{"id":1642,"date":"2015-12-19T14:29:20","date_gmt":"2015-12-19T14:29:20","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1642"},"modified":"2024-08-20T15:43:41","modified_gmt":"2024-08-20T15:43:41","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-linhas-de-forca-campo-e-potencial-eletrostatico","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletrostatica\/linhas-de-forca-de-campo-de-um-campo-eletrico-potencial-eletrostatico\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-linhas-de-forca-campo-e-potencial-eletrostatico\/","title":{"rendered":"Linhas de for\u00e7a e Potencial eletrost\u00e1tico"},"content":{"rendered":"

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

Linhas de for\u00e7a (campo) e Potencial eletrost\u00e1tico<\/b><\/span><\/span><\/span><\/p>\n

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

02-<\/b><\/span><\/span><\/span> R- C\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Nas proximidades de q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>, pois nela \u00e9 maior a concentra\u00e7\u00e3o de linhas de campo.<\/b><\/span><\/span><\/p>\n

b) Como q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u00e9 positiva (linhas de campo divergentes)\u00a0 e q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0negativa (linhas de campo convergentes), o sinal do produto q<\/b><\/span><\/span>1.<\/b><\/span><\/span><\/sub>q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u00e9 negativo.<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span> R- D\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Cargas positivas s\u00e3o fontes de campo el\u00e9trico enquanto cargas negativas s\u00e3o sorvedouros. Pela an\u00e1lise da figura, como as linhas de campo el\u00e9trico saem de B e chegam a A, conclui-se que A \u00e9 negativa e B \u00e9 positiva.<\/b><\/span><\/span><\/p>\n

b) Da figura, percebemos que da carga B saem o dobro de linhas de campo que chegam na carga A, portanto: |Q<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>| = 2|Q<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>|.<\/b><\/span><\/span><\/p>\n

c) N\u00e3o. Pois caso fosse poss\u00edvel, haveria diferentes vetores\u00a0\"\"\u00a0em cada ponto de cruzamento das linhas de campo.<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe a figura abaixo, onde em cada espa\u00e7o entre as placas foram colocados todos os<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

vetores, com as intensidades de cada\u00a0placa, pois as placas s\u00e3o grandes.<\/b><\/span><\/span><\/p>\n

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

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>(01) Verdadeira\u00a0 —\u00a0 veja figura abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

(02) Verdadeira\u00a0 —\u00a0 veja que na extremidade A existe maior concentra\u00e7\u00e3o de linhas de campo<\/b><\/span><\/span><\/p>\n

(04) Verdadeira\u00a0 —\u00a0 na extremidade A, nas cargas negativas a for\u00e7a \u00e9 contr\u00e1ria ao campo<\/b><\/span><\/span><\/p>\n

(08) Verdadeira\u00a0 —\u00a0 na extremidade B, nas cargas positivas campo e for\u00e7a tem mesma dire\u00e7\u00e3o e mesmo sentido<\/b><\/span><\/span><\/p>\n

(16) Verdadeira\u00a0 —\u00a0 a for\u00e7a de atra\u00e7\u00e3o \u00e9 maior que a for\u00e7a de repuls\u00e3o<\/b><\/span><\/span><\/p>\n

(32) Falsa\u00a0 —\u00a0 veja (04) e (08)<\/b><\/span><\/span><\/p>\n

(64) Verdadeira\u00a0 —\u00a0 o campo el\u00e9trico \u00e9 uniforme e consequentemente a for\u00e7a el\u00e9trica \u00e9 constante.\u00a0<\/b><\/span><\/span><\/p>\n

R- (01 + 02 + 04 + 08 + 16 + 64)=95<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span> R- B<\/b><\/span><\/span>\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe que na regi\u00e3o do ponto 2 as linhas de campo el\u00e9trico s\u00e3o mais pr\u00f3ximas\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Saem da A e da C (positivas) e chegam a B (negativa)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>I. Correta\u00a0 —\u00a0 Observe na figura abaixo que a carga da esquerda (positiva) produz num ponto P<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

campo de afastamento (\"\"), a carga da direita (negativa) produz no mesmo ponto P uma campo de aproxima\u00e7\u00e3o (\"\") e a soma vetorial desses dois campos produz um vetor campo el\u00e9trico resultante (\"\") que \u00e9 tangente ao ponto.<\/b><\/span><\/span><\/p>\n

II. Correta\u00a0 —\u00a0 A quantidade de linhas de for\u00e7a (de campo) que saem das cargas positivas \u00e9 a mesma das que chegam \u00e0s negativas.<\/b><\/span><\/span><\/p>\n

III. Falsa\u00a0 —\u00a0 a intensidade de cada campo independe da intensidade do outro (E=KQ\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>), e o campo resultante \u00e9 a soma vetorial desses dois campos.<\/b><\/span><\/span><\/p>\n

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

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

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>I. Falsa\u00a0 —\u00a0 \u00e9 maior na regi\u00e3o C onde a concentra\u00e7\u00e3o de linhas de campo \u00e9 maior.<\/b><\/span><\/span><\/p>\n

II. Correta\u00a0 —\u00a0 quando a carga \u00e9 negativa for\u00e7a e campo tem sentidos contr\u00e1rios.<\/b><\/span><\/span><\/p>\n

III. Correta\u00a0 —\u00a0 quando a carga \u00e9 positiva for\u00e7a e campo tem mesmo sentido.<\/b><\/span><\/span><\/p>\n

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

14-<\/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

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

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>V=Kq\/d=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.(-10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>)\/3,8.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V= -2,368.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Falsa\u00a0 —\u00a0 a carga Q \u00e9 positiva e origina em P campo el\u00e9trico de afastamento e fornecido por E=KQ\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) Falsa\u00a0 —\u00a0 veja a anterior.<\/b><\/span><\/span><\/p>\n

c) Correta\u00a0 —\u00a0 V=KQ\/d=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.2.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>\/6.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=3.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>V<\/b><\/span><\/span><\/p>\n

d) Falsa\u00a0 —\u00a0 veja a anterior<\/b><\/span><\/span><\/p>\n

e) Falsa\u00a0 —\u00a0 veja a e c.<\/b><\/span><\/span><\/p>\n

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

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A em rela\u00e7\u00e3o ao \u221e – U<\/b><\/span><\/span>A\u221e<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0– V<\/b><\/span><\/span>\u221e<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 150=V<\/b><\/span><\/span>A\u00ad<\/b><\/span><\/span><\/sub>\u00a0\u2013 0\u00a0 —\u00a0 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=150V\u00a0 —\u00a0 B em rela\u00e7\u00e3o ao \u221e – U<\/b><\/span><\/span>B\u221e<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0– V<\/b><\/span><\/span>\u221e<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0—\u00a0 100=V<\/b><\/span><\/span>B\u00ad<\/b><\/span><\/span><\/sub>\u00a0\u2013 0\u00a0 —V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=100V\u00a0 —\u00a0 A em rela\u00e7\u00e3o a B – U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=150 \u2013 100=50V ou A em rela\u00e7\u00e3o a B com V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

50=V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 0\u00a0 —\u00a0 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=50V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>I. Falsa\u00a0 —\u00a0 potencial el\u00e9trico \u00e9 grandeza escalar, n\u00e3o tem dire\u00e7\u00e3o nem sentido.
\nII. Correta\u00a0 —\u00a0 observe que nos tr\u00eas casos o potencial em 0 vale V<\/b><\/span><\/span>tota<\/b><\/span><\/span><\/sub>=Kq\/R + Kq\/R + Kq\/R\u00a0 — V<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=3Kq\/R<\/b><\/span><\/span><\/p>\n

III. Falsa\u00a0 —\u00a0 veja a anterior<\/b><\/span><\/span><\/p>\n

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

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>O potencial no ponto 0, que \u00e9 uma grandeza escalar \u00e9 a soma alg\u00e9brica do potencial em 0 originado pelos \u201cn\u201d pr\u00f3tons\u00a0 —\u00a0 V<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sub>=nKq\/r\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Um hex\u00e1gono regular \u00e9 composto por 6 tri\u00e2ngulos equil\u00e1teros\u00a0 —\u00a0 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=Kq\/d + Kq\/d + KQ\/2d\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

0=2Kq\/d + KQ\/2d\u00a0 —\u00a0Q=-4q\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Se os dois \u00e1tomos devem ficar em equil\u00edbrio, a resultante das outras for\u00e7as internas deve anular a for\u00e7a el\u00e9trica entre as cargas\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=KQ.q\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.\u25021,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>\u2502 . \u2502-1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>\u2502\/(2.10<\/b><\/span><\/span>-10<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0F=5,76.10<\/b><\/span><\/span>-9<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

b) O m\u00f3dulo da energia potencial el\u00e9trica associada ao sistema vale\u00a0 —\u00a0 E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=Kq.q\/d=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.\u25021,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>\u2502 . \u2502-1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>\u2502\/(2.10<\/b><\/span><\/span>-10<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=1,152.10<\/b><\/span><\/span>-18<\/b><\/span><\/span><\/sup>J.<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Para que essas cargas se aproximem \u00e9 preciso que um agente externo aplique for\u00e7as, for\u00e7ando a aproxima\u00e7\u00e3o, e fornecendo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

energia ao sistema\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>V=KQ\/d\u00a0 —\u00a0 30=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>Q\/3.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Q=10<\/b><\/span><\/span>-10<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=KQ\/d=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.(-2,0.10<\/b><\/span><\/span>-9<\/b><\/span><\/span><\/sup>)\/1\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=-18V\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=KQ\/d=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.(-2,0.10<\/b><\/span><\/span>-9<\/b><\/span><\/span><\/sup>)\/2\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=-9V\u00a0 —\u00a0 U<\/b><\/span><\/span>12<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0=<\/b><\/span><\/span><\/p>\n

(-18) \u2013 (-9)\u00a0 —\u00a0 U<\/b><\/span><\/span>12<\/b><\/span><\/span><\/sub>=-9V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=KQq\/d\u00a0 —\u00a0 observe que se voc\u00ea dobrar cada carga E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0ficar\u00e1 quadruplicada e que se voc\u00ea multiplicar a dist\u00e2ncia por 4, E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0 ficar\u00e1 4 vezes menor\u00a0 —<\/b><\/span><\/span>\u00a0 R- E<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>E<\/b><\/span><\/span>p+Q<\/b><\/span><\/span><\/sub>=K(+Q)\/3a\u00a0 —\u00a0 E<\/b><\/span><\/span>p-Q<\/b><\/span><\/span><\/sub>=K(-Q)\/a\u00a0 —\u00a0 E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=+KQ\/3a \u2013 KQ\/a\u00a0 —\u00a0 E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=(KQ \u2013 3KQ)\/3a\u00a0 —\u00a0 E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=-2KQ\/3a\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

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

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe na figura abaixo que o deslocamento da part\u00edcula \u00e9 para a direita e \u00e9 espont\u00e2neo,<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

portanto a energia potencial eletrost\u00e1tica do sistema est\u00e1 diminuindo. A for\u00e7a faz surgir sobre a part\u00edcula uma acelera\u00e7\u00e3o, portanto sua velocidade e consequentemente sua energia cin\u00e9tica est\u00e1 aumentando\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe que para que o potencial criado pelas duas cargas seja nula nos pontos, como elas tem mesmo m\u00f3dulo e sinais contr\u00e1rios, esses pontos devem ser eq\u00fcidistantes de cada carga\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Os pontos devem estar eq\u00fcidistantes de Q\u00a0 —\u00a0 veja figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Lembre-se de que cargas positivas originam campos de afastamento e negativas, campos de aproxima\u00e7\u00e3o\u00a0 —\u00a0 veja a figura\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

32-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>O potencial el\u00e9trico \u00e9 uma grandeza escalar e, no centro do quadrado \u00e9 a soma alg\u00e9brica dos potenciais criados por cada carga\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 A \u2013 E<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=Kq\/d + Kq\/d + Kq\/d + Kq\/d=<\/b><\/span><\/span>4Kq\/d\u00a0<\/b><\/span><\/span>\u00a0\u00a0—\u00a0 B \u2013 E<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>= Kq\/d + Kq\/d – Kq\/d – Kq\/d<\/b><\/span><\/span>=0\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 C \u2013 E<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>= Kq\/d + Kq\/d – Kq\/d – Kq\/d<\/b><\/span><\/span>=0<\/b><\/span><\/span>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Como a carga \u00e9 positiva ela ficar\u00e1 sujeita a uma for\u00e7a que tem a mesma dire\u00e7\u00e3o e sentido que o campo el\u00e9trico e intensidade<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F=q.E, que a far\u00e1 acelerar para a direita (figura) sujeita a uma acelera\u00e7\u00e3o a, aumentando sua velocidade e consequentemente sua energia cin\u00e9tica e diminuindo sua energia potencial el\u00e9trica de modo que sua energia mec\u00e2nica seja constante (a for\u00e7a el\u00e9trica \u00e9 conservativa)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- 05<\/b><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0V=0=K.(+2q)\/d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ K(-q)\/d<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sub>\u00a0\u00a0—\u00a0 d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1, d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=2 (K)\u00a0 —\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=3, d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=6 (L)\u00a0 —<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=0\u00a0 — \u00a0KQ\/(104 \u2013 x) \u2013 K3Q\/x=0\u00a0 —\u00a0 1\/(104 \u2013 x) = 3\/x\u00a0 —\u00a0 x=312\/4\u00a0 —\u00a0 x=78cm<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

36-<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>Situa\u00e7\u00e3o inicial<\/b><\/span><\/span><\/p>\n

\"\"
\nE<\/b><\/span><\/span>P1<\/b><\/span><\/span><\/sub>=Kq<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/d + Kq<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/d + Kq<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/d\u00a0 —\u00a0 E<\/b><\/span><\/span>p1<\/b><\/span><\/span><\/sub>=3kq<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/d<\/b><\/span><\/span><\/p>\n

Situa\u00e7\u00e3o final<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

E<\/b><\/span><\/span>p2<\/b><\/span><\/span><\/sub>=Kq<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/d + K(2q)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/d + K.(2q)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/d\u00a0 —\u00a0 E<\/b><\/span><\/span>p2<\/b><\/span><\/span><\/sub>=5Kq<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/d\u00a0 —\u00a0 E<\/b><\/span><\/span>p2<\/b><\/span><\/span><\/sub>=(5\/3)E<\/b><\/span><\/span>p1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) As duas cargas originam em x campos de afastamento que devem se anular\u00a0 —\u00a0 E<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=Kq<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/(6 \u2013 x)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=Kq<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/x<\/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 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

K.4.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\/(6 \u2013 x)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=K.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\/x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 4\/(6 \u2013 x)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=1x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/<\/b><\/span><\/span><\/sub>(6 \u2013 x)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=1\/4\u00a0 —\u00a0 x\/(6 \u2013 x)=1\/2\u00a0 —\u00a0 x=6\/3\u00a0 —\u00a0 x=2\u00a0 —\u00a0 o campo el\u00e9trico \u00e9 nulo a 2m de q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0e a 4m de q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.= ou na posi\u00e7\u00e3o x=4m<\/b><\/span><\/span><\/p>\n

b) V=Kq<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/3 + Kq<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/3=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.4.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\/3 + 9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\/3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=15,0.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>V<\/b><\/span><\/span><\/p>\n

c) E<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.4.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\/9=4.10<\/b><\/span><\/span>4<\/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>-5<\/b><\/span><\/span><\/sup>\/9=10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N\/C \u00a0—\u00a0 E<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=4.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0\u2013 10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=3.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N\/C\u00a0 —\u00a0 E=F\/q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 3.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>=F\/10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F=3.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>N\u00a0 —\u00a0 F=ma\u00a0 —\u00a0 3.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>=1\u00aa\u00a0 —\u00a0<\/b><\/span><\/span>a=0,3m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

38-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>E=KQ\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 9.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=KQ\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 KQ=9.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=KQ\/d\u00a0 —\u00a0 18.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=KQ\/d\u00a0 —\u00a0 KQ=18.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>d\u00a0 —\u00a0 9.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=18.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>d\u00a0 —\u00a0 d=2m\u00a0 —\u00a0 18.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>Q\/2\u00a0 —\u00a0 Q=36.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q=4.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>C ou Q=4\u03bcC<\/b><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Quando x=0,6m \u2013 E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=6.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0 V=E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\/Q\u00a0 —\u00a0 V=6.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\/10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=6.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>V<\/b><\/span><\/span><\/p>\n

40-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Todo ponto \u201caterrado\u201d possui potencial nulo\u00a0 — na figura 2\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>3,1<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>2,1<\/b><\/span><\/span><\/sub>\u00a0= 0\u00a0 —\u00a0 KQ<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/a KQ\/b + KQ\/b=0\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/a=-2Q\/b\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=-2Qa\/b\u00a0 —\u00a0 na figura 3\u00a0 —\u00a0 KQ<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/a + KQ<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/b + KQ\/b=0\u00a0 — Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/a=Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/b \u2013 Q\/b\u00a0 —\u00a0 substituindo Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/a=+2Qa\/b<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 Q\/b\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=Qa\/b(2a\/b \u2013 1)\u00a0 —\u00a0\u00a0\u00a0 na figura 4\u00a0 —\u00a0 KQ<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\/a + KQ<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/b + KQ<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/b=0\u00a0 —\u00a0 Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\/a + Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/b + q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/b=0\u00a0 —\u00a0 usando as express\u00f5es de Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\/a=2Qa\/b<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 Qa\/b<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(2a\/b \u2013 1)\u00a0 —\u00a0 Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\/a=Qa\/b<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(2 \u2013 2a\/b + 1)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=Qa<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(3 \u2013 2a\/b)<\/b><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Como o sistema \u00e9 sim\u00e9trico quanto a energia potencial el\u00e9trica, o que ocorrer\u00e1 \u00e9 a transforma\u00e7\u00e3o de energia potencial gravitacional em energia cin\u00e9tica\u00a0 —\u00a0 m.g.h = m.v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 g.h = v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 g.6 = v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 \u00a0\u00a0—\u00a0 \u00a0v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 12g \u00a0\u00a0—\u00a0<\/b><\/span><\/span>\u00a0v = \u221a12g<\/b><\/span><\/span><\/p>\n

42-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A despolariza\u00e7\u00e3o ocorre na fase em que o potencial sobe, que \u00e9 a fase 0. A repolariza\u00e7\u00e3o ocorre quando o potencial est\u00e1 voltando ao potencial de repouso, o que ocorre na fase 3.\u00a0<\/b><\/span><\/span><\/p>\n

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

43-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Dados: Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0= 1,44.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0C; k = 1,3.10<\/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>.<\/b><\/span><\/span><\/p>\n

C\u00e1lculo da dist\u00e2ncia (L) da carga Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0ao ponto (0;0;c), que \u00e9 o ponto (0;0;3), (figura).<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Aplicando Pit\u00e1goras\u00a0 —\u00a0 L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 4<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 3<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 12<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 169\u00a0 —\u00a0 L = 13 m.\u00a0 —\u00a0 Calculando o potencial el\u00e9trico\u00a0 —\u00a0 V=KQ<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\/L=1,3.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.1,44.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\/13\u00a0 —\u00a0 V=14,4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>V\u00a0 —\u00a0 V=14,4kV\u00a0 —<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

45-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe na figura abaixo as superf\u00edcies equipotenciais com as respectivas dist\u00e2ncias\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=kQ\/r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0(I)\u00a0 —\u00a0 \u00a0V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=kQ\/r<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0(II)\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

V<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=kQ\/r<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0(III)\u00a0 —\u00a0 (II)\/(I)\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=kQ\/r<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0x r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/kQ\u00a0 —\u00a0 r<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2\/3m\u00a0 —\u00a0 r<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=0,67m\u00a0 —\u00a0 (III)\/(I)\u00a0 —\u00a0 V<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=kQ\/r<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>x r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/kQ\u00a0 —\u00a0 r<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=2m\u00a0 —<\/b><\/span><\/span><\/p>\n

Dist\u00e2ncia d pedida\u00a0 —\u00a0 d=r<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0\u2013 r<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2,00 \u2013 0,67\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0d=1,33m<\/b><\/span><\/span><\/p>\n

46-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 V = 600 V\u00a0 —\u00a0\u00a0 E = 200 V\/m\u00a0 —\u00a0 \u00a0k = 9.10<\/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>.<\/b><\/span><\/span><\/p>\n

A carga Q \u00e9 positiva, pois o potencial el\u00e9trico por ela gerado \u00e9 positivo\u00a0 —\u00a0 potencial\u00a0 —\u00a0 V=kQ\/r\u00a0 —\u00a0 E=kQ\/r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V\/E=kQ\/r x r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/kQ\u00a0 —\u00a0 600\/200=r\u00a0 —\u00a0 r=3m\u00a0 —\u00a0 V=kQ\/r\u00a0 —\u00a0 600=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.Q\/3\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0Q=2,0.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

47-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>I. Correta\u00a0 —\u00a0 quanto mais concentradas as linhas de for\u00e7a, mais intenso \u00e9 o campo el\u00e9trico.<\/b><\/span><\/span><\/p>\n

II. Falsa\u00a0 —\u00a0 no sentido das linhas de for\u00e7a o potencial el\u00e9trico \u00e9 decrescente, portanto V<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0> V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

III. Falsa\u00a0 —\u00a0 part\u00edculas com carga negativa sofrem for\u00e7a em sentido oposto ao do vetor campo el\u00e9trico, movimentando-se espontaneamente para regi\u00f5es de maior potencial el\u00e9trico.<\/b><\/span><\/span><\/p>\n

IV. Correta\u00a0 —\u00a0 part\u00edculas positivamente carregadas movimentam-se espontaneamente no mesmo sentido dos menores potenciais, ganhando energia cin\u00e9tica, consequentemente, diminuindo sua energia potencial.\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

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

 <\/p>\n

48-<\/b><\/span><\/span><\/span>Veja as caracter\u00edsticas do potencial originado nesse caso:\u00a0\"\"<\/b><\/span><\/span><\/p>\n

Do gr\u00e1fico, quando d=0,15m, V=300V\u00a0 —\u00a0 V=k.Q\/d\u00a0 —\u00a0 300=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.Q\/0,15\u00a0 —\u00a0 Q=45\/9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q=5.10<\/b><\/span><\/span>-9<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>=<\/b><\/span><\/span>5nC<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

49<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> <\/b>Num campo el\u00e9trico uniforme o vetor campo el\u00e9trico tem em todos os seus infinitos pontos mesma intensidade, mesma<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

dire\u00e7\u00e3o e mesmo sentido e \u00e9\u00a0obtido entre duas placas condutoras id\u00eanticas e paralelas e eletrizadas com cargas de mesmo m\u00f3dulo, mas de sinais contr\u00e1rios\u00a0 —\u00a0 observe que nesse caso as superf\u00edcies equipotenciais (todos os seus pontos possuem o mesmo potencial el\u00e9trico)\u00a0 s\u00e3o planas e paralelas entre si\u00a0 —\u00a0 assim, os pontos A e C possuem o mesmo potencial el\u00e9trico (est\u00e3o numa mesma superf\u00edcie equipotencial)\u00a0 —\u00a0 observe tamb\u00e9m que o potencial diminui no sentido das linhas de for\u00e7a, que se afastam da placa positiva\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

50-<\/b><\/span><\/span><\/span> <\/b>Lembre-se de que o potencial\u00a0 el\u00e9trico diminui\u00a0 ao longo das linhas de for\u00e7a (de campo)\u00a0 —\u00a0\u00a0R- A\u00a0 —\u00a0 ou, do enunciado 0 < VA\u00a0\u2013 VC\u00ad\u00a0 —\u00a0 VA> VC\u00a0\u00a0—\u00a0 0 < VB\u00a0\u2013 VC\u00a0 —\u00a0 VB\u00a0> VC\u00a0\u00a0—\u00a0 VA\u00a0\u2013 VC\u00a0< VB\u00a0– VC\u00a0 —\u00a0 VA\u00a0< VB\u00a0\u00a0—\u00a0 VC\u00a0< VA\u00a0< VB\u00a0 —\u00a0 R- A<\/b><\/span><\/span><\/p>\n

51-<\/b><\/span><\/span><\/span> Campo el\u00e9trico de intensidade E criado por uma carga de m\u00f3dulo Q a uma dist\u00e2ncia d dessa carga\u00a0 —\u00a0 E=k.Q\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0(I) —\u00a0 potencial el\u00e9trico de intensidade V criado por uma carga de m\u00f3dulo Q a uma dist\u00e2ncia d dessa carga\u00a0 —\u00a0 V=k.Q\/d (II)\u00a0 —<\/b><\/span><\/span><\/p>\n

Dividindo (I) por (II)\u00a0 —\u00a0 (I)\/(II) = E\/V=(kQ\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>)x(d\/kQ)=1d\u00a0 —\u00a0 E\/V=1\/d\u00a0 —\u00a0 V = E.d\u00a0 —\u00a0 180=360.d\u00a0 —\u00a0 d=0,5m\u00a0 —\u00a0 substituindo d=0,5m em (II)\u00a0 —\u00a0 180=9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.Q\/0,5\u00a0 —\u00a0 Q=90\/9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Q=10.10<\/b><\/span><\/span>-9<\/b><\/span><\/span><\/sup>=10nC\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

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

\u00a0<\/span><\/p>\n

52-<\/b><\/span><\/span><\/span> Inicialmente voc\u00ea deve se lembrar que as linhas de for\u00e7a de um campo magn\u00e9tico s\u00e3o linhas fechadas e de um campo eletrost\u00e1tico s\u00e3o linhas abertas\u00a0 —\u00a0 A figura I mostra as linhas de for\u00e7a (de campo) de um campo el\u00e9trico (eletrost\u00e1tico) criado por uma placa plana, muito extensa e\u00a0 uniformemente eletrizada por cargas el\u00e9tricas positivas (campos de afastamento)\u00a0 —\u00a0 a figura II sugere o campo eletrost\u00e1tico originado por duas cargas pontuais eletrizadas com cargas positivas (campo de afastamento), colocadas pr\u00f3ximas\u00a0 uma da outra, mas pode representar tamb\u00e9m o campo magn\u00e9tico originado por dois p\u00f3los norte de im\u00e3s diferentes (linhas de indu\u00e7\u00e3o saem dos p\u00f3los norte), quando colocados pr\u00f3ximos (veja figura abaixo), apenas que, nesse caso as<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

linhas s\u00e3o fechadas (saem do p\u00f3lo norte de cada im\u00e3 e chegam ao p\u00f3lo sul do mesmo im\u00e3)\u00a0 —\u00a0 a figura III representa campo magn\u00e9tico, pois s\u00e3o linhas fechadas\u00a0 —\u00a0 pode representar as linhas de indu\u00e7\u00e3o de uma espira percorrida<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

por corrente el\u00e9trica ou por dois fios retil\u00edneos, percorridos por correntes el\u00e9tricas de sentidos opostos (veja figuras acima)\u00a0 —\u00a0 a figura IV tamb\u00e9m mostra um campo magn\u00e9tico (linhas fechadas)\u00a0 —\u00a0 esse campo magn\u00e9tico \u00e9 gerado por um fio retil\u00edneo, perpendicular ao plano dessa folha, com a corrente el\u00e9trica saindo da mesma.<\/b><\/span><\/span><\/p>\n

R-\u00a0<\/b><\/span><\/span>A ou E<\/b><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/span>\u00a0<\/a><\/h3>\n

 <\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Linhas de for\u00e7a (campo) e Potencial eletrost\u00e1tico 01- R- A\u00a0 —\u00a0 veja teoria 02- R- C\u00a0\u00a0—\u00a0 veja teoria 03-\u00a0a) Nas proximidades de q1, pois nela \u00e9 maior a concentra\u00e7\u00e3o de linhas de campo. b) Como q1\u00a0\u00e9 positiva (linhas de campo divergentes)\u00a0 e q2\u00a0negativa (linhas de campo convergentes), o sinal do produto q1.q2\u00a0\u00e9 negativo.<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1638,"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-1642","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1642","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=1642"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1642\/revisions"}],"predecessor-version":[{"id":10621,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1642\/revisions\/10621"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1638"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1642"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}