{"id":1604,"date":"2015-12-02T14:38:50","date_gmt":"2015-12-02T14:38:50","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1604"},"modified":"2024-08-20T10:57:19","modified_gmt":"2024-08-20T10:57:19","slug":"1604-2","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletrostatica\/forca-eletrica-lei-de-coulomb\/1604-2\/","title":{"rendered":"For\u00e7a el\u00e9trica \u2013 Lei de Coulomb"},"content":{"rendered":"

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

For\u00e7a el\u00e9trica \u2013 Lei de Coulomb<\/span><\/span><\/span><\/p>\n

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

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>F=KQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>.4.10<\/b><\/span><\/span><\/span>-16<\/b><\/span><\/span><\/sup><\/span>.6.10<\/b><\/span><\/span><\/span>-16<\/b><\/span><\/span><\/sup><\/span>\/(3.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F=9.24.10<\/b><\/span><\/span><\/span>-23<\/b><\/span><\/span><\/sup><\/span>\/9.10<\/b><\/span><\/span><\/span>-18<\/b><\/span><\/span><\/sup><\/span>\u00a0 — F=24.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>\u00a0N\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>F=KQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>=9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>.2.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>.2.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>=36.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=36.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 d=6.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

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

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=Q\u00a0 —\u00a0 F=KQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 3,6.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>=9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>.Q.Q\/(1)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=3,6.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\/9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>=0,4.10<\/b><\/span><\/span><\/span>-11<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=4.10<\/b><\/span><\/span><\/span>-12<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Q=2.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>C\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na express\u00e3o F=KQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0que a intensidade da for\u00e7a F \u00e9 inversamente proporcional ao quadrado da dist\u00e2ncia\u00a0 —\u00a0 se a dist\u00e2ncia ficou 3 vezes maior, a for\u00e7a dever\u00e1 ficar 3<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=9 vezes menor\u00a0 —\u00a0 3,6.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>\/9\u00a0 —\u00a0 F=0,4.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R-B<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Observe na express\u00e3o F=KQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0que a intensidade da for\u00e7a F \u00e9 diretamente proporcional ao m\u00f3dulo de cada carga\u00a0 — se voc\u00ea triplicar o valor de uma das cargas a for\u00e7a el\u00e9trica ficar\u00e1 3 vezes maior.<\/b><\/span><\/span><\/span><\/p>\n

b) Observe na express\u00e3o F=KQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0que a for\u00e7a F \u00e9 inversamente proporcional ao quadrado da dist\u00e2ncia entre as cargas\u00a0 —\u00a0 se voc\u00ea duplicar a dist\u00e2ncia, a for\u00e7a el\u00e9trica ficar\u00e1 2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=4 vezes menor.<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que a for\u00e7a \u00e9 de repuls\u00e3o o que significa que as cargas t\u00eam mesmo sinal, ou seja, (+).(+)>0 e (-).(-)>0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=Kq<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>>0 (q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0tem mesmo sinal)\u00a0 —\u00a0 -F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=K.q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\/(2r)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=(Kq<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/3)\/4r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 -F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=Kq<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/12r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span><0 (q<\/b><\/span><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub><\/span>e q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>tem sinais opostos)\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>= Kq<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0X 12r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/Kq<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=12\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- (02 + 16)=18<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>F=KQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 720=9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>.Q.2Q\/1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=720\/18.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=40.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Q=\u221a(4.10<\/b><\/span><\/span><\/span>-8<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>Q=2.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>C e Q\u2019=4.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>For\u00e7a el\u00e9trica\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=KQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 for\u00e7a gravitacional\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=GMm\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 igualando-as\u00a0 —\u00a0 KQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= GMm\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 KQ<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/d<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup><\/span>= GM<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 M<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=KG\u00a0 —\u00a0 MQ=\u221a(KG)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Antes do contato\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=KQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=KQ.9Q\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=9KQ<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 ap\u00f3s colocadas em contato e separadas cada uma fica com carga Q\u2019 tal que Q\u2019= (-Q + 9Q)\/2\u00a0 —\u00a0 Q\u2019=\u00a0 4Q\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=KQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=K4Q.4Q(2d)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=16KQ<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/4d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=4KQ<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\/F<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=4KQ<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0X d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/9KQ<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\/F<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=4\/9\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quando r=3m\u00a0 —\u00a0 F=2,5.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0 F=KQq\/r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 2,5.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>=9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>.Q.Q\/3<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2,5.10<\/b><\/span><\/span><\/span>-13<\/b><\/span><\/span><\/sup><\/span>=25.10<\/b><\/span><\/span><\/span>-14<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Q=5.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>C\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que, como a carga de uma part\u00edcula dobrou, a for\u00e7a tamb\u00e9m dobrar\u00e1 (2F) e como a dist\u00e2ncia dobrou, a for\u00e7a ficar\u00e1 4 vezes menor (2F\/4=F\/2)\u00a0 —\u00a0 F\u2019=15\/2\u00a0 —\u00a0 F\u2019=7,5N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Cada esfera ficar\u00e1 com cargas de sinais opostos e de m\u00f3dulo\u00a0 —\u00a0 Q=n.e=5.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>.1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Q=8.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>C\u00a0 —\u00a0 F=KQq\/d<\/b><\/span><\/span><\/span>2\u00a0\u00a0<\/b><\/span><\/span><\/sup><\/span>—\u00a0 F=9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>.8.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>.8.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>\/(8.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F=9.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\/10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F=9.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe as figuras abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

\"\"<\/p>\n

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

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na figura abaixo que a carga \u2013Q n\u00e3o pode estar \u00e0 esquerda de Q e nem a direita de 4Q, pois nessas posi\u00e7\u00f5es a<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

resultante n\u00e3o seria nula\u00a0 —\u00a0 ela tem que estar entre Q e 4Q na posi\u00e7\u00e3o da figura\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>,<\/b><\/span><\/span><\/span>-Q<\/b><\/span><\/span><\/sub><\/span>=KQQ\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>4Q,-Q<\/b><\/span><\/span><\/sub><\/span>=K4QQ\/(3 \u2013 m)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 KQQ\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>= K4QQ\/(3 \u2013 m)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 1\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=4\/(3 \u2013 m)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 (3 \u2013 m)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=4\u00a0 —\u00a0 (3 \u2013 m)\/m=\u221a4\u00a0 —\u00a0 2m=3 \u2013 m\u00a0 —\u00a0 m=1\u00a0 —\u00a0 est\u00e1 na abscissa x=2 + 1=3\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) A intensidade da for\u00e7a el\u00e9trica \u00e9 igual \u00e0 intensidade da for\u00e7a centr\u00edpeta\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 KQq\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V=\u221a(KQq\/mR)<\/b><\/span><\/span><\/span><\/p>\n

b) V=2\u03c0R\/T\u00a0 —\u00a0 2\u03c0R\/T=<\/b><\/span><\/span><\/span>\u221a<\/b><\/span><\/span><\/span>(KQq\/mR)\u00a0 —\u00a0 4<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0=KQq\/mR\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=4\u03c0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>mR\u00a0 —\u00a0<\/b><\/span><\/span><\/span>T=2<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>R\u221a(mR\/KQq)<\/b><\/span><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Ou as duas s\u00e3o positivas ou negativas<\/b><\/span><\/span><\/span><\/p>\n

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

b) A\u00e7\u00e3o de q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e de q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0sobre q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>q1q2<\/b><\/span><\/span><\/sub><\/span>=Kq<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>q3q2<\/b><\/span><\/span><\/sub><\/span>=K.q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Kq<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= K.q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=q\u00a0 —\u00a0 a\u00e7\u00e3o de q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0e de q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0sobre q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>q3q1<\/b><\/span><\/span><\/sub><\/span>=Kq<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/(2r)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>q2q1<\/b><\/span><\/span><\/sub><\/span>=KQ<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Kq<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/(2r)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= KQ<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/r<\/b><\/span><\/span><\/span>2\u00a0\u00a0<\/b><\/span><\/span><\/sup><\/span>—\u00a0 q=4.2,7.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>q=q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=10,8.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span><\/p>\n

c) Se voc\u00ea retirar um pouquinho a carga q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0da posi\u00e7\u00e3o de equil\u00edbrio, ela n\u00e3o retorna \u00e0 mesma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>equil\u00edbrio inst\u00e1vel.<\/b><\/span><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo as cargas el\u00e9tricas id\u00eanticas, a for\u00e7a exercida por 3 sobre 2 deve ser 4 vezes maior\u00a0 que a for\u00e7a exercida por 1 sobre 2 (metade da dist\u00e2ncia)\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>12<\/b><\/span><\/span><\/sub><\/span>=4.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>32<\/b><\/span><\/span><\/sub><\/span>=4.4.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>32<\/b><\/span><\/span><\/sub><\/span>=16.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=16.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 4.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=12.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A carga q\u2019 deve estar entre as duas cargas a uma dist\u00e2ncia m de +q\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=Kq\u2019.q\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 se estiver \u00e0 dist\u00e2ncia n de +4q\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>= Kq\u2019.4q\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Kq\u2019.q\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= Kq\u2019.4q\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 (n\/m)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=4\u00a0 —\u00a0 n=2m\u00a0 —\u00a0 n + m = 6\u00a0 —\u00a0 m=2\u00a0 —\u00a0 n=4\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na figura abaixo que a carga pontual para ficar em equil\u00edbrio n\u00e3o pode estar entre as cargas e que deve estar no ponto<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

S, pois a carga positiva tem intensidade maior que a negativa\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que F<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 9 vezes maior que F<\/b><\/span><\/span><\/span>CB<\/b><\/span><\/span><\/sub><\/span>=3.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>N (inversamente proporcional ao quadrado da dist\u00e2ncia)\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=9.3.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=27.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=27.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 3.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=24.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>For\u00e7a el\u00e9trica\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=KQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 For\u00e7a gravitacional\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=GMm\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0 —\u00a0 quando a dist\u00e2ncia \u00e9 2km F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 KQq\/2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=GMm\/2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 KQq=GMm\u00a0 —\u00a0 quando a dist\u00e2ncia \u00e9 5km as massas M e m e as cargas Q e q continuam iguais, ent\u00e3o F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u00a0continua igual a F<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\u00a0e as part\u00edculas continuam equilibradas\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>For\u00e7a el\u00e9trica que 1 exerce em 3\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>13<\/b><\/span><\/span><\/sub><\/span>=Kq<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\/3<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0 —\u00a0\u00a0 F<\/b><\/span><\/span><\/span>13<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0 como a dist\u00e2ncia e as cargas s\u00e3o as mesmas a for\u00e7a que 2 exerce em 3 tamb\u00e9m ser\u00e1 a mesma\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>23<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>N=F<\/b><\/span><\/span><\/span>13<\/b><\/span><\/span><\/sub><\/span>=F\u00a0 —\u00a0 a for\u00e7a que q<\/b><\/span><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub><\/span>exerce em q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 de atra\u00e7\u00e3o e a que q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0exerce em q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 de repuls\u00e3o (figura).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe que os tri\u00e2ngulos s\u00e3o eq\u00fcil\u00e1teros, ent\u00e3o a intensidade de F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0ser\u00e1\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=F=10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>N<\/b><\/span><\/span><\/span>,<\/b><\/span><\/span><\/span>dire\u00e7\u00e3o horizontal<\/b><\/span><\/span><\/span>\u00a0e<\/b><\/span><\/span><\/span>\u00a0sentido para esquerda.<\/b><\/span><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Como elas est\u00e3o no ar, com o tempo as esferas ir\u00e3o se descarregando e se aproximando\u00a0 —\u00a0 as for\u00e7as que agem nas esferas, T intensidade da for\u00e7a de tra\u00e7\u00e3o no fio,\u00a0 P intensidade da for\u00e7a peso e F for\u00e7a el\u00e9trica de repuls\u00e3o entre as esferas com cargas de mesmo sinal, para qualquer posi\u00e7\u00e3o das esferas tem sempre a mesma intensidade em ambos os fios e as decomposi\u00e7\u00f5es dessas for\u00e7as ser\u00e3o sempre sim\u00e9tricas,e os \u00e2ngulos ser\u00e3o sempre iguais.<\/b><\/span><\/span><\/span><\/p>\n

b) Colocando as for\u00e7as que agem sobre uma das esferas e que s\u00e3o\u00a0 —\u00a0\u00a0\"\"– for\u00e7a de tra\u00e7\u00e3o no fio\u00a0 —\u00a0\u00a0\"\"\u00a0– for\u00e7a peso\u00a0 —\u00a0\u00a0\"\"– for\u00e7a<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

El\u00e9trica\u00a0 —\u00a0 considerando m como a dist\u00e2ncia entre as duas cargas —\u00a0 sen\u03b1=(m\/2)\/L\u00a0 —\u00a0 0,6=m\/2.0,09\u00a0 —\u00a0 m=0,108m\u00a0 —\u00a0 no tri\u00e2ngulo de for\u00e7as\u00a0 —\u00a0 tg\u03b1=F\/P\u00a0 —\u00a0 0,75=F\/P\u00a0 —\u00a0 F=0,75P\u00a0 —\u00a0 KQ.Q\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=0,75.m.g\u00a0 —\u00a0 9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>.Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/(0,108)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=0,048.10\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>Q=2,5.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>\u00a0C<\/b><\/span><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na figura, onde todas as for\u00e7as que agem sobre a carga Q acima, da direita foram<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

colocas, que, para que\u00a0\"\"\u00a0anule\u00a0\"\"\u00a0as cargas q e Q devem ter sinais opostos\u00a0 —\u00a0\u00a0\"\"\u00a0=\"\"\u00a0+\u00a0\"\"\u00a0\u00a0—\u00a0 (F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= (F<\/b><\/span><\/span><\/span>qQ<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ (F<\/b><\/span><\/span><\/span>qQ<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 (F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=KqQ\/L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ KqQ\/L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=\u221a2.K.q.Q\/L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 L \u2013 lado do quadrado\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>QQ<\/b><\/span><\/span><\/sub><\/span>=K.Q.Q\/(\u221a2L)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>QQ<\/b><\/span><\/span><\/sub><\/span>=K.Q.Q\/2L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 resultante nula\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>QQ\u00ad<\/b><\/span><\/span><\/sub><\/span>\u00a0 — \u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u221a2.K.q.Q\/L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= K.Q.Q\/2L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 \u221a2q=Q\/2\u00a0 —\u00a0 Q\/q=2\u221a2\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Figura A:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=2Kqq<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/(d\/2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=8Kqq<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

Figura B:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=2Kqq<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/(d\/2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.cos60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0= 2Kqq<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/(d\/2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.(1\/2)\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=Kqq<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=8Kqq<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0X d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/Kqq<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=8<\/b><\/span><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Depois do contato cada corpo ter\u00e1 carga de (1 + 5)\/2= 3 C<\/b><\/span><\/span><\/span><\/p>\n

F = k.q.Q\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>.3.3\/3<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0N<\/b><\/span><\/span><\/span><\/p>\n

A for\u00e7a ser\u00e1 repulsiva, pois os dois corpos apresentam a mesma natureza el\u00e9trica (s\u00e3o cargas positivas).<\/b><\/span><\/span><\/span><\/p>\n

\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Na dire\u00e7\u00e3o horizontal para qualquer uma das esferas \u00e9 verdadeiro afirmar que\u00a0 —\u00a0 k.Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>= T.senq\u00a0 —\u00a0 na dire\u00e7\u00e3o vertical<\/b><\/span><\/span><\/span><\/p>\n

m.g = T.cosq\u00a0 —\u00a0 dividindo as duas express\u00f5es\u00a0 —\u00a0 k.Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/(m.g.d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>) = senq\/cosq = 1 (pois q = 45\u00b0)\u00a0 —\u00a0 d=Q\u221ak\/(mg)\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

d=30.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/sub><\/span>\u221a9.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 d=90.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 d=90cm\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

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

35-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Das informa\u00e7\u00f5es iniciais sabemos que\u00a0 —\u00a0 \u00a0F = k.q.q\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F = k.(q\/d)<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0 na configura\u00e7\u00e3o apresentada a for\u00e7a resultante sobre q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9\u00a0 —\u00a0 Fr = \u00d6[F21<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ F31<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>]<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0 Fr = \u00d6[(k.3q.q\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ (k.4q.q\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)]<\/b><\/span><\/span><\/span>2\u00a0\u00a0<\/b><\/span><\/span><\/sup><\/span>—\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup><\/span>Fr = \u00d6[9k<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.q<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\/d<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 16.k<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.q<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\/d<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>]<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0 Fr = \u00d6[25k<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.q<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\/d<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>] = 5.k.(q\/d)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Fr = 5.F\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Em cada uma das extremidades das quatro diagonais que passam pelo centro do cubo h\u00e1 duas cargas de mesmo m\u00f3dulo e de<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0mesmo sinal. Elas exercem na carga central (tamb\u00e9m de mesmo sinal e mesmo m\u00f3dulo que as dos v\u00e9rtices) for\u00e7as de mesma intensidade e de sentidos opostos. Portanto, essas for\u00e7as se equilibram, sendo ent\u00e3o nula a resultante dessas for\u00e7as.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

Na situa\u00e7\u00e3o inicial, as cargas negativas (-q), nas extremidades, repelem-se com for\u00e7as de intensidade F, sendo 2 d a dist\u00e2ncia entre elas\u00a0 —\u00a0 como as cargas negativas est\u00e3o em equil\u00edbrio, elas trocam for\u00e7as, tamb\u00e9m, de intensidade F com a carga positiva (+Q) central, sendo d a dist\u00e2ncia do centro \u00e0s extremidades\u00a0 —\u00a0 a lei de Coulomb nos afirma que a intensidade das for\u00e7as eletrost\u00e1ticas entre duas cargas varia com o inverso do quadrado da dist\u00e2ncia entre essas cargas\u00a0 —\u00a0 F=kQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 na situa\u00e7\u00e3o final, a dist\u00e2ncia entre as cargas negativas foi reduzida \u00e0 metade (de 2 d para d) logo, as for\u00e7as de repuls\u00e3o entre elas passam a ter intensidade 4 F\u00a0 —\u00a0 por\u00e9m,\u00a0 a dist\u00e2ncia de cada carga negativa \u00e0 carga central tamb\u00e9m \u00e9 reduzida \u00e0 metade (de d para d\/2) quadruplicando, tamb\u00e9m, as for\u00e7as de atra\u00e7\u00e3o entre elas, ou seja, 4 F\u00a0 —\u00a0 portanto o equil\u00edbrio \u00e9 mantido com Q\u2019 = 1 Q\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

Na primeira situa\u00e7\u00e3o, as for\u00e7as s\u00e3o atrativas e t\u00eam intensidade\u00a0 —\u00a0 F=kQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 (I)\u00a0 —\u00a0 na segunda situa\u00e7\u00e3o, as for\u00e7as s\u00e3o repulsivas e t\u00eam intensidade\u00a0 —\u00a0 F\u2019=k.4Q.3q\/(2d)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F\u2019=3kQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(II)\u00a0 —\u00a0 comparando I com II voc\u00ea conclui que\u00a0 F\u2019=3F e que as for\u00e7as que eram atrativas passam a ser repulsivas\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como n\u00e3o foi especificada a massa da barra \u00e9 considerada desprez\u00edvel, como tamb\u00e9m a massa da carga suspensa\u00a0 —\u00a0 as for\u00e7as eletrost\u00e1ticas entre as cargas t\u00eam a mesma dire\u00e7\u00e3o da reta que passa pelos seus centros\u00a0 —\u00a0 para que o sistema fique em equil\u00edbrio as for\u00e7as eletrost\u00e1ticas devem ser de atra\u00e7\u00e3o\u00a0 —\u00a0 \u00a0as intensidades da for\u00e7a de tra\u00e7\u00e3o no fio e das for\u00e7as eletrost\u00e1ticas s\u00e3o iguais (T = F), como ilustrado na figura\u00a0 —\u00a0 da figura\u00a0 —\u00a0 cos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=d\/r\u00a0 —\u00a0 r=d\/cos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 r=d\/(\u221a3\/2) (I)\u00a0 —\u00a0 lei de Coulomb\u00a0 —\u00a0 F=k<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>Qq\/r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

(II)\u00a0 —\u00a0 (I) em (II)\u00a0 — F = k<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>Qq\/(2d\/\u221a3)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F=3k<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>Qq\/4d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(III)\u00a0 —\u00a0 para que a barra esteja em equil\u00edbrio a somat\u00f3ria dos momentos deve ser nula\u00a0 —\u00a0 adotando o p\u00f3lo no ponto O mostrado na figura\u00a0 —\u00a0 Fcos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u2113<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=mg\u2113<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0(IV)\u00a0 —\u00a0 (III) em (IV)\u00a0 —\u00a0 3k<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>Qq\u221a3 \u2113<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/4d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.2=mg \u2113<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 3k<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>Qq\u221a3 \u2113<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/8d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=mg \u2113<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 q=8mg\u2113<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/3\u221a3k<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>Q\u2113<\/b><\/span><\/span><\/span>1\"\"<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 como as for\u00e7as entre Q e q s\u00e3o de atra\u00e7\u00e3o e, se Q \u00e9 positiva, q s\u00f3 pode ser negativa ou vice versa\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

40-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe as figuras abaixo\u00a0 —\u00a0 no tri\u00e2ngulo hachurado da figura 1\u00a0 —\u00a0 Pit\u00e1goras\u00a0 —\u00a0 L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ z<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

L=\u221a(d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ z<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>) (I)\u00a0 —\u00a0cos\u03b8=z\/L\u00a0 —\u00a0 as for\u00e7as de repuls\u00e3o mostradas t\u00eam intensidade dada pela lei de Coulomb\u00a0 —\u00a0 F=kq3q\/L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F=k3q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(II)\u00a0 —\u00a0 (I) em (II)\u00a0 —\u00a0\u00a0 F=k3q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/(\u221ad<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ y<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F= k3q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/(d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ y<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)\u00a0 (III)\u00a0 —\u00a0 observe na figura 2 que, como a massa m est\u00e1 em equil\u00edbrio a for\u00e7a resultante sobre ela \u00e9 nula\u00a0 —\u00a0 P=2F<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mg=2Fcos\u03b8\u00a0 —\u00a0 cos\u03b8=z\/L\u00a0 —\u00a0 m=2F(z\/L)\/g (IV)\u00a0 —\u00a0 (III) em (IV)\u00a0 —\u00a0 m=(2\/g).(k3q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>).(z\/L)\u00a0 —\u00a0 m=6kq<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>z\/gL<\/b><\/span><\/span><\/span>3\u00a0\u00a0<\/b><\/span><\/span><\/sup><\/span>—\u00a0 m=6kq<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>z\/g(d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ z<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2\/3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe as figuras abaixo que mostram as duas situa\u00e7\u00f5es descritas no enunciado\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

F=kQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(I)\u00a0 —\u00a0 F\u2019=kQq\/(2d)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=KQq\/4d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 (I)\/(II)\u00a0 —\u00a0 F\/F\u2019=kQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0x 4d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/kQq\u00a0 —\u00a0 F\/F\u2019=4\u00a0 —\u00a0 F\u2019=F\/4\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

42-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Observe na figura abaixo a imagem da carga \u2013Q (-Q\u2019) e a decomposi\u00e7\u00e3o de suas velocidades e das velocidades do espelho nas dire\u00e7\u00f5es paralelas e perpendiculares ao espelho:\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe na figura acima que v.sen\u03b1 \u00e9 a velocidade com que a imagem da carga \u2013Q se aproxima perpendicularmente do espelho e que v.cos\u03b1 \u00e9 a velocidade com que a imagem da carga \u2013Q se move paralelamente ao espelho\u00a0 —\u00a0 mas, tamb\u00e9m, como o espelho se aproxima da carga \u2013Q com velocidade v.sen\u03b1, a imagem de \u2013Q se aproximar\u00e1 do espelho com velocidade 2v.sen\u03b1\u00a0 —\u00a0 como o espelho se aproxima de \u2013Q paralelamente ao eixo x, a imagem de \u2013Q n\u00e3o sofrer\u00e1 deslocamento na dire\u00e7\u00e3o paralela ao espelho\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

Se voc\u00ea somar vetorialmente os vetores velocidades da imagem \u2013Q\u2019 e os decompondo em x e y, obter\u00e1\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=(3vsen\u03b1).sen\u03b1 \u2013<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

(vcos\u03b1).cos\u03b1\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=v(3sen<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u03b1 \u2013 cos<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u03b1)\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>= – (3vsen\u03b1).cos\u03b1 \u2013 (vcos\u03b1).sen\u03b1\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>= – 4vsen\u03b1.cos\u03b1\u00a0 —\u00a0 mas sen<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u03b1=(1 \u2013 cos2\u03b1)\/2 e sen2\u03b1=2sen\u03b1.cos\u03b1\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=v(3sen<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u03b1 \u2013 (1 \u2013 sen<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u03b1))\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=v(1 \u2013 2cos2\u03b1)\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 v<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>= – 4vsen2\u03b1\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>= – 2vsen2\u03b1<\/b><\/span><\/span><\/span><\/p>\n

b) Acelera\u00e7\u00e3o centr\u00edpeta da carga negativa\u00a0 —\u00a0 intensidade\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/r\u00a0 —\u00a0 dire\u00e7\u00e3o e sentido\u00a0 —\u00a0 vertical e para cima<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 a figura mostra as for\u00e7as que agem sobre a carga negativa\u00a0 —\u00a0 a acelera\u00e7\u00e3o tangencial tem a dire\u00e7\u00e3o do eixo x\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>ex<\/b><\/span><\/span><\/sub><\/span>\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0ma=F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>cos\u03b2\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>==kQQ\/r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 ma=kQQ\/r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 r=d\/sen\u03b2\u00a0 —\u00a0 ma=kQQ\/(d\/sen\u03b2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.cos\u03b2\u00a0 —\u00a0 a=kQ<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>sen<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u03b2cos\u03b2\/md<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o tangencial\u00a0 —\u00a0 intensidade \u2013 a<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>= kQ<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>sen<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u03b2cos\u03b2\/md<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 dire\u00e7\u00e3o \u2013 eixo x\u00a0 —\u00a0 sentido \u2013 contr\u00e1rio a x.<\/b><\/span><\/span><\/span><\/p>\n

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

43-<\/b><\/span><\/span><\/span> <\/b>a) Do texto\u00a0 —\u00a0 \u201cSegundo todos os c\u00e1lculos, as futuras usinas de fus\u00e3o nuclear poder\u00e3o extrair de 1 metro c\u00fabico de \u00e1gua uma quantidade de energia igual \u00e0 de 2 mil barris de petr\u00f3leo\u201d\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0barris\u00a0 —\u00a0 100 m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 n barris\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

n=2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 n=2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0barris de petr\u00f3leo\u00a0 —\u00a0 como cada barril cont\u00e9m 1,5.10<\/b><\/span><\/span><\/span>6\u00a0<\/b><\/span><\/span><\/sup><\/span>kcal, 2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0barris conter\u00e3o\u00a0 —\u00a0 W=2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0barris x<\/b><\/span><\/span><\/span><\/p>\n

1,5.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>kcal\/barril\u00a0 —\u00a0 W=3,0.10<\/b><\/span><\/span><\/span>11<\/b><\/span><\/span><\/sup><\/span>\u00a0kcal=3,0.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>cal\u00a0 —\u00a0 do enunciado\u00a0 —\u00a0 1 BEP (Barril Equivalente de Petr\u00f3leo), equivale a 1,45.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0cal\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 BEP \u2013 1,45.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0cal\u00a0 —\u00a0 n\u2019 BEP \u2013 3.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0cal\u00a0 —\u00a0 n\u2019=3.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0cal\/1,45.10<\/b><\/span><\/span><\/span>9\u00a0<\/b><\/span><\/span><\/sup><\/span>cal\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

n\u2019\u22482,07.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0BEP<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

b) Do texto: \u201cOs centros dos n\u00facleos dos \u00e1tomos de hidrog\u00eanio devem estar a 1.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>\u00a0metros um do outro para que ocorra a fus\u00e3o\u201d\u00a0 —\u00a0 ainda do texto\u00a0 —\u00a0 \u201cessa fus\u00e3o \u00e9 o processo no qual dois n\u00facleos de \u00e1tomos leves (por exemplo, o hidrog\u00eanio \u2013 cujo n\u00facleo \u00e9 constitu\u00eddo por 1 pr\u00f3ton com carga el\u00e9trica elementar \u00e9 1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>C) se combinam, ou se fundem, constituindo um elemento mais pesado. Os n\u00facleos, ent\u00e3o, carregados positivamente, devem se aproximar suficientemente um do outro, ou seja, vencer a for\u00e7a de repuls\u00e3o eletrost\u00e1tica entre eles\u201d\u00a0 —\u00a0 portanto, s\u00e3o dados\u00a0 —\u00a0 d=1.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 .|Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>|= |Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>|=1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>C\u00a0 —\u00a0 k=9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>N.m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/C<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Lei de Coulomb\u00a0 —\u00a0\u00a0 F=k.|Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>|. |Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>|.\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>.1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>.1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>\/(1.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0 —\u00a0 F=23,04.10<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0F=230,4 N<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Voltar para os exerc\u00edcios<\/a><\/span><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre For\u00e7a el\u00e9trica \u2013 Lei de Coulomb \u00a0 01-\u00a0F=KQq\/d2=9.109.4.10-16.6.10-16\/(3.10-9)2\u00a0 —\u00a0 F=9.24.10-23\/9.10-18\u00a0 — F=24.10-5\u00a0N\u00a0 —\u00a0\u00a0R- D 02-\u00a0F=KQq\/d2\u00a0 —\u00a0 10-1=9.109.2.10-6.2.10-6\/d2\u00a0 —\u00a0 10-1=36.10-3\/d2\u00a0 —\u00a0 d2=36.10-2\u00a0 —\u00a0 d=6.10-1m\u00a0 —\u00a0\u00a0R- B 03- R- E\u00a0 —\u00a0 veja teoria 04-\u00a0Q1=Q2=Q\u00a0 —\u00a0 F=KQq\/d2\u00a0 —\u00a0 3,6.10-2=9.109.Q.Q\/(1)2\u00a0 —\u00a0 Q2=3,6.10-2\/9.109=0,4.10-11\u00a0 —\u00a0 Q2=4.10-12\u00a0 —\u00a0 Q=2.10-6C\u00a0 —\u00a0\u00a0R- E 05-\u00a0Observe na express\u00e3o F=KQq\/d2\u00a0que a intensidade da for\u00e7a F<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":200,"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-1604","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1604","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=1604"}],"version-history":[{"count":6,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1604\/revisions"}],"predecessor-version":[{"id":10608,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1604\/revisions\/10608"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/200"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1604"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}