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/(1.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0E<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\/1.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=18.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>N\/C\u00a0 —\u00a0 E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=9.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>N\/C\u00a0 —\u00a0 E<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=E<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>+ E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=(18.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ (9.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 — \u00a0intensidade\u00a0 -E<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=9\u221a5.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>N\/C\u00a0 —\u00a0 dire\u00e7\u00e3o – tg\u03b1=E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/E<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=1\/2, sendo \u03b1 o \u00e2ngulo que E<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\u00a0forma com o eixo x\u00a0 —\u00a0 sentido – se afastando da origem a partir do ponto A.<\/b><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A \u00fanica alternativa em que o objeto com carga positiva n\u00e3o \u00e9 desviado (for\u00e7a resultante nula) durante a queda \u00e9 a E<\/b><\/span><\/span><\/p>\n

R- E<\/b><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Quando E=400N\/C \u2013 d=0,10m\u00a0 —\u00a0 E=KQ\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 400=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.Q\/(10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Q=4.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\/9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Q\u22480,45.10<\/b><\/span><\/span>-9<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

Q \u2248 4,5.10<\/b><\/span><\/span>-10\u00a0<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

b) E=KQ\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.4,5.10<\/b><\/span><\/span>-10<\/b><\/span><\/span><\/sup>\/(3.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E=40,5.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\/9.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E=4,5.10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>E=45 N\/C<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Se existe campo el\u00e9trico uniforme, sobre a part\u00edcula surgir\u00e1 uma for\u00e7a e consequentemente uma acelera\u00e7\u00e3o constantes\u00a0 —<\/b><\/span><\/span><\/p>\n

R – C<\/b><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>E=F\/q\u00a0 —\u00a0 E=ma\/q\u00a0 —\u00a0 nesta express\u00e3o, como a massa e a carga s\u00e3o constantes, se voc\u00ea quadruplicar a intensidade do campo el\u00e9trico voc\u00ea estar\u00e1 quadruplicando a acelera\u00e7\u00e3o\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>Como<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>o<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>campo<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>el\u00e9trico<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>\u00e9<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>uniforme,<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>a<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>carga<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>fica<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>sujeita<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>a<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>uma<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>for\u00e7a<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>resultante<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>(el\u00e9trica)<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>constante,<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>adquirindo<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>um<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>movimento<\/b><\/span><\/span>\u2009<\/b><\/span><\/span>retil\u00edneo<\/b><\/span><\/span>\u2009<\/b><\/span><\/span><\/p>\n

Acelerado\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A carga deve ter sinal\u00a0<\/b><\/span><\/span>negativo<\/b><\/span><\/span>\u00a0para que a for\u00e7a el\u00e9trica seja contr\u00e1ria ao campo el\u00e9trico e possa equilibrar a for\u00e7a peso\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P=m.g=0,5.10\u00a0 —\u00a0 P=5N\u00a0 —\u00a0 E=F\/q\u00a0 —\u00a0 100=F\/q\u00a0 —\u00a0 100=P\/q+\u00a0 —\u00a0 100=5\/q\u00a0 —\u00a0<\/b><\/span><\/span>q=5.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0C<\/b><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A dire\u00e7\u00e3o e o sentido do a for\u00e7a el\u00e9trica est\u00e3o relacionadas com a dire\u00e7\u00e3o e o sentido do campo el\u00e9trico, no caso a for\u00e7a el\u00e9trica \u00e9 vertical e para cima (el\u00e9tron tem carga negativa)\u00a0 —<\/b><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>01. Falsa\u00a0 —\u00a0 um corpo eletrizado pode atrair um corpo neutro, por indu\u00e7\u00e3o eletrost\u00e1tica, nunca repelir.<\/b><\/span><\/span><\/p>\n

02. Correta\u00a0 —\u00a0 veja teoria de campo el\u00e9trico.<\/b><\/span><\/span><\/p>\n

04- Falsa\u00a0 —\u00a0 o campo el\u00e9trico gerado por uma esfera condutora \u00e9 de afastamento (Q>0) ou de aproxima\u00e7\u00e3o (Q<0), em suas vizinhan\u00e7as externas.<\/b><\/span><\/span><\/p>\n

08- Falsa\u00a0 —\u00a0 Observe na express\u00e3o F=KQq\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0que a for\u00e7a \u00e9 inversamente proporcional ao quadrado da dist\u00e2ncia.<\/b><\/span><\/span><\/p>\n

R- 02<\/b><\/span><\/span><\/p>\n

38-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0A placa da esquerda tem carga positiva\u00a0(campo de afastamento) fazendo com que ao penetrar no espa\u00e7o entre as placas, a carga positiva ser\u00e1 acelerada pela for\u00e7a el\u00e9trica em dire\u00e7\u00e3o \u00e0 placa negativa e, ao sair, a for\u00e7a deixar\u00e1 de existir e ela seguir\u00e1 com velocidade constante V, tal que V>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Como o el\u00e9tron tem carga negativa, quando ele penetrar na regi\u00e3o do campo el\u00e9trico surgir\u00e1 sobre ele uma for\u00e7a el\u00e9trica com sentido contr\u00e1rio ao do campo e esta for\u00e7a ir\u00e1 aceler\u00e1-lo at\u00e9 abandonar a regi\u00e3o do campo com velocidade V tal que V>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

R- A<\/b><\/span><\/span><\/p>\n

40-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Os n\u00eautrons n\u00e3o sofrem desvio e o campo el\u00e9trico desvia os pr\u00f3tons para baixo\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Para que a gota atinja P ela deve ser desviada pela for\u00e7a el\u00e9trica para cima e, como a gota tem carga negativa o campo el\u00e9trico deve ter sentido contr\u00e1rio ao da for\u00e7a, ou seja, para baixo\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

42-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

43-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A for\u00e7a resultante \u00e9 para baixo e a gota cai com acelera\u00e7\u00e3o a, tal que F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 P – F=m.0,2g\u00a0 —\u00a0 m.g- Q.E=m0,2g\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

mg \u2013 0,2mg=QE\u00a0 —\u00a0 Q=0,8mg\/Q\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

44-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>gota em equil\u00edbrio\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 F=P\u00a0 —\u00a0 q.E=m.g\u00a0 —\u00a0 1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>.E=1,6.10<\/b><\/span><\/span>-25<\/b><\/span><\/span><\/sup>.10\u00a0 —\u00a0 E=1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>\/1,6.10<\/b><\/span><\/span>-24<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>E=1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0N\/C<\/b><\/span><\/span><\/p>\n

45-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) F=ma\u00a0 —\u00a0 qE=ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=qE\/m<\/b><\/span><\/span><\/p>\n

b) Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a.\u0394S\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.qE\/m.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=\u221a(V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.q.(E\/m).a)
\n<\/b><\/span><\/span><\/p>\n

46-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) carga negativa\u00a0 —\u00a0 a for\u00e7a el\u00e9trica tem que estar para a esquerda em sentido oposto ao do campo el\u00e9trico.<\/b><\/span><\/span><\/p>\n

b) Colocando as for\u00e7as sobre a esfera que s\u00e3o: a tra\u00e7\u00e3o\"\"\u00a0no fio\u00a0 —\u00a0 a for\u00e7a el\u00e9trica\u00a0\"\"\u00a0horizontal e para a esquerda para manter a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0esfera em equil\u00edbrio\u00a0 —\u00a0 o peso\"\"\u00a0, vertical e para baixo\u00a0 —\u00a0 tg\u03b8=F\/P=qE\/mg\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03b8=arc tg qE\/mg<\/b><\/span><\/span><\/p>\n

47-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

48-<\/b><\/span><\/span><\/span> R- C\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 os dipolos est\u00e3o flutuando\u00a0 —\u00a0 as for\u00e7as peso e el\u00e9trica se anulam\u00a0 —\u00a0 observe que os dipolos ficar\u00e3o alinhados predominantemente na dire\u00e7\u00e3o vertical com as cargas negativas voltadas para baixo (repuls\u00e3o) e as positivas para cima (atraa\u00e7\u00e3o)<\/b><\/span><\/span><\/p>\n

49-<\/b><\/span><\/span><\/span> R- A\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

50-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>For\u00e7a resultante sobre a part\u00edcula dentro do campo el\u00e9trico \u00e9\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\u00a0= F<\/b><\/span><\/span>el\u00e9trica<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>gravitacional<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 m.a = q.E \u2013 m.g \u00a0—\u00a0 q.E = m.a + m.g \u00a0—\u00a0 E = (m\/q).(a + g)\u00a0 —\u00a0 na horizontal devido a aus\u00eancia de for\u00e7as o movimento \u00e9 uniforme ent\u00e3o\u00a0 —\u00a0 S = S<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ v.t\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

L = v.t\u00a0 —\u00a0 t = L\/v onde t \u00e9 o tempo para a part\u00edcula percorrer a dist\u00e2ncia L\u00a0 —\u00a0 na vertical devido \u00e0s for\u00e7as atuantes o movimento \u00e9 uniformemente variado\u00a0 —\u00a0 V<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>\u00a0= at = aL\/v\u00a0 —\u00a0 como a part\u00edcula dever\u00e1 sair do campo com alcance m\u00e1ximo fora dele, ela dever\u00e1 sair do campo em \u00e2ngulo de 45\u00b0, o que significa que a velocidade v<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>\u00a0= v\u00a0 —\u00a0 V<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>\u00a0= aL\/v = v\u00a0 —\u00a0 a = v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/L\u00a0 —\u00a0 E = (m\/q). (a + g)\u00a0 —\u00a0 \u00a0<\/b><\/span><\/span>E=(m\/q).(g + v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/L)<\/b><\/span><\/span><\/p>\n

51-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) F<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=20F<\/b><\/span><\/span>E<\/b><\/span><\/span><\/sub>=20Kq<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=20.9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.(1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/1,6.10<\/b><\/span><\/span>-15<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=180.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=1.800N\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0F<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=1,8.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N
\n<\/b><\/span><\/span>b)F = q.E = 1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>.2.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0= 3,2.10<\/b><\/span><\/span>-13<\/b><\/span><\/span><\/sup>\u00a0N<\/b><\/span><\/span><\/p>\n

52-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A carga positiva colocada em P ser\u00e1 mais repelida pelo canto superior direito do que pelo canto inferior esquerdo. Al\u00e9m disso, ser\u00e1 mais atra\u00edda pelo canto superior esquerdo do que pelo canto inferior direito. Assim a resultante dever\u00e1 estar apontando para a esquerda\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

53-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) As for\u00e7as pedidas est\u00e3o no esquema a seguir<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Dada a simetria da figura devido ao \u00e2ngulo a figura \u00e9 um quadrado e desta forma a for\u00e7a el\u00e9trica ter\u00e1 mesmo m\u00f3dulo que o peso da carga, ou seja, F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0= P = m.g = 0,015.10 = 0,15 N\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=P=0,15N\u00a0<\/b><\/span><\/span><\/p>\n

b) Como o campo \u00e9 uniforme F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0= q.E\u00a0 —\u00a0 \u00a0q = F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\/E = 0,15 \/ (500.103) = 0,0000003 = 3.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>\u00a0C\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0q =0,3\u00a0mC<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

54-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Se a part\u00edcula \u00e9 carregada negativamente e est\u00e1 se movendo na dire\u00e7\u00e3o e sentido do campo el\u00e9trico existe uma for\u00e7a constante de sentido oposto atuando sobre a part\u00edcula. Desta forma a part\u00edcula apresentar\u00e1 uma acelera\u00e7\u00e3o constante e negativa, o que est\u00e1 caracterizado no diagrama de apar\u00eancia parab\u00f3lica de concavidade para baixo\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

55-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) De acordo com a lei de Newton da gravita\u00e7\u00e3o, toda massa gera em torno de si um campo gravitacional. E de acordo com a lei de Coulomb, toda carga el\u00e9trica gera em torno de si um campo el\u00e9trico.<\/b><\/span><\/span><\/p>\n

b) Conforme a experi\u00eancia de Oersted, quando uma part\u00edcula eletrizada est\u00e1 em movimento, al\u00e9m dos campos gravitacional e el\u00e9trico, ela gera, tamb\u00e9m, campo magn\u00e9tico. Quando dois, ou mais, desses campos de mesma natureza se superp\u00f5em, o campo resultante depende da dire\u00e7\u00e3o e do sentido de cada um. Portanto, esses campos s\u00e3o grandezas vetoriais.<\/b><\/span><\/span><\/p>\n

Obs: se a part\u00edcula eletrizada sofrer acelera\u00e7\u00e3o, h\u00e1 emiss\u00e3o de onda eletromagn\u00e9tica, que \u00e9 a propaga\u00e7\u00e3o do campo eletromagn\u00e9tico atrav\u00e9s do espa\u00e7o.\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

56-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Com as gotas neutras n\u00e3o haver\u00e1 a atra\u00e7\u00e3o eletrost\u00e1tica.<\/b><\/span><\/span><\/p>\n

b) A folha ter\u00e1 a indu\u00e7\u00e3o de cargas opostas ao da gota.<\/b><\/span><\/span><\/p>\n

c) A for\u00e7a de atra\u00e7\u00e3o \u00e9 tanto maior quanto mais pr\u00f3ximas estiverem as gotas da folha.<\/b><\/span><\/span><\/p>\n

d) Correta<\/b><\/span><\/span><\/p>\n

e) A forma\u00e7\u00e3o de campos el\u00e9tricos \u00e9 sempre no sentido do positivo para o negativo e neste caso ser\u00e1 ent\u00e3o das gotas para a folha<\/b><\/span><\/span><\/p>\n

R- D.<\/b><\/span><\/span><\/p>\n

57-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Devido \u00e0 blindagem eletrost\u00e1tica, o vetor campo el\u00e9trico no interior de cada uma das cavidades \u00e9 nulo. Logo, a for\u00e7a entre as cargas q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0e q<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0tamb\u00e9m \u00e9 nula, pois F=qE=q.0=0\u00a0 —<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

58-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Coment\u00e1rio: materiais met\u00e1licos apresentam maior condutividade el\u00e9trica, por isso s\u00e3o mais facilmente polarizados e atra\u00eddos por campos el\u00e9tricos externos.\u00a0<\/b><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/p>\n

59-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Chamemos de A, B, C e D esses v\u00e9rtices\u00a0 —\u00a0 as cargas s\u00e3o positivas ent\u00e3o criam campos el\u00e9tricos de afastamento\u00a0 —\u00a0 como se mostra na figura a seguir, os campos\u00a0\"\"e\u00a0\"\"\u00a0t\u00eam mesma dire\u00e7\u00e3o e sentidos opostos anulando-se\u00a0 —\u00a0\u00a0 restam os campos\u00a0\"\"\u00a0e\u00a0\"\"\u00a0que ,somados vetorialmente, t\u00eam campo resultante\u00a0\"\", horizontal e para esquerda.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- B<\/b><\/span><\/span><\/p>\n

60-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Quando se coloca um isolante entre as cargas, mudam-se as condi\u00e7\u00f5es do meio entre as cargas. Por isso a lei de Coulomb para cargas puntiformes \u00e9:<\/b><\/span><\/span><\/p>\n

E=(1\/4.\u03c0\u03b5).Q\/r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0A constante\u00a0e\u00a0\u00e9 \u00e1 permissividade el\u00e9trica do meio. Portanto, ao se colocar um isolante entre as placas, o campo el\u00e9trico entre elas varia. A figura abaixo ilustra um exemplo para duas placas carregadas com cargas de sinais opostos.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R-C
\n<\/b><\/span><\/span><\/p>\n

61-<\/b><\/span><\/span><\/span> R- A<\/b><\/span><\/span>\u00a0 —\u00a0 E = F\/q ou F = q.E<\/b><\/span><\/span><\/p>\n

 <\/p>\n

62-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Como o coeficiente de x<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0na equa\u00e7\u00e3o dada \u00e9 igual a 1, e a, b e c s\u00e3o ra\u00edzes da equa\u00e7\u00e3o, podemos escrev\u00ea-la como:<\/b><\/span><\/span><\/p>\n

(x \u2013 a)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>(x \u2013 b)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>(x \u2013 c) = 0. Desenvolvendo esses produtos, obtemos:<\/b><\/span><\/span><\/p>\n

x<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0\u2013 (a + b + c)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ (a<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>b + a<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>c + b<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>c)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>x<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u2013 a<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>b<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>c<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>= 0<\/b><\/span><\/span><\/p>\n

Comparando os coeficientes, podemos montar o sistema:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como podemos notar, n\u00e3o \u00e9 um sistema f\u00e1cil de ser resolvido. Atentando, por\u00e9m, para a equa\u00e7\u00e3o (III) notamos que o produto dessas ra\u00edzes \u00e9 igual a 144. Podemos, ent\u00e3o, fazer uma pesquisa de solu\u00e7\u00e3o, com os divisores de 144 que s\u00e3o: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, …,144), para encontrar uma primeira raiz.<\/b><\/span><\/span><\/p>\n

Notamos que x<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= 3 \u00e9 uma raiz. Dividimos, ent\u00e3o, o polin\u00f4mio dado por (x \u2013 3) para baixar o grau.<\/b><\/span><\/span><\/p>\n

Fazendo a divis\u00e3o pelo dispositivo pr\u00e1tico de Briot Ruffini (ou pelo m\u00e9todo da chave):<\/b><\/span><\/span><\/p>\n

 <\/p>\n\n\n\n<\/colgroup>\n\n\n
\"\"<\/td>\n <\/p>\n

 <\/p>\n

 <\/p>\n

(x<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0\u2013 19<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 96<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>x \u2013 144)\"\"(x \u2013 3) = x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 16<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>x<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>+ 48<\/b><\/span><\/span><\/p>\n

 <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

 <\/p>\n

Igualando a zero, calculamos as outras duas ra\u00edzes:<\/b><\/span><\/span><\/p>\n

x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 16<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>x<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>+ 48 = 0. Essas duas ra\u00edzes tem soma 16 e produto 48. Ent\u00e3o: x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 4<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>e x<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0= 12.<\/b><\/span><\/span><\/p>\n

Do enunciado (c < a < b). Ent\u00e3o: c = 3; a = 4; b = 12; ou seja: (a;b;c) = (4;12;3).<\/b><\/span><\/span><\/p>\n

Para melhor visualiza\u00e7\u00e3o, observemos a figura a seguir, que mostra a localiza\u00e7\u00e3o das cargas.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

\"\"<\/p>\n

 <\/p>\n

Dados: k = 1,3<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00b4<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>\u00a0N.m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/C<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0= 1,44<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00b4<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>10<\/b><\/span><\/span>\u20134<\/b><\/span><\/span><\/sup>\u00a0C\u00a0 —\u00a0 Lembrando que a intensidade do vetor campo el\u00e9trico em um ponto \u00e0 dist\u00e2ncia d de uma carga puntiforme \u00e9 dado por:\u00a0\"\"\u00a0temos:<\/b><\/span><\/span><\/p>\n

E<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0=\u00a0\"\"\u00a0E<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= 1,17<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00b4<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0N\/C.<\/b><\/span><\/span><\/p>\n

E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0=\u00a0<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/sub>\u00a0E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 1,3<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00b4<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0N\/C.<\/b><\/span><\/span><\/p>\n

E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0=\u00a0\"\"\u00a0E<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0= 2,08<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00b4<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0N\/C.<\/b><\/span><\/span><\/p>\n

Escrevendo na forma vetorial, o vetor resultante no ponto (a,b,c) \u00e9:<\/b><\/span><\/span><\/p>\n

\"\".<\/b><\/span><\/span><\/p>\n

Para encontrar o vetor (\"\"\u00a0de menor intensidade, paralelo ao vetor campo el\u00e9trico, dividamos todos os componentes pelo componente de menor valor, ou seja, por 1,3<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00b4<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>. Assim:<\/b><\/span><\/span><\/p>\n

\"\"\u00de\u00a0\"\"<\/b><\/span><\/span><\/p>\n

Portanto, o vetor paralelo ao vetor campo el\u00e9trico no ponto (a;b;c) \u00e9 o vetor (9;1;16)\u00a0 — \u00a0R- E<\/b><\/span><\/span><\/p>\n

63-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Apenas para ilustrar a resolu\u00e7\u00e3o, suponhamos que a placa inferior esteja eletrizada positivamente e, a superior, negativamente.<\/b><\/span><\/span><\/p>\n

A figura mostra o vetor campo el\u00e9trico de cada uma das placas em tr\u00eas pontos: A e C, fora delas, e B, entre elas.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como se trata de placas infinitas, o campo el\u00e9trico criado por cada uma delas \u00e9 uniforme\u00a0 —\u00a0 assim\u00a0 —\u00a0 E<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0= 0; EB<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0= 2 E e<\/b><\/span><\/span><\/p>\n

E<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0= 0\u00a0 —<\/b><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/p>\n

64-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Como o campo el\u00e9trico (\"\") \u00e9 horizontal, e a for\u00e7a el\u00e9trica (\"\") s\u00f3 age na dire\u00e7\u00e3o do campo, ela n\u00e3o influi no tempo de queda\u00a0 —\u00a0 como o enunciado n\u00e3o informou o sinal da carga, considere-a\u00a0 positiva\u00a0 —\u00a0 por isso, a for\u00e7a el\u00e9trica est\u00e1 no mesmo sentido do campo el\u00e9trico.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

a) Como indicado na figura, quando a part\u00edcula atingir a altura h\u2019 = 5<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>H, ela ter\u00e1 ca\u00eddo 4<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>H\u00a0 —\u00a0 como na vertical o movimento \u00e9 uniformemente variado\u00a0 —\u00a0 4<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>H =\u00a0\"\"\u00a0\u00a0\u00de\u00a0\u00a0 8<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>H = g<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>t\u2019\u00a0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00a0\u00a0\u00de\u00a0\"\".<\/b><\/span><\/span><\/p>\n

b) Analogamente ao item anterior, para o tempo total\u00a0 —\u00a0 9<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>H =\u00a0\"\"\u00a0\u00a0\u00de\u00a0\u00a0 18<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>H = g\"\"\u00a0\u00a0\u00a0\u00de\u00a0\"\"<\/b><\/span><\/span><\/p>\n

c) o tempo que a part\u00edcula permanece sob a\u00e7\u00e3o do campo el\u00e9trico \u00e9\u00a0\"\"<\/b><\/span><\/span><\/p>\n

d) A dist\u00e2ncia horizontal percorrida pela part\u00edcula, como mostrado na figura \u00e9\u00a0 —\u00a0 D = d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u00a0\u00a0<\/b><\/span><\/span>(I)<\/b><\/span><\/span>\u00a0 —\u00a0 dist\u00e2ncia horizontal\u00a0<\/b><\/span><\/span>d<\/b><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub>\u00e9 percorrida em movimento uniformemente acelerado pela a\u00e7\u00e3o do campo el\u00e9trico, durante o tempo\u00a0<\/b><\/span><\/span>t\u2019<\/b><\/span><\/span>\u00a0 —\u00a0 \u00a0acelera\u00e7\u00e3o horizontal da part\u00edcula\u00a0 —\u00a0 F = m<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>a<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u00a0\u00a0\u00a0q<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>E = m<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>a<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 a<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>\u00a0=\u00a0\"\"\u00a0\u00a0—\u00a0 ent\u00e3o\u00a0 —\u00a0 d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0=\u00a0\"\"\u00a0\u00a0—\u00a0 d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0=\u00a0\"\"\"\"\u00a0—\u00a0 d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0=\u00a0\"\"\"\"\u00a0\u00a0—\u00a0 \u00a0\u00a0d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= 4\"\"\u00a0\u00a0\u00a0<\/b><\/span><\/span>(II)<\/b><\/span><\/span><\/p>\n

A partir da\u00ed, o movimento da part\u00edcula \u00e9 um lan\u00e7amento obl\u00edquo para baixo, sob a\u00e7\u00e3o exclusiva do campo gravitacional. O tempo para atingir o solo \u00e9\u00a0 —\u00a0 t<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sub>\u00a0= t<\/b><\/span><\/span>T<\/b><\/span><\/span><\/sub>\u00a0\u2013 t\u2019\u00a0\u00a0\u00de\u00a0 t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0=\"\"\u00a0\u00a0\u00a0<\/b><\/span><\/span>(III)<\/b><\/span><\/span>\u00a0 —\u00a0 ao abandonar o campo el\u00e9trico, a velocidade horizontal da part\u00edcula \u00e9\u00a0<\/b><\/span><\/span>v<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>, como indicado na figura, adquirida durante o tempo (<\/b><\/span><\/span>t\u2019<\/b><\/span><\/span>) em que sofreu a\u00e7\u00e3o do campo el\u00e9trico, sendo, ent\u00e3o\u00a0 —\u00a0 v<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>\u00a0= a<\/b><\/span><\/span>x\u00a0<\/b><\/span><\/span><\/sub>t\u2019\u00a0\u00a0\u00de\u00a0 v<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>\u00a0=\u00a0\"\"\"\"<\/b><\/span><\/span>(IV)<\/b><\/span><\/span><\/p>\n

A dist\u00e2ncia horizontal\u00a0<\/b><\/span><\/span>d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span>\u00e9 percorrida em movimento uniforme, com a velocidade\u00a0<\/b><\/span><\/span>v<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span>durante o tempo\u00a0<\/b><\/span><\/span>t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 assim, de\u00a0<\/b><\/span><\/span>(III)\u00a0<\/b><\/span><\/span>e\u00a0<\/b><\/span><\/span>(IV)\u00a0<\/b><\/span><\/span>vem\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= v<\/b><\/span><\/span>x\u00a0<\/b><\/span><\/span><\/sub>t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u00a0\u00a0\u00de\u00a0\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0=\u00a0\"\"\u00a0—\u00a0 aplicando a distributiva\u00a0 —\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0=\u00a0\"\"\u00a0—\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0=\u00a0\"\"\u00a0—\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 4\"\"\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>(V)<\/b><\/span><\/span>\u00a0 —\u00a0 substituindo\u00a0<\/b><\/span><\/span>(II)\u00a0<\/b><\/span><\/span>e<\/b><\/span><\/span>\u00a0(V)\u00a0<\/b><\/span><\/span>em<\/b><\/span><\/span>\u00a0(I)<\/b><\/span><\/span>\u00a0 —\u00a0 D = 4\"\"\u00a0+ 4\"\"\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0D = 8\"\".\u00a0<\/b><\/span><\/span><\/p>\n

65-<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span>Veja a figura abaixo, onde as cargas el\u00e9tricas devem ter sinais contr\u00e1rios, para que um campo seja de aproxima\u00e7\u00e3o e o outro de afastamento e, s\u00f3 assim podem se anular:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

66-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A figura mostra o campo el\u00e9trico e as for\u00e7as que agem na part\u00edcula\u00a0 —\u00a0 observe que a carga deve ser negativa para que a for\u00e7a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

magn\u00e9tica\u00a0\"\"\u00a0tenha sentido para cima\u00a0 —\u00a0 como a carga est\u00e1 em equil\u00edbrio a for\u00e7a resultante sobre ela \u00e9 nula\u00a0 —\u00a0 F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=P\u00a0 —<\/b><\/span><\/span><\/p>\n

q.E=m.g\u00a0 —\u00a0 q=mg\/E=2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.10\/5.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 q=4.10<\/b><\/span><\/span>-5\u00a0<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 q=40\u03bcC\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

67-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 E = 3.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0V\/m\u00a0 —\u00a0 V = 9 kV = 9.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0V\u00a0 —\u00a0 V=E.d\u00a0 —\u00a0 9.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=3.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>.d\u00a0 —\u00a0 d=3.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0 d=3mm\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

68-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

69-<\/b><\/span><\/span><\/span> H\u00e1 situa\u00e7\u00f5es na natureza que s\u00e3o imposs\u00edveis de ocorrer. Com base nessa afirma\u00e7\u00e3o, assinale, abaixo, a alternativa em que se apresenta um fen\u00f4meno f\u00edsico que n\u00e3o ocorre.<\/b><\/span><\/span><\/p>\n

A) Uma massa, ao ser abandonada numa regi\u00e3o do espa\u00e7o onde h\u00e1 um campo gravitacional, passa a se movimentar no sentido do campo gravitacional.<\/b><\/span><\/span><\/p>\n

B) Uma carga el\u00e9trica, ao ser abandonada numa regi\u00e3o do espa\u00e7o onde h\u00e1 um campo el\u00e9trico, passa a se movimentar em sentido contr\u00e1rio ao campo el\u00e9trico.<\/b><\/span><\/span><\/p>\n

C) Dois corpos, a temperaturas diferentes, s\u00e3o colocados em contato e isolados da vizinhan\u00e7a. O calor flui do corpo de temperatura mais baixa para o de temperatura mais alta.<\/b><\/span><\/span><\/p>\n

D) Uma carga el\u00e9trica, ao ser abandonada numa regi\u00e3o do espa\u00e7o onde h\u00e1 um campo el\u00e9trico, passa a se movimentar no sentido do campo el\u00e9trico.<\/b><\/span><\/span><\/p>\n

A) Correta\u00a0 —\u00a0 toda massa abandonada no interior de um campo gravitacional fica sujeito a uma for\u00e7a que tem o mesmo sentido do campo gravitacional.<\/b><\/span><\/span><\/p>\n

B) Correta\u00a0 — Observe nas informa\u00e7\u00f5es abaixo, que se a carga for negativa a for\u00e7a el\u00e9trica sobre ela a desloca em<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

sentido contr\u00e1rio ao do campo el\u00e9trico.<\/b><\/span><\/span><\/p>\n

C) Falsa\u00a0 — calor \u00e9 a transfer\u00eancia de energia entre dois ou mais corpos devido \u00e0 diferen\u00e7a de temperatura entre ele e essa transfer\u00eancia ocorre semptre do corpo de maior para o de menor temperatura\u00a0 —\u00a0 essa\u00a0transfer\u00eancia de energia (calor)\u00a0ocorre at\u00e9 que as mol\u00e9culas de todas as subst\u00e2ncias que est\u00e3o em contato t\u00e9rmico apresentem\u00a0a mesma temperatura\u00a0que \u00e9 denominada\u00a0temperatura de equil\u00edbrio t\u00e9rmico.<\/b><\/span><\/span><\/p>\n

D) Correta\u00a0 —\u00a0 observe nas informa\u00e7\u00f5es fornecidas em B, que se a carga for positiva a for\u00e7a el\u00e9trica sobre ela a desloca<\/b><\/span><\/span><\/p>\n

no mesmo sentido que o do campo el\u00e9trico.<\/b><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

70-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe abaixo as caracter\u00edsticas do vetor campo el\u00e9trico, nesse caso:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Intensidade do campo el\u00e9trico que q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0cria na posi\u00e7\u00e3o onde est\u00e1 q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 E<\/b><\/span><\/span>12<\/b><\/span><\/span><\/sub>=k.q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.16.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\/2<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E<\/b><\/span><\/span>12<\/b><\/span><\/span><\/sub>=36.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N\/C e \u00e9 de afastamento de q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>, pois q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub><0\u00a0 —\u00a0 intensidade do campo el\u00e9trico que q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0cria na posi\u00e7\u00e3o onde est\u00e1 q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 E<\/b><\/span><\/span>32<\/b><\/span><\/span><\/sub>=kq<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.4.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\/1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E<\/b><\/span><\/span>32<\/b><\/span><\/span><\/sub>=36.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N\/C e \u00e9 de afastamento de q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>, pois q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub><0\u00a0 —\u00a0 observe na figura que esses dois<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

vetores tem mesma intensidade, mesma dire\u00e7\u00e3o mas sentidos contr\u00e1rios, ent\u00e3o se anulam e\u00a0<\/b><\/span><\/span>E<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

71-<\/b><\/span><\/span><\/span> a) Orientando a trajet\u00f3ria da primeira placa para a segunda placa (velocidade positiva para a direita e negativa para a esquerda):<\/b><\/span><\/span><\/p>\n

1<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0trecho\u00a0 —\u00a0 o el\u00e9tron \u00e9 acelerado a partir do repouso (V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0) em movimento progressivo (velocidade positiva) at\u00e9 atingir a velocidade m\u00e1xima (V<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>);<\/b><\/span><\/span><\/p>\n

2<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0trecho\u00a0 —\u00a0 a for\u00e7a contr\u00e1ria agora desacelera o el\u00e9tron e sua velocidade m\u00e1xima V<\/b><\/span><\/span>max<\/b><\/span><\/span><\/sub>\u00a0come\u00e7a a diminuir, ainda em movimento<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0progressivo (velocidade positiva), ou seja, indo para a direita, at\u00e9 parar (V=0);<\/b><\/span><\/span><\/p>\n

3<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0trecho\u00a0 — o el\u00e9tron, a partir de V=0, acelera para a esquerda em movimento retr\u00f3grado (velocidade negativa) at\u00e9 atingir velocidade de \u2013 V\u2019, com que retorna \u00e0 placa.<\/b><\/span><\/span><\/p>\n

Representando gr\u00e1ficamente:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) A dist\u00e2ncia m\u00ednima entre as placas, para que ele n\u00e3o atinja a segunda placa ocorre quando ele estiver na imin\u00eancia de atingi-la, ele pare (V=0)\u00a0 —\u00a0 em todo gr\u00e1fico VxT a dist\u00e2ncia percorrida \u00e9 numericamente igual a \u00e1rea, no caso \u00e1rea de um tri\u00e2ngulo\u00a0 —\u00a0 V<\/b><\/span><\/span>max<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

a.t=0 +6,4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>=12,8m\/s\u00a0 —\u00a0 \u00e1rea do tri\u00e2ngulo 1\u00a0 —\u00a0 A<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=b.h\/2=2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.12,8\/2=12,8.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0 observe que A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=A<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 A<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=d=2.12,8.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>= 25,6.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>d<\/b><\/span><\/span>m\u00ednima<\/b><\/span><\/span><\/sub>=25,6.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span><\/p>\n

c) O tempo total do movimento vale\u00a0 —\u00a0 t<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=4.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0+ t\u2019, sendo t\u2019 o tempo que ele demora no retorno at\u00e9 retornar a placa onde chega com velocidade V\u2019 e, neste trecho em que retorna ele percorre a mesma dist\u00e2ncia que voc\u00ea j\u00e1 calculou no item anterior, que \u00e9 de \u2206S=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.t\u2019 + at\u2019<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 25,6.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>=0.t\u2019 + 6,4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.t\u2019<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 t\u2019<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 25,6.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\/3,2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0=8.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 t\u2019\u22482,8.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>s\u00a0 —\u00a0 t<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=4.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0+ 2,8.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>t<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>\u22486,8.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>s<\/b><\/span><\/span><\/p>\n

d) dados\u00a0 —\u00a0 m=9.10<\/b><\/span><\/span>-31<\/b><\/span><\/span><\/sup>kg<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/sup>—\u00a0 q=1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 a=6,4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>ms<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 princ\u00edpio fundamental da din\u00e2mica\u00a0 —\u00a0 F=m.a=9.10<\/b><\/span><\/span>-31<\/b><\/span><\/span><\/sup>.6,4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=<\/b><\/span><\/span><\/p>\n

57,6.10<\/b><\/span><\/span>-28<\/b><\/span><\/span><\/sup>N\u00a0 —\u00a0 E=F\/q=57,6.10<\/b><\/span><\/span>-28<\/b><\/span><\/span><\/sup>\/1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>=36.10<\/b><\/span><\/span>-9<\/b><\/span><\/span><\/sup>N\/C\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0E=3,6.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>N\/C (V\/m).<\/b><\/span><\/span><\/p>\n

72-<\/b><\/span><\/span><\/span> Como elas t\u00eam as mesmas dimens\u00f5es (mesmo raio), quando conectadas por um fio met\u00e1lico condutor (como se estivessem em contato) elas se comportar\u00e3o como um \u00fanico corpo com carga total Q<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=(Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)=[20\u03bcC + (-4 \u03bcC)] —\u00a0 Q<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=16\u03bcC\u00a0 —\u00a0 ap\u00f3s desconectadas, cada uma ficar\u00e1 com carga el\u00e9trica\u00a0 Q\u2019=16\u03bcC\/2=8\u03bcC\u00a0 —\u00a0 como todas as cargas s\u00e3o positivas, E<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0repele E<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0com for\u00e7a para a direita e E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0tamb\u00e9m repele E<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0com for\u00e7a para a direita\u00a0 —\u00a0 assim, a for\u00e7a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

resultante sobre E<\/b><\/span><\/span>3\u00a0\u00a0<\/b><\/span><\/span><\/sub>ter\u00e1 a dire\u00e7\u00e3o e sentido indicados na figura acima.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios!<\/span><\/a><\/h3>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Campo el\u00e9trico 01-\u00a0I. Certo\u00a0 —\u00a0 s\u00f3 surge for\u00e7a sobre a carga se existir campo el\u00e9trico II. Certo\u00a0 —\u00a0 se n\u00e3o tiver carga, pode existir o campo el\u00e9trico sem existir for\u00e7a el\u00e9trica. III. Errado —\u00a0 n\u00e3o surgir\u00e1 for\u00e7a sobre a carga se n\u00e3o houver campo el\u00e9trico R- D 02-\u00a0II e IV s\u00e3o falsas\u00a0 —\u00a0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":221,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-1704","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1704","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=1704"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1704\/revisions"}],"predecessor-version":[{"id":10591,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1704\/revisions\/10591"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/221"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1704"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}