{"id":4845,"date":"2020-01-09T14:07:14","date_gmt":"2020-01-09T14:07:14","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=4845"},"modified":"2024-09-04T14:12:29","modified_gmt":"2024-09-04T14:12:29","slug":"vestibulares-recentes-resolucao-comentada-eletrostatica-2018-2019","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/eletricidade\/eletrostatica-2018-2019\/vestibulares-recentes-resolucao-comentada-eletrostatica-2018-2019\/","title":{"rendered":"Vestibulares Recentes \u2013 Resolu\u00e7\u00e3o Comentada \u2013 Eletrost\u00e1tica \u2013 2018\/2019"},"content":{"rendered":"

Vestibulares Recentes \u2013 Resolu\u00e7\u00e3o Comentada \u2013 Eletrost\u00e1tica \u2013 2018\/2019<\/span><\/span><\/span><\/p>\n

01-<\/span><\/span><\/span><\/p>\n

Aqui, a \u00fanica alternativa incorreta \u00e9 alternativa (C), pois a carga el\u00e9trica \u00e9 transferida\u00a0<\/b><\/span><\/span>exclusivamente<\/b><\/span><\/span>\u00a0pelo movimento dos\u00a0<\/b><\/span><\/span>el\u00e9trons<\/b><\/span><\/span>, deixando os\u00a0<\/b><\/span><\/span>pr\u00f3tons fixos<\/b><\/span><\/span>. A carga\u00a0<\/b><\/span><\/span>positiva<\/b><\/span><\/span>\u00a0acaba sendo ent\u00e3o dada pela\u00a0<\/b><\/span><\/span>aus\u00eancia de el\u00e9trons<\/b><\/span><\/span>. Logo, a\u00a0<\/b><\/span><\/span>(C) deve ser a alternativa marcada.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

 <\/p>\n

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

Eletriza\u00e7\u00e3o por contato<\/b><\/span><\/span><\/span><\/p>\n

Pode ocorrer<\/b><\/span><\/span><\/span>\u00a0entre\u00a0<\/b><\/span><\/span><\/span>dois condutores<\/b><\/span><\/span><\/span>\u00a0(cargas el\u00e9tricas se distribuem em suas superf\u00edcie externas) ou entre um\u00a0<\/b><\/span><\/span><\/span>condutor e um isolante<\/b><\/span><\/span><\/span>\u00a0(no isolante as cargas el\u00e9tricas ficam somente no local do contato).<\/b><\/span><\/span><\/span><\/p>\n

Considere\u00a0<\/b><\/span><\/span><\/span>um condutor A eletrizado com carga positiva (QA\u00a0= + Q)<\/b><\/span><\/span><\/span>\u00a0e outro\u00a0<\/b><\/span><\/span><\/span>condutor B,
\n<\/b><\/span><\/span><\/span>eletricamente\u00a0<\/b><\/span><\/span><\/span>neutro<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>(QB\u00a0= 0)<\/b><\/span><\/span><\/span>\u00a0presos a\u00a0<\/b><\/span><\/span><\/span>suportes isolantes<\/b><\/span><\/span><\/span>\u00a0(para que n\u00e3o descarreguem),<\/b><\/span><\/span><\/span>inicialmente<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>separados.<\/b><\/span><\/span><\/span><\/p>\n

Quando s\u00e3o\u00a0<\/b><\/span><\/span><\/span>colocados em contato<\/b><\/span><\/span><\/span>, os dois se comportam como se fossem\u00a0<\/b><\/span><\/span><\/span>um \u00fanico corpo e haver\u00e1 uma distribui\u00e7\u00e3o de cargas nas superf\u00edcies de A e de B.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Ap\u00f3s essa distribui\u00e7\u00e3o de cargas\u00a0<\/b><\/span><\/span><\/span>s\u00e3o separados e A ficar\u00e1 com carga QA\u2019\u00a0e B com carga QB\u2019<\/b><\/span><\/span><\/span>, tal que\u00a0<\/b><\/span><\/span><\/span>QA\u2019\u00a0+ QB\u2019 = + Q (princ\u00edpio da conserva\u00e7\u00e3o das cargas el\u00e9tricas).<\/b><\/span><\/span><\/span><\/p>\n

Observe que, se\u00a0<\/b><\/span><\/span><\/span>os\u00a0condutores forem id\u00eanticos<\/b><\/span><\/span><\/span>\u00a0e com as\u00a0<\/b><\/span><\/span><\/span>mesmas dimens\u00f5es<\/b><\/span><\/span><\/span>, voc\u00ea teria\u00a0<\/b><\/span><\/span><\/span>QA\u2019 = QB\u2019= Q\u2019\u00a0<\/b><\/span><\/span><\/span>\u00a0Q\u2019 + Q\u2019 = + Q\u00a0\u00a0<\/b><\/span><\/span><\/span>Q\u2019=Q\/2<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Se os dois corpos\u00a0<\/b><\/span><\/span><\/span>forem\u00a0id\u00eanticos<\/b><\/span><\/span><\/span>\u00a0e estiverem\u00a0<\/b><\/span><\/span><\/span>inicialmente eletrizados com cargas\u00a0QA\u00a0e QB,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>ap\u00f3s\u00a0<\/b><\/span><\/span><\/span>o contato eles ter\u00e3o cargas id\u00eanticas Q\u2019 tal que\u00a0<\/b><\/span><\/span><\/span>Q\u2019= (QA\u00a0+ QB)\/2.<\/b><\/span><\/span><\/span><\/p>\n

No caso do exerc\u00edcio:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

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

Corpos eletrizados<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Q = n.e\u00a0<\/b><\/span><\/span><\/span>\u00a01 = n.1,6.10-19\u00a0\u00a0<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/span>\u00a0= 1\/1,6.10-19\u00a0= 0,625.1019<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 6,25.1018\u00a0el\u00e9trons.<\/b><\/span><\/span><\/span><\/p>\n

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

04-<\/span><\/span><\/span><\/p>\n

Eletriza\u00e7\u00e3o por contato<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Se os dois corpos\u00a0<\/b><\/span><\/span><\/span>forem\u00a0id\u00eanticos<\/b><\/span><\/span><\/span>\u00a0e estiverem\u00a0<\/b><\/span><\/span><\/span>inicialmente eletrizados com cargas\u00a0QA\u00a0e QB,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>ap\u00f3s\u00a0<\/b><\/span><\/span><\/span>o contato eles ter\u00e3o cargas id\u00eanticas Q\u2019 tal que\u00a0<\/b><\/span><\/span><\/span>Q\u2019= (QA\u00a0+ QB)\/2.<\/b><\/span><\/span><\/span><\/p>\n

Esfera de carga Q em contato com a primeira, neutra<\/b><\/span><\/span><\/span>\u00a0\u00a0Q\u2019<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>=(Q + 0)\/2\u00a0\u00a0<\/b><\/span><\/span><\/span>Q\u2019<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= Q\/2.<\/b><\/span><\/span><\/span><\/p>\n

Esfera de carga Q\u2019 em contato com a segunda, neutra\u00a0<\/b><\/span><\/span><\/span>\u00a0Q\u2019\u2019<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>=(Q\/2 + 0)\/2\u00a0\u00a0<\/b><\/span><\/span><\/span>Q\u2019\u2019<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= Q\/4.<\/b><\/span><\/span><\/span><\/p>\n

Esfera de carga Q\u2019\u2019 em contato com a terceira, neutra\u00a0<\/b><\/span><\/span><\/span>\u00a0Q\u2019\u2019\u2019<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>=(Q\/4 + 0)\/2\u00a0\u00a0<\/b><\/span><\/span><\/span>Q\u2019\u2019\u2019<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= Q\/8.<\/b><\/span><\/span><\/span><\/p>\n

Esfera de carga Q\u2019\u2019\u2019 em contato com a quarta, neutra\u00a0<\/b><\/span><\/span><\/span>\u00a0Q\u2019\u2019\u2019\u2019<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>=(Q\/8 + 0)\/2\u00a0\u00a0<\/b><\/span><\/span><\/span>Q\u2019\u2019\u2019\u2019<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= Q\/16.<\/b><\/span><\/span><\/span><\/p>\n

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

05-<\/span><\/span><\/span><\/p>\n

Se voc\u00ea n\u00e3o domina a teoria, ela \u00e9 extensa e est\u00e1 a seguir:<\/b><\/span><\/span><\/span><\/p>\n

Eletriza\u00e7\u00e3o por atrito<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Para que ocorra\u00a0<\/b><\/span><\/span><\/span>eletriza\u00e7\u00e3o por atrito<\/b><\/span><\/span><\/span>\u00a0os\u00a0<\/b><\/span><\/span><\/span>corpos atritados<\/b><\/span><\/span><\/span>\u00a0devem ser de\u00a0<\/b><\/span><\/span><\/span>materiais diferentes<\/b><\/span><\/span><\/span>\u00a0e a distribui\u00e7\u00e3o de cargas obedece \u00e0 s\u00e9rie\u00a0<\/b><\/span><\/span><\/span>triboel\u00e9trica abaixo (fornecida).<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Atritando\u00a0<\/b><\/span><\/span><\/span>quaisquer materiais dessa s\u00e9rie<\/b><\/span><\/span><\/span>, o que\u00a0<\/b><\/span><\/span><\/span>estiver \u00e0 esquerda ficar\u00e1 com carga positiva e o da direita, negativa.<\/b><\/span><\/span><\/span><\/p>\n


\n<\/b><\/span><\/span><\/span>Eletriza\u00e7\u00e3o por contato<\/b><\/span><\/span><\/span><\/p>\n

Pode ocorrer<\/b><\/span><\/span><\/span>\u00a0entre\u00a0<\/b><\/span><\/span><\/span>dois condutores<\/b><\/span><\/span><\/span>\u00a0(cargas el\u00e9tricas se distribuem em suas superf\u00edcies externas) ou entre um\u00a0<\/b><\/span><\/span><\/span>condutor e um isolante<\/b><\/span><\/span><\/span>\u00a0(no isolante as cargas el\u00e9tricas ficam somente no local do contato).<\/b><\/span><\/span><\/span><\/p>\n

Considere\u00a0<\/b><\/span><\/span><\/span>um condutor A eletrizado com carga positiva (QA\u00a0= + Q)<\/b><\/span><\/span><\/span>\u00a0e outro\u00a0<\/b><\/span><\/span><\/span>condutor B,
\n<\/b><\/span><\/span><\/span>eletricamente\u00a0<\/b><\/span><\/span><\/span>neutro<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>(QB\u00a0= 0)<\/b><\/span><\/span><\/span>\u00a0presos a\u00a0<\/b><\/span><\/span><\/span>suportes isolantes<\/b><\/span><\/span><\/span>\u00a0(para que n\u00e3o descarreguem,\u00a0<\/b><\/span><\/span><\/span>inicialmente<\/b><\/span><\/span><\/span><\/p>\n

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

Quando s\u00e3o\u00a0<\/b><\/span><\/span><\/span>colocados em contato<\/b><\/span><\/span><\/span>, os dois se comportam como se fossem\u00a0<\/b><\/span><\/span><\/span>um \u00fanico corpo e haver\u00e1 uma distribui\u00e7\u00e3o de cargas nas superf\u00edcies de A e de B.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Ap\u00f3s essa distribui\u00e7\u00e3o de cargas\u00a0<\/b><\/span><\/span><\/span>s\u00e3o separados e A ficar\u00e1 com carga QA\u2019\u00a0e B com carga QB\u2019<\/b><\/span><\/span><\/span>, tal que\u00a0<\/b><\/span><\/span><\/span>QA\u2019\u00a0+ QB\u2019 = + Q (princ\u00edpio da conserva\u00e7\u00e3o das cargas el\u00e9tricas).<\/b><\/span><\/span><\/span><\/p>\n

Observe que, se\u00a0<\/b><\/span><\/span><\/span>os\u00a0condutores forem id\u00eanticos<\/b><\/span><\/span><\/span>\u00a0e com as\u00a0<\/b><\/span><\/span><\/span>mesmas dimens\u00f5es<\/b><\/span><\/span><\/span>, voc\u00ea teria\u00a0<\/b><\/span><\/span><\/span>QA\u2019 = QB\u2019= Q\u2019\u00a0<\/b><\/span><\/span><\/span>\u00a0Q\u2019 + Q\u2019 = + Q\u00a0\u00a0<\/b><\/span><\/span><\/span>Q\u2019= Q\/2<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Se os dois corpos\u00a0<\/b><\/span><\/span><\/span>forem\u00a0id\u00eanticos<\/b><\/span><\/span><\/span>\u00a0e estiverem\u00a0<\/b><\/span><\/span><\/span>inicialmente eletrizados com cargas\u00a0QA\u00a0e QB,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>ap\u00f3s\u00a0<\/b><\/span><\/span><\/span>o contato eles ter\u00e3o cargas id\u00eanticas Q\u2019 tal que\u00a0<\/b><\/span><\/span><\/span>Q\u2019= (QA\u00a0+ QB)\/2.<\/b><\/span><\/span><\/span><\/p>\n

Exemplos:<\/b><\/span><\/span><\/span><\/p>\n

Em todos os casos a seguir as\u00a0<\/b><\/span><\/span><\/span>esferas s\u00e3o id\u00eanticas,<\/b><\/span><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span><\/span>depois de colocadas em contato s\u00e3o separadas e cada uma delas ficar\u00e1 com carga Q\u2019:<\/b><\/span><\/span><\/span><\/p>\n

a)<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>QA\u00a0= 0 e QB= \u2013 8\u03bcC<\/b><\/span><\/span><\/span>\u00a0\u00a0Q\u2019 = (QA\u00a0+ QB)\/2 = (0 \u2013 8\u03bc)\/2\u00a0\u00a0<\/b><\/span><\/span><\/span>Q\u2019= \u2013 4\u03bcC (carga de cada uma ap\u00f3s o contato)<\/b><\/span><\/span><\/span><\/p>\n

b) QA= 2C e QB= \u2013 6C<\/b><\/span><\/span><\/span>\u00a0\u00a0Q\u2019 = (QA\u00a0+ QB)\/2 = (2 \u2013 6)\/2\u00a0\u00a0<\/b><\/span><\/span><\/span>Q\u2019 = -2C (carga de cada uma ap\u00f3s o contato)<\/b><\/span><\/span><\/span><\/p>\n

c)<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>QA\u00a0= 6\u03bcC, QB\u00a0= -8 \u03bcC e QC= 12 \u03bcC\u00a0<\/b><\/span><\/span><\/span>\u00a0Q\u2019 = (QA\u00a0+ QB\u00a0+ QC)\/3=(6 \u03bc \u2013 8 \u03bc + 12 \u03bc)\/3\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Q\u2019= (10\/3) \u03bcC.<\/b><\/span><\/span><\/span><\/p>\n


\n<\/b><\/span><\/span><\/span>Eletriza\u00e7\u00e3o por indu\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

Induzir eletricamente<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>significa provocar\u00a0<\/b><\/span><\/span><\/span>uma separa\u00e7\u00e3o de cargas, sem que haja contato. Observe dois processos:<\/b><\/span><\/span><\/span><\/p>\n

Primeiro processo<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>Considere duas esferas condutoras, inicialmente descarregadas (neutras)<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>e encostadas uma na outra (figura 1).\u00a0<\/b><\/span><\/span><\/span>Aproxima-se, sem encostar,<\/b><\/span><\/span><\/span>\u00a0uma\u00a0<\/b><\/span><\/span><\/span>terceira\u00a0<\/b><\/span><\/span><\/span>esfera\u00a0<\/b><\/span><\/span><\/span><\/p>\n

condutora\u00a0<\/b><\/span><\/span><\/span>eletrizada.<\/b><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span><\/span>esfera eletrizada<\/b><\/span><\/span><\/span>\u00a0provoca uma\u00a0<\/b><\/span><\/span><\/span>indu\u00e7\u00e3o el\u00e9trica (movimenta\u00e7\u00e3o de cargas) nas outras duas at\u00e9 que elas apresentem a distribui\u00e7\u00e3o indicada na figura 3.<\/b><\/span><\/span><\/span><\/p>\n

Em seguida,\u00a0<\/b><\/span><\/span><\/span>sem tirar do lugar a esfera eletrizada,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>afasta-se um pouco uma esfera da outra.<\/b><\/span><\/span><\/span>\u00a0Finalmente,\u00a0<\/b><\/span><\/span><\/span>sem mexer mais nas esferas<\/b><\/span><\/span><\/span>,\u00a0<\/b><\/span><\/span><\/span>remove-se a esfera eletrizada<\/b><\/span><\/span><\/span>, levando-a para\u00a0<\/b><\/span><\/span><\/span>muito longe<\/b><\/span><\/span><\/span>\u00a0das esferas.<\/b><\/span><\/span><\/span><\/p>\n

Nessa\u00a0<\/b><\/span><\/span><\/span>situa\u00e7\u00e3o final, as<\/b><\/span><\/span><\/span>\u00a0cargas se\u00a0<\/b><\/span><\/span><\/span>espalham ficando na superf\u00edcie externa de cada carga, ficando eletrizadas conforme a figura 4.<\/b><\/span><\/span><\/span><\/p>\n

Segundo processo<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>Aproxima-se um condutor eletrizado<\/b><\/span><\/span><\/span>, por exemplo,\u00a0<\/b><\/span><\/span><\/span>negativamente<\/b><\/span><\/span><\/span>\u00a0(indutor) de um\u00a0<\/b><\/span><\/span><\/span>condutor neutro que se deseja eletrizar (induzido), sem encostar, mas bem pr\u00f3ximos (figura 1).<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Os\u00a0<\/b><\/span><\/span><\/span>el\u00e9trons livres do indutor<\/b><\/span><\/span><\/span>\u00a0ser\u00e3o\u00a0<\/b><\/span><\/span><\/span>repelidos,<\/b><\/span><\/span><\/span>\u00a0ficando\u00a0<\/b><\/span><\/span><\/span>o lado direito do induzido com excesso de el\u00e9trons<\/b><\/span><\/span><\/span>\u00a0e o lado\u00a0<\/b><\/span><\/span><\/span>esquerdo com falta de el\u00e9trons,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>fen\u00f4meno\u00a0<\/b><\/span><\/span><\/span>que recebe o nome de\u00a0<\/b><\/span><\/span><\/span>indu\u00e7\u00e3o el\u00e9trica\u00a0(figura 2).<\/b><\/span><\/span><\/span><\/p>\n

As\u00a0<\/b><\/span><\/span><\/span>cargas el\u00e9tricas<\/b><\/span><\/span><\/span>\u00a0que se concentram nas\u00a0<\/b><\/span><\/span><\/span>duas extremidades opostas\u00a0<\/b><\/span><\/span><\/span>(denominadas\u00a0cargas induzidas) s\u00e3o de\u00a0<\/b><\/span><\/span><\/span>mesmo m\u00f3dulo, mas de sinais opostos<\/b><\/span><\/span><\/span>\u00a0e, por esse motivo o\u00a0<\/b><\/span><\/span><\/span>induzido continua neutro.<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Em seguida,\u00a0<\/b><\/span><\/span><\/span>ainda na presen\u00e7a do indutor<\/b><\/span><\/span><\/span>\u00a0voc\u00ea liga\u00a0<\/b><\/span><\/span><\/span>o induzido \u00e0 terra\u00a0<\/b><\/span><\/span><\/span>(fio terra ou com seu dedo) e\u00a0<\/b><\/span><\/span><\/span>observa que el\u00e9trons se deslocam do induzido para a terra (figura 3).<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Se o\u00a0<\/b><\/span><\/span><\/span>indutor tivesse cargas positivas<\/b><\/span><\/span><\/span>, ao\u00a0<\/b><\/span><\/span><\/span>final de todo o processo o induzido ficaria eletrizado com cargas negativas.<\/b><\/span><\/span><\/span><\/p>\n

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

06-<\/span><\/span><\/span><\/p>\n

s\u00a0<\/b><\/span><\/span><\/span>for\u00e7as el\u00e9tricas trocadas entre A e B s\u00e3o de repuls\u00e3o<\/b><\/span><\/span><\/span>\u00a0(qA\u00a0e qB\u00a0se repelem\u00a0<\/b><\/span><\/span><\/span>\u201cmesmo sinal\u201d<\/b><\/span><\/span><\/span>) e possuem a\u00a0<\/b><\/span><\/span><\/span>mesma intensidade (m\u00f3dulo) pelo princ\u00edpio da a\u00e7\u00e3o e rea\u00e7\u00e3o<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>FA\u00a0= FB.<\/b><\/span><\/span><\/span><\/p>\n

Como as setas representam as acelera\u00e7\u00f5es, a intensidade (m\u00f3dulo) de B \u00e9 maior que a de A\u00a0\u00a0aB\u00a0> aA.<\/b><\/span><\/span><\/span><\/p>\n

Segunda lei de Newton\u00a0\u00a0FA\u00a0= mA.aA\u00a0e FB\u00a0= mB.aB\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>FA\u00a0= FB\u00a0\u00a0mA.aA\u00a0= mB.aB\u00a0<\/b><\/span><\/span><\/span>\u00a0como\u00a0<\/b><\/span><\/span><\/span>aB\u00a0> aA, mB\u00a0deve ser menor que mA\u00a0\u00a0mB\u00a0< mA.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Campo El\u00e9trico Uniforme<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

R \u2013 C<\/b><\/span><\/span><\/span><\/p>\n

08-<\/span><\/span><\/span><\/p>\n

Representa\u00e7\u00e3o esquem\u00e1tica das linhas de campo ou linhas de for\u00e7a de um campo el\u00e9trico<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

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

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

<\/p>\n

Como a\u00a0<\/b><\/span><\/span><\/span>carga q \u00e9 negativa e o campo el\u00e9trico tem dire\u00e7\u00e3o vertical e sentido para baixo<\/b><\/span><\/span><\/span>\u00a0ent\u00e3o a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a el\u00e9trica sobre q t ter\u00e1 dire\u00e7\u00e3o vertical e sentido para cima.<\/b><\/span><\/span><\/span><\/p>\n

Na\u00a0<\/b><\/span><\/span><\/span>figura abaixo est\u00e3o\u00a0<\/b><\/span><\/span><\/span>a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a el\u00e9trica e o peso que agem sobre a carga q.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

R- (02 + 04 + 08) = 14<\/b><\/span><\/span><\/span><\/p>\n

10-<\/span><\/span><\/span><\/p>\n

Se o campo el\u00e9trico \u00e9 uniforme, temos que:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Vamos retomar a imagem do exerc\u00edcio:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Vamos\u00a0<\/b><\/span><\/span><\/span>analisar a diferen\u00e7a de potencial U na dimens\u00e3o X e Y<\/b><\/span><\/span><\/span>, para isso vamos\u00a0<\/b><\/span><\/span><\/span>isolar o E\u00a0<\/b><\/span><\/span><\/span>na equa\u00e7\u00e3o de cima:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Agora que sabemos os campos em X e Y\u00a0<\/b><\/span><\/span><\/span>precisamos achar o seu resultante<\/b><\/span><\/span><\/span>, como eles s\u00e3o ortogonais (est\u00e3o a 90\u00b0 um do outro),\u00a0<\/b><\/span><\/span><\/span>podemos usar Pit\u00e1goras:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Alternativa correta \u00e9 a D.<\/b><\/span><\/span><\/span><\/p>\n

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

Como o enunciado pede o\u00a0<\/b><\/span><\/span><\/span>campo el\u00e9trico fora da casca esf\u00e9rica<\/b><\/span><\/span><\/span>, podemos usar o\u00a0<\/b><\/span><\/span><\/span>princ\u00edpio da superposi\u00e7\u00e3o,\u00a0<\/b><\/span><\/span><\/span>que diz que o\u00a0<\/b><\/span><\/span><\/span>campo E1<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>fora da casca esf\u00e9rica ser\u00e1 igual a soma\u00a0<\/b><\/span><\/span><\/span>de um campo E4a<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>gerado por uma esfera\u00a0<\/b><\/span><\/span><\/span>de raio 4a\u00a0<\/b><\/span><\/span><\/span>carregada\u00a0<\/b><\/span><\/span><\/span>positivamente<\/b><\/span><\/span><\/span>\u00a0e de outro\u00a0<\/b><\/span><\/span><\/span>campo Ea<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>gerado por uma outra esfera\u00a0<\/b><\/span><\/span><\/span>de raio a\u00a0<\/b><\/span><\/span><\/span>carregada\u00a0<\/b><\/span><\/span><\/span>negativamente<\/b><\/span><\/span><\/span>, ambas posicionadas\u00a0<\/b><\/span><\/span><\/span>no mesmo centro<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n


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

No qual k0\u00a0\u00e9 a constante el\u00e9trica no v\u00e1cuo. Caso o campo d\u00ea positivo,\u00a0<\/b><\/span><\/span><\/span>ele \u00e9 radial para fora<\/b><\/span><\/span><\/span>, enquanto que se for negativo, \u00e9\u00a0<\/b><\/span><\/span><\/span>radial para dentro<\/b><\/span><\/span><\/span>. Essa mesma f\u00f3rmula pode ser usada para as esferas de carga Q e de qualquer raio R, por\u00e9m foi dada a\u00a0<\/b><\/span><\/span><\/span>densidade volum\u00e9trica de carga \u03c1<\/b><\/span><\/span><\/span>\u00a0ao em vez da\u00a0<\/b><\/span><\/span><\/span>carga el\u00e9trica<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>, devemos ent\u00e3o fazer a convers\u00e3o utilizando a\u00a0<\/b><\/span><\/span><\/span>defini\u00e7\u00e3o de densidade de carga<\/b><\/span><\/span><\/span>, que \u00e9 a\u00a0<\/b><\/span><\/span><\/span>carga dividida pelo volume<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

O\u00a0<\/b><\/span><\/span><\/span>campo total E\u00a0<\/b><\/span><\/span><\/span>ser\u00e1 ent\u00e3o a\u00a0<\/b><\/span><\/span><\/span>soma de todos os campos calculados<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

12-<\/span><\/span><\/span><\/p>\n

\u00a0a) A\u00a0<\/b><\/span><\/span>corrente m\u00e9dia<\/b><\/span><\/span>\u00a0\u00e9 definida como a\u00a0<\/b><\/span>varia\u00e7\u00e3o de carga sobre um intervalo de tempo<\/b><\/span><\/span>. A\u00a0<\/b><\/span>varia\u00e7\u00e3o da carga el\u00e9trica total \u00e9 o n\u00famero de el\u00e9trons que fluiu de uma esfera at\u00e9 a outra multiplicada pela carga el\u00e9trica fundamental do el\u00e9tron<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Substituindo a \u00faltima equa\u00e7\u00e3o na primeira, e colocando os valores dados no enunciado, chegamos em:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

b) O\u00a0<\/b><\/span><\/span>campo el\u00e9trico\u00a0<\/b><\/span><\/span>\u00e9 dado pela rela\u00e7\u00e3o:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Em que k \u00e9 a constante eletrost\u00e1tica do meio (neste caso \u00e9 a do ar), Q \u00e9 o valor da carga geradora e d \u00e9 a dist\u00e2ncia da carga geradora ao ponto em que queremos conhecer o valor do campo.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Repare que teremos\u00a0<\/b><\/span><\/span>dois campos atuando no ponto M\u00a0<\/b><\/span><\/span>relacionados \u00e0s\u00a0<\/b><\/span><\/span>esferas 1 e 2<\/b><\/span><\/span>. Utilizando a f\u00f3rmula do campo el\u00e9trico para as esferas teremos:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Para encontrar o campo el\u00e9trico total no ponto M, devemos observar o vetor dos campos el\u00e9tricos das esferas individualmente, respeitando a seguinte constata\u00e7\u00e3o:<\/b><\/span><\/span><\/span><\/span><\/p>\n

    \n
  1. Objetos com carga positiva possuem suas linhas de campo apontando para fora.<\/b><\/span><\/span><\/span><\/span><\/li>\n
  2. Objetos com carga negativa possuem suas linhas de campo apontando para dentro.<\/b><\/span><\/span><\/span><\/span><\/li>\n<\/ol>\n

    Logo abaixo h\u00e1 uma ilustra\u00e7\u00e3o de como as linhas de campo se comportam para objetos redondos carregados.<\/b><\/span><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Aplicando este conceito na figura, temos:<\/b><\/span><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Como os campos est\u00e3o opostos um ao outro no ponto M, o m\u00f3dulo do campo total nesse ponto ser\u00e1 dado pela diferen\u00e7a entre os dois, tal que:<\/b><\/span><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    13-<\/span><\/span><\/span><\/p>\n

    <\/p>\n

    14-<\/span><\/span><\/span><\/p>\n

    Como, pelo\u00a0<\/b><\/span><\/span><\/span>enunciado o el\u00e9tron se desloca de A para B,<\/b><\/span><\/span><\/span>\u00a0ent\u00e3o a\u00a0<\/b><\/span><\/span><\/span>placa A \u00e9 negativa\u00a0<\/b><\/span><\/span><\/span>(menor potencial), pois deve\u00a0<\/b><\/span><\/span><\/span>repelir o el\u00e9tron (carga negativa)\u00a0<\/b><\/span><\/span><\/span>e a\u00a0<\/b><\/span><\/span><\/span>placa B \u00e9 positiva\u00a0<\/b><\/span><\/span><\/span>(maior potencial)\u00a0<\/b><\/span><\/span><\/span>devendo atrair o el\u00e9tron.<\/b><\/span><\/span><\/span><\/p>\n

    I.\u00a0Correta\u00a0<\/b><\/span><\/span><\/span>\u00a0Considerando a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a el\u00e9trica\u00a0<\/b><\/span><\/span><\/span>como a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a resultante sobre o el\u00e9tron\u00a0<\/b><\/span><\/span><\/span>(\u00fanica for\u00e7a que age sobre ele) e lembrando que o\u00a0<\/b><\/span><\/span><\/span>campo el\u00e9trico entre as placas \u00e9 uniforme,<\/b><\/span><\/span><\/span>\u00a0vamos calcular o\u00a0<\/b><\/span><\/span><\/span>trabalho da for\u00e7a el\u00e9trica entre A e B (WAB),\u00a0<\/b><\/span><\/span><\/span>em\u00a0<\/b><\/span><\/span><\/span>fun\u00e7\u00e3o da diferen\u00e7a de potencial entre A e B sabendo que o el\u00e9tron possui carga negativa e que VA\u00a0< VB:<\/b><\/span><\/span><\/span><\/p>\n

    WBA\u00a0= q.(VA\u00a0\u2013 VB)\u00a0<\/b><\/span><\/span><\/span>= (-1,6.10-19).(- 2,0.104)\u00a0<\/b><\/span><\/span><\/span>= 3,2.10-15\u00a0J.\u00a0<\/b><\/span><\/span><\/span>(Observe que esse\u00a0<\/b><\/span><\/span><\/span>trabalho independe da trajet\u00f3ria dependendo apenas de q e de (VA\u00a0\u2013 VB)\u00a0<\/b><\/span><\/span><\/span>que\u00a0<\/b><\/span><\/span><\/span>s\u00e3o os mesmos para as duas trajet\u00f3rias)<\/b><\/span><\/span><\/span><\/p>\n

    Esse trabalho tamb\u00e9m\u00a0<\/b><\/span><\/span><\/span>\u00e9 o mesmo\u00a0<\/b><\/span><\/span><\/span>quando\u00a0<\/b><\/span><\/span><\/span>calculado pela varia\u00e7\u00e3o da energia cin\u00e9tica\u00a0<\/b><\/span><\/span><\/span>(Teorema da energia cin\u00e9tica)\u00a0\u00a0<\/b><\/span><\/span><\/span>WAB\u00a0= EcB\u00a0\u2013 EcA\u00a0<\/b><\/span><\/span><\/span>\u00a03,2.10-15\u00a0= EcB\u00a0\u2013 mVA2\/2 = EcB\u00a0\u2013 m.02\/2 = EcB\u00a0\u2013 0\u00a0\u00a0<\/b><\/span><\/span><\/span>EcB\u00a0= 3,2.10-15\u00a0J.(energia cin\u00e9tica com que o el\u00e9tron atinge a placa B).<\/b><\/span><\/span><\/span><\/p>\n

    II.\u00a0Correta\u00a0\u00a0Como j\u00e1 vimos em I.<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    III.\u00a0Falsa\u00a0\u00a0Veja I e II.<\/b><\/span><\/span><\/span><\/p>\n

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

    15-<\/span><\/span><\/span><\/p>\n

    \u00a0Essa\u00a0<\/b><\/span><\/span><\/span>descarga entre a nuvem (cargas negativas) e o solo (cargas positivas)\u00a0<\/b><\/span><\/span><\/span>atrav\u00e9s do ar (diel\u00e9trico) que,\u00a0<\/b><\/span><\/span><\/span>devido \u00e0 elevada ddp U entre as nuvens e o solo\u00a0<\/b><\/span><\/span><\/span>se transforma em condutor, tem o mesmo comportamento el\u00e9trico que o de um\u00a0<\/b><\/span><\/span><\/span>capacitor de capacidade C = Q\/U.<\/b><\/span><\/span><\/span><\/p>\n

    Supondo essa capacidade constante, o gr\u00e1fico\u00a0<\/b><\/span><\/span><\/span>UxQ est\u00e1 representado abaixo:<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Mas observe que\u00a0<\/b><\/span><\/span><\/span>n\u00e3o temos a diferen\u00e7a de potencial.<\/b><\/span><\/span><\/span>\u00a0Ent\u00e3o vamos encontr\u00e1-la. O enunciado informa que\u00a0<\/b><\/span><\/span><\/span>temos um campo el\u00e9trico uniforme:<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Substituindo U = E.d em Ep\u00a0= Q.U\/d\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Ep\u00a0=\u00a0<\/b><\/span><\/span><\/span><\/p>\n

    Q\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>carga el\u00e9trica\u00a0<\/b><\/span><\/span><\/span>transportada\u00a0<\/b><\/span><\/span><\/span>Q = 2.102\u00a0C.<\/b><\/span><\/span><\/span><\/p>\n

    E<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0intensidade do\u00a0<\/b><\/span><\/span><\/span>campo el\u00e9trico\u00a0<\/b><\/span><\/span><\/span>entre a base da nuvem e o solo\u00a0\u00a0<\/b><\/span><\/span><\/span>E = 5.106\u00a0V\/m.<\/b><\/span><\/span><\/span><\/p>\n

    d\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>dist\u00e2ncia e<\/b><\/span><\/span><\/span>ntre a base da nuvem e o solo\u00a0\u00a0<\/b><\/span><\/span><\/span>d = 500 m.<\/b><\/span><\/span><\/span><\/p>\n

    Substituindo esses valores em Ep\u00a0=\u00a0\u00a0Ep\u00a0=\u00a0\u00a0= 2500.108\u00a0\u00a0<\/b><\/span><\/span><\/span>Ep\u00a0=\u00a0\u00a0J.<\/b><\/span><\/span><\/span><\/p>\n

    R- C<\/b><\/span><\/span><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

    Vestibulares Recentes \u2013 Resolu\u00e7\u00e3o Comentada \u2013 Eletrost\u00e1tica \u2013 2018\/2019 01- Aqui, a \u00fanica alternativa incorreta \u00e9 alternativa (C), pois a carga el\u00e9trica \u00e9 transferida\u00a0exclusivamente\u00a0pelo movimento dos\u00a0el\u00e9trons, deixando os\u00a0pr\u00f3tons fixos. A carga\u00a0positiva\u00a0acaba sendo ent\u00e3o dada pela\u00a0aus\u00eancia de el\u00e9trons. Logo, a\u00a0(C) deve ser a alternativa marcada.   02- Eletriza\u00e7\u00e3o por contato Pode ocorrer\u00a0entre\u00a0dois condutores\u00a0(cargas el\u00e9tricas se distribuem em suas superf\u00edcie externas) ou<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":4843,"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-4845","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4845","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=4845"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4845\/revisions"}],"predecessor-version":[{"id":4846,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4845\/revisions\/4846"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4843"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=4845"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}