{"id":3325,"date":"2017-10-01T20:34:14","date_gmt":"2017-10-01T20:34:14","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3325"},"modified":"2024-09-04T14:05:07","modified_gmt":"2024-09-04T14:05:07","slug":"resolucao-eletrostatica-2014-2013","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/eletricidade\/eletrostatica-2014-2013\/resolucao-eletrostatica-2014-2013\/","title":{"rendered":"Resolu\u00e7\u00e3o Eletrost\u00e1tica – 2014 – 2013"},"content":{"rendered":"

Resolu\u00e7\u00e3o Comentada \u2013 2014- 2013 <\/span><\/span><\/span><\/p>\n

ELETROST\u00c1TICA<\/span><\/span><\/span><\/p>\n

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

Lei de Coulomb:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Cargas el\u00e9tricas de mesmo sinal se repelem e de sinais opostos se atraem — analise atentamente a figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe que pela lei de Coulomb F=KQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0que — F<\/b><\/span><\/span><\/span>4A<\/b><\/span><\/span><\/sub><\/span>\u00a0= F<\/b><\/span><\/span><\/span>2A<\/b><\/span><\/span><\/sub><\/span>\u00a0e que F<\/b><\/span><\/span><\/span>1A\u00a0<\/b><\/span><\/span><\/sub><\/span>= F<\/b><\/span><\/span><\/span>3A<\/b><\/span><\/span><\/sub><\/span>\u00a0— adicionando vetorialmente essas for\u00e7as voc\u00ea obt\u00e9m a for\u00e7a resultante\u00a0\"\"\u00a0pedida, cuja dire\u00e7\u00e3o e sentido<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

est\u00e3o indicados na figura — <\/b><\/span><\/span><\/span>R – B<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Por conven\u00e7\u00e3o, as linhas de for\u00e7a do campo el\u00e9trico e, consequentemente, o pr\u00f3prio campo el\u00e9trico\u00a0\"\"se\u00a0afastam das cargas positivas e se aproximam das negativas\u00a0— assim, a placa\u00a0A est\u00e1 eletrizada com cargas positivas e a B com cargas negativas<\/b><\/span><\/span><\/span><\/p>\n

Quando a part\u00edcula com\u00a0carga positiva q\u00a0\u00e9 abandonada em M ela ser\u00e1\u00a0repelida pela placa A,\u00a0positiva\u00a0e\u00a0atra\u00edda pela placa B,\u00a0negativa, se deslocando para a\u00a0direita at\u00e9 atingir a placa B.<\/b><\/span><\/span><\/span><\/p>\n

Como s\u00e3o desprezados os efeitos gravitacionais (for\u00e7a peso) e existe o v\u00e1cuo entre as placas (n\u00e3o existe atrito) a for\u00e7a resultante (F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a) sobre a carga\u00a0q\u00a0ser\u00e1 a for\u00e7a el\u00e9trica de intensidade\u00a0F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=q.E, tal que:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Ent\u00e3o a part\u00edcula de carga positiva\u00a0q\u00a0\u00e9 acelerada a partir de M (de onde sai do repouso\u00a0V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0) e, ap\u00f3s percorres\u00a0d\/2\u00a0atinge a placa B com velocidade\u00a0V, fornecida pela equa\u00e7\u00e3o de\u00a0Torricelli:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo\u00a0n\u00a0as esferas, quando separadas em agrupamentos com quantidades iguais de esferas, cada agrupamentos A, B e C ter\u00e1 n\/3 esferas id\u00eanticas conforme a figura:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Quando voc\u00ea coloca a esfera\u00a0P\u00a0eletrizada com carga\u00a0Q\u00a0em contato com as\u00a0n\/3\u00a0esferas id\u00eanticas e neutras,<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

do agrupamento A, essa carga\u00a0Q\u00a0ser\u00e1\u00a0igualmente distribu\u00edda\u00a0entre as (1 + n\/3)\u00a0esferas e a nova carga\u00a0Q\u2019\u00a0da esfera P ser\u00e1:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Agora voc\u00ea vai colocar essa nova carga\u00a0Q\u2019\u00a0da esfera\u00a0P\u00a0em contato com as\u00a0(1 + n\/3)\u00a0esferas id\u00eanticas e<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

neutras do agrupamento B e a nova carga Q\u2019\u2019 da esfera P ser\u00e1:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

04-<\/b><\/span><\/span><\/span> <\/b>F<\/b><\/span><\/span><\/span>q1Q<\/b><\/span><\/span><\/sub><\/span>=kq<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.Q\/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>-6<\/b><\/span><\/span><\/sup><\/span>.5.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\/(5.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=180.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\/25.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0F<\/b><\/span><\/span><\/span>q1Q<\/b><\/span><\/span><\/sub><\/span>\u00a0= F<\/b><\/span><\/span><\/span>q2Q<\/b><\/span><\/span><\/sub><\/span>\u00a0= 7,2.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0= 0,72N.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe na\u00a0figura acima que a for\u00e7a resultante \u00e9 a soma vetorial\u00a0das duas for\u00e7a achadas —\u00a0F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 0,72<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 0,72<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2.0,72<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0— F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=0,72.\u221a2=0,72.1,4 —\u00a0F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=1,008×10=10N<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Condutores: Aproxima-se um condutor eletrizado, por exemplo, negativamente de um condutor neutro, sem<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

encostar, mas bem pr\u00f3ximos (figura 1).Os el\u00e9trons livres do indutor ser\u00e3o repelidos, ficando o lado direito do induzido com excesso de el\u00e9trons e o lado esquerdo com falta de el\u00e9trons, fen\u00f4meno que recebe o nome de\u00a0indu\u00e7\u00e3o el\u00e9trica\u00a0(figura 2). Observe na figura 2 que a\u00a0for\u00e7a de atra\u00e7\u00e3o\u00a0entre as cargas negativas e positivas \u00e9\u00a0maior\u00a0que a\u00a0for\u00e7a de repuls\u00e3o\u00a0entre as cargas negativasporque a dist\u00e2ncia entre as cargas\u00a0que se atraem\u00a0\u00e9 menor\u00a0e, quanto\u00a0menor a dist\u00e2nciaentre cargas de mesmo m\u00f3dulo,\u00a0maior a for\u00e7a\u00a0entre elas.<\/b><\/span><\/span><\/span><\/p>\n

<\/a> Isolantes:\u00a0Polariza\u00e7\u00e3o de cargas num isolante\u00a0 — nos isolantes a polariza\u00e7\u00e3o acontece nas mol\u00e9culas ou os \u00e1tomos da superf\u00edcie do corpo.<\/b><\/span><\/span><\/span><\/p>\n

Polariza\u00e7\u00e3o at\u00f4mica<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n\n\n<\/colgroup>\n\n\n
Nos isolantes a polariza\u00e7\u00e3o acontece nas mol\u00e9culas ou os \u00e1tomos da superf\u00edcie do corpo.<\/b><\/span><\/span><\/span><\/p>\n

Polariza\u00e7\u00e3o at\u00f4mica<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n\n\n\n\n<\/colgroup>\n\n\n\n
\"\"<\/td>\n\"\"<\/td>\n\"\"<\/td>\n<\/tr>\n
A nuvem eletr\u00f4nica distribu\u00edda<\/b><\/span><\/span><\/span><\/p>\n

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

\u00c1tomo n\u00e3o polarizado.<\/b><\/span><\/span><\/span><\/td>\n

Deforma\u00e7\u00e3o da nuvem eletr\u00f4nica pela a\u00e7\u00e3o de cargas indutoras negativas<\/b><\/span><\/span><\/span><\/p>\n

\u00c1tomo polarizado. Dipolo el\u00e9trico.<\/b><\/span><\/span><\/span><\/td>\n

Deforma\u00e7\u00e3o da nuvem eletr\u00f4nica<\/b><\/span><\/span><\/span><\/p>\n

pela a\u00e7\u00e3o de cargas indutoras positivas.<\/b><\/span><\/span><\/span><\/p>\n

\u00c1tomo polarizado. Dipolo el\u00e9trico.<\/b><\/span><\/span><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Nos \u00e1tomos, em torno do n\u00facleo positivo, os el\u00e9trons se distribuem numa “nuvem eletr\u00f4nica” com\u00a0 densidades vari\u00e1veis; onde a densidade for maior, maior \u00e9 a probabilidade de se encontrar os el\u00e9trons.<\/b><\/span><\/span><\/span><\/p>\n

Essa nuvem pode ser deformada, deslocando os centros de cargas positivas e negativas em dois p\u00f3los.<\/b><\/span><\/span><\/span><\/p>\n

Com as mol\u00e9culas – combina\u00e7\u00e3o de \u00e1tomos – tamb\u00e9m ocorre o fen\u00f4meno da polariza\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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

\u00a006-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Os\u00a0vetores em azul\u00a0representam as for\u00e7as trocadas entre\u00a0cada esfera\u00a0e observe que <\/b><\/span><\/span><\/span><\/p>\n

\"\" a\u00a0incompat\u00edvel\u00a0\u00e9\u00a0a a\u00a0alternativa E, pois se a\u00a0for\u00e7a entre dois pares \u00e9 de atra\u00e7\u00e3o (sinais opostos),\u00a0a\u00a0for\u00e7a entre o outro par dever\u00e1 ser de repuls\u00e3o (mesmo sinal).<\/b><\/span><\/span><\/span><\/p>\n

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

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Os\u00a0dois\u00a0geradores se comportam como se fossem\u00a0um s\u00f3 de \u03b5=5V, eletrizando o capacitor C com essa ddp. de\u00a0U=5V.<\/b><\/span><\/span><\/span><\/p>\n

Energia\u00a0no capacitor —\u00a0W=Q.U\/2\u00a0— 125.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0= Q.5\/2 —\u00a0Q=50.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>C.<\/b><\/span><\/span><\/span><\/p>\n

Capacit\u00e2ncia\u00a0do capacitor —\u00a0C=Q\/U=50.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\/5 —\u00a0C=10.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0F.<\/b><\/span><\/span><\/span><\/p>\n

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

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Se\u00a0B tem carga negativa e repele A ent\u00e3o A ter\u00e1 tamb\u00e9m carga negativa\u00a0(cargas de mesmo sinal se repelem).<\/b><\/span><\/span><\/span><\/p>\n

Se, quando o objeto met\u00e1lico \u00e9\u00a0aproximado de A que possui carga negativa\u00a0\u00e9 atra\u00eddo,\u00a0ou ele possui carga positiva ou est\u00e1 neutro e \u00e9 atra\u00eddo por indu\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

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

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>I. Falsa\u00a0— o campo \u00e9\u00a0mais intenso\u00a0onde ocorre\u00a0maior concentra\u00e7\u00e3o de linhas de campo el\u00e9trico, no caso, na regi\u00e3o B.<\/b><\/span><\/span><\/span><\/p>\n

II. Falsa\u00a0—\u00a0veja III.<\/b><\/span><\/span><\/span><\/p>\n

III. Correta\u00a0— veja na figura abaixo que quem\u00a0origina esse campo \u00e9 uma carga negativa\u00a0e por isso<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

o\u00a0potencial el\u00e9trico \u00e9 maior em A, conforme voc\u00ea pode observar na parte inferior do gr\u00e1fico acima que fornece os valores alg\u00e9bricos do potencial de uma carga negativa.<\/b><\/span><\/span><\/span><\/p>\n

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

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo a carga el\u00e9trica\u00a0q negativa e o campo el\u00e9trico vertical e para baixo,\u00a0a for\u00e7a el\u00e9trica\u00a0sobre a\u00a0carga q ter\u00e1 dire\u00e7\u00e3o vertical e para cima\u00a0e intensidade\u00a0—\u00a0F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=q.E=q.50 —\u00a0F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=50q.<\/b><\/span><\/span><\/span><\/p>\n

Peso do bal\u00e3o — P=mg=10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.10 — P=10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>N.<\/b><\/span><\/span><\/span><\/p>\n

Como ele\u00a0flutua\u00a0a resultante das for\u00e7as que agem sobre ele \u00e9 nula e\u00a0F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=P\u00a0— 50q = 10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0q=<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\/50=0,02.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>=2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0q=200.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>=200 \u03bcC.<\/b><\/span><\/span><\/span><\/p>\n

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

Carga el\u00e9trica e processos de eletriza\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

70-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01. Falsa\u00a0 —\u00a0 ficar\u00e1 com carga positiva, mas \u00e9 porque ceder\u00e1 el\u00e9trons ao teflon.<\/b><\/span><\/span><\/span><\/p>\n

02. Correta\u00a0 —\u00a0 Na figura abaixo voc\u00ea disp\u00f5e de uma placa de vidro e de um pano de l\u00e3 presos em suportes isolantes<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

para n\u00e3o descarregar), ambos inicialmente neutros. Em seguida, s\u00e3o atritados (esfregados) e depois separados e isolados sem influ\u00eancia el\u00e9trica externa de outros corpos. Nessas condi\u00e7\u00f5es, a quantidade de cargas el\u00e9tricas (el\u00e9trons livres) que um cede \u00e9 a mesma que o outro recebe, ou seja, o vidro ceder\u00e1 el\u00e9trons e adquirir\u00e1 carga positiva +Q e a l\u00e3 receber\u00e1 el\u00e9trons e ficar\u00e1 com carga de mesmo m\u00f3dulo, mas negativa \u2013Q.<\/b><\/span><\/span><\/span><\/p>\n

04. Correta\u00a0 —\u00a0 veja (02).<\/b><\/span><\/span><\/span><\/p>\n

08. Correta\u00a0 —\u00a0 quanto mais afastados os elementos atritados estiverem na s\u00e9rie triboel\u00e9trica, maior ser\u00e1 a facilidade de transfer\u00eancia de el\u00e9trons maior quantidade de eletricidade est\u00e1tica adquirida.<\/b><\/span><\/span><\/span><\/p>\n

16. Falsa\u00a0 —\u00a0 os dois bast\u00f5es de vidro quando atritados com pele de gato ficam com cargas positivas e, quando aproximados ir\u00e3o se repelir, pois possuem cargas el\u00e9tricas de mesmo sinal.<\/b><\/span><\/span><\/span><\/p>\n

32. Falsa\u00a0 —\u00a0 s\u00f3 haver\u00e1 transfer\u00eancia de el\u00e9trons se os materiais atritados forem diferentes.<\/b><\/span><\/span><\/span><\/p>\n

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

71-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo as esferas id\u00eanticas, ap\u00f3s colocadas em contato e depois separas ambas ficar\u00e3o com cargas el\u00e9tricas de mesmo sinal e de valor Q=(Q1\u00a0+ Q2)\/2\u00a0 —\u00a0 A com B\u00a0 —\u00a0 Q\u2019=(+ 3e – 5e)\/2\u00a0 —\u00a0 Q\u2019A=Q\u2019B= – 1e\u00a0 —\u00a0 A com C\u00a0 —\u00a0 Q\u2019\u2019=(-1e + 3e)\/2\u00a0 —\u00a0 QA\u2019\u2019=QB\u2019\u2019= + 1e\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C.<\/b><\/span><\/span><\/span><\/p>\n

72-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Leia a teoria a seguir:<\/b><\/span><\/span><\/span><\/p>\n

Aproxima-se um condutor eletrizado, por exemplo, negativamente de um condutor neutro, sem <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

encostar, mas bem pr\u00f3ximos (figura 1).Os el\u00e9trons livres do indutor ser\u00e3o repelidos, ficando o lado direito do induzido com excesso de el\u00e9trons e o lado esquerdo com falta de el\u00e9trons, fen\u00f4meno que recebe o nome de\u00a0indu\u00e7\u00e3o el\u00e9trica\u00a0(figura 2)\u00a0 —\u00a0 observe na figura 2 que a\u00a0for\u00e7a de atra\u00e7\u00e3o\u00a0entre as cargas negativas e positivas \u00e9\u00a0maior\u00a0que a\u00a0for\u00e7a de repuls\u00e3o\u00a0entre as cargas negativas porque a dist\u00e2ncia entre as cargas\u00a0que se atraem\u00a0\u00e9 menor\u00a0e, quanto\u00a0menor a dist\u00e2ncia entre cargas de mesmo m\u00f3dulo,\u00a0maior a for\u00e7a\u00a0entre elas\u00a0 —\u00a0 o corpo neutro continua neutro (mesmo n\u00famero de pr\u00f3tons e de el\u00e9trons)\u00a0 —\u00a0 portanto,\u00a0um corpo eletrizado pode atrair um corpo neutro,\u00a0figura 3.<\/b><\/span><\/span><\/span><\/p>\n

O mesmo ocorre se o bast\u00e3o estiver eletrizado com carga positiva, s\u00f3 que agora voc\u00ea ter\u00e1, na esfera neutra, cargas negativas do lado esquerdo e positivas do lado direito\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0\u00a0R- E.<\/b><\/span><\/span><\/span><\/p>\n

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

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

44-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>S\u00e3o dados esfera superior, positiva de m\u00f3dulo\u00a0 —\u00a0 q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=0,001C=10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>C\u00a0 — esfera inferior negativa de m\u00f3dulo q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2,5.<\/b><\/span><\/span><\/span><\/p>\n

10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>C\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=9g=9.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0 peso de m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.g=9.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.10\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=9.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0 sendo as cargas de sinais opostos a for\u00e7a el\u00e9trica de intensidade F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u00a0entre elas \u00e9 de atra\u00e7\u00e3o e, estando as part\u00edculas em equil\u00edbrio a for\u00e7a resultante<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

sobre a part\u00edcula 2 \u00e9 nula e, sobre ela\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/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>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 9.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.2,5.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 9.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=22,5. 10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\/9.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2,5.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 d=\u221a0,25\u00a0 —\u00a0 d=0,5m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B.<\/b><\/span><\/span><\/span><\/p>\n

45-\u00a0Trata-se de um movimento circular uniforme onde, em cada ponto a for\u00e7a el\u00e9trica sobre (-q) de dire\u00e7\u00e3o radial e para o centro da circunfer\u00eancia tem intensidade 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 essa for\u00e7a \u00e9 igual \u00e0 for\u00e7a resultante centr\u00edpeta sobre (-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

q) de intensidade F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u00a0= 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>\u00a0= mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 kQq\/R = mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(I)\u00a0 —\u00a0 por outro lado, como Q est\u00e1 em repouso,a energia total do sistema \u00e9 a cin\u00e9tica de (-q), E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 mais a energia potencial el\u00e9trica do sistema fornecida por E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=kQ(-q)\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 kQq\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>(II)\u00a0 —\u00a0 (I) em (II)\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0= – mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

\u00a046-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Carga el\u00e9trica que foi transferida de um para outro\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1e \u2013 1,6.10-19C\u00a0 —\u00a0 5.103e \u2013 Q\u00a0 —\u00a0 Q=5.1013×1,6.10-19\u00a0 —\u00a0 Q=8.10-6C\u00a0 —\u00a0 um deles, que recebeu el\u00e9trons, ficou com excesso de el\u00e9trons (carga negativa de Q= – 8.10-6C0 e o outro, que cedeu el\u00e9trons, ficou com carga positiva de Q=8.10-6C\u00a0 —\u00a0 como os dois corpos ficaram com cargas de sinais opostos a for\u00e7a entre eles ser\u00e1 de atra\u00e7\u00e3o\u00a0 —\u00a0 a intensidade dessa for\u00e7a \u00e9 fornecida por F=K.Q.Q\/d2, onde as cargas Q est\u00e3o em m\u00f3dulo\u00a0 —\u00a0 F=9.109.8.10-6.8.10-6\/(10-2)2=576.10-3\/10-4\u00a0 —\u00a0 F=5760N=5,76kN\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E.<\/b><\/span><\/span><\/span><\/p>\n

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

47-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) A velocidade dos neutrinos \u00e9 a mesma que a da luz no v\u00e1cuo(e aproximadamente, no ar)\u00a0 —\u00a0 V=3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>m\/s\u00a0 —\u00a0 dist\u00e2ncia do Sol \u00e0 Terra\u00a0 —\u00a0 \u2206S=1,5.10<\/b><\/span><\/span><\/span>11<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 V=\u2206S\/\u2206t\u00a0 —\u00a0 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>=1,5.10<\/b><\/span><\/span><\/span>11<\/b><\/span><\/span><\/sup><\/span>\/\u2206t\u00a0 —\u00a0 \u2206t=1,5.10<\/b><\/span><\/span><\/span>11<\/b><\/span><\/span><\/sup><\/span>\/3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>=0,5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 — \u2206t=500s=8min2s.<\/b><\/span><\/span><\/span><\/p>\n

b) Observe a seq\u00fc\u00eancia das figuras abaixo, onde a tra\u00e7\u00e3o\u00a0\"\"\u00a0foi decomposta em suas componentes <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

horizontal\u00a0\"\"\u00a0e vertical\u00a0\"\"\u00a0\u00a0—\u00a0 T<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/sub><\/span>=Tcos45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=(\u221a2\/2).T\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>=Tsen45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=(\u221a2\/2).T\u00a0 —\u00a0 equil\u00edbrio na vertical\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>=mg\u00a0 —\u00a0 (\u221a2\/2).T= mg\u00a0 —\u00a0 T=\u221a2.mg\u00a0 —\u00a0 equil\u00edbrio na horizontal\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=T<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=(\u221a2\/2).T=(\u221a2\/2).\u221a2.mg\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=mg=0,004.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.10\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

m=4.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=kq<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 4.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>\u00a0= 9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>.q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/(3.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.9.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\/9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 q=\u221a(4.10<\/b><\/span><\/span><\/span>-18<\/b><\/span><\/span><\/sup><\/span>)\u00a0\u00a0—\u00a0\u00a0q=2.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>C\u00a0\u00a0—\u00a0 (essas duas cargas s\u00e3o ambas positivas ou ambas negativas, pois as hastes se repelem).<\/b><\/span><\/span><\/span><\/p>\n

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

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

73-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dado\u00a0 —\u00a0 q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=(1 \u2013 q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=kq<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=k2q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/(1 \u2013 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 campo el\u00e9trico resultante nulo\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0=<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=kq<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= k2q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/(1 \u2013 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 1\/d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 2\/(1 \u2013 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 (1 \u2013 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2\u00a0 —\u00a0 {(1 \u2013 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)\/d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>}<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2\u00a0 —\u00a0\u00a0 {(1 \u2013 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)\/d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>}=\u221a2\u00a0 —\u00a0 (1 \u2013 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)\/d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=1,4\u00a0 —\u00a0 1,4d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= 1 \u2013 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=1\/2,4\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=0,4m\u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B.<\/b><\/span><\/span><\/span><\/p>\n

74-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Cargas positivas criam, a\u00a0 certa dist\u00e2ncia d delas, campos de afastamento e negativas, campos de aproxima\u00e7\u00e3o\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

A intensidade desses campos \u00e9 fornecida por E=kq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>, onde q \u00e9 o m\u00f3dulo da carga\u00a0 —\u00a0 para que o campo el\u00e9trico seja<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

nulo no ponto C, tendo as duas cargas mesmo sinal (positivo ou negativo), a condi\u00e7\u00e3o \u00e9\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 kq<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= kq<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/(3d)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 kq<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= kq<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/9d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=9q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E.<\/b><\/span><\/span><\/span><\/p>\n

75-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Enquanto estiverem entre as placas ficam sujeitos a um campo el\u00e9trico E e consequentemente \u00e0 uma for\u00e7a el\u00e9trica\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u00a0relacionadas por E=F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\/q=F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\/e\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=e.E\u00a0 —\u00a0 desprezadas as a\u00e7\u00f5es gravitacionais, a for\u00e7a el\u00e9trica \u00e9 a for\u00e7a resultante F<\/b><\/span><\/span><\/span>R\u00a0<\/b><\/span><\/span><\/sub><\/span>sobre esses el\u00e9trons e, pela segunda lei de Newton F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u00a0= F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 e.E = m.a\u00a0 —\u00a0\u00a0a=e.E\/m.<\/b><\/span><\/span><\/span><\/p>\n

b) A parcela horizontal da velocidade (eixo x) permanece constante (lan\u00e7amento obl\u00edquo) e igual V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=constante e trata-se de um movimento retil\u00edneo uniforme de equa\u00e7\u00e3o x = x<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>.t\u00a0 —\u00a0 entre as placas x=L\u00a0 —\u00a0 L = 0 + V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t\u00a0 —\u00a0t=L\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

c) Enquanto os el\u00e9trons est\u00e3o entre as placas surge sobre eles uma for\u00e7a el\u00e9trica, vertical e para cima (F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u00a0e E tem mesma dire\u00e7\u00e3o mas sentidos contr\u00e1rios pois el\u00e9trons possuem cargas negativas)\u00a0 —\u00a0 essa for\u00e7a age sobre os el\u00e9trons<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

enquanto estiverem entre as placa durante t=L\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 como num lan\u00e7amento obliquo, o movimento vertical para cima \u00e9 um movimento uniformemente variado (acelerado com a=eE\/m) de equa\u00e7\u00e3o hor\u00e1ria\u00a0 —\u00a0 y = y<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>t + at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 y=\u2206y=0 + 0.t + (e.E\/m).(L\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0\u2206y=e.E.L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2mV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

<\/a>d) Equa\u00e7\u00e3o da velocidade do MUV, na dire\u00e7\u00e3o vertical\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>\u00a0= V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a.t\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0 + (e.E\/m).(L\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0\u00a0—\u00a0\u00a0V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=e.E.L\/m.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

Linhas de for\u00e7a (de campo) \u2013 Potencial eletrost\u00e1tico<\/b><\/span><\/span><\/span><\/p>\n

53-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A) Para quebrar a rigidez diel\u00e9trica do ar e ocorrer descarga el\u00e9trica a maior intensidade do campo el\u00e9trico deve superar E=3.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>Vm\u00a0 —\u00a0 d=3km=3.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 E=Vd\u00a0 —\u00a0 3.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>=V\/3.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0V=9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>V ou V=9nV.<\/b><\/span><\/span><\/span><\/p>\n

B) Se cada raio gera energia de 10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>J, 3 raios liberar\u00e3o uma energia de W=3.109J\u00a0 — pot\u00eancia em \u2206t=1s\u00a0 —\u00a0 P=W\/\u2206t=<\/b><\/span><\/span><\/span><\/p>\n

3.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\/1\u00a0 —\u00a0 P=3.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>W ou P=3nW.<\/b><\/span><\/span><\/span><\/p>\n

C) n=P<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>\/P<\/b><\/span><\/span><\/span>1 raio<\/b><\/span><\/span><\/sub><\/span>=1,5.10<\/b><\/span><\/span><\/span>10<\/b><\/span><\/span><\/sup><\/span>\/10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>=1,5.10=15 raios.<\/b><\/span><\/span><\/span><\/p>\n

54-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Equil\u00edbrio eletrost\u00e1tico entre dois condutores\u00a0\u2013 Sejam dois condutores esf\u00e9ricos A e B, de raios R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0e R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0respectivamente, com cargas el\u00e9tricas Q<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0e Q<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0e potenciais el\u00e9tricos diferentes V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0e V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0(fig. I)\u00a0 —\u00a0 unindo-os por um fio condutor ou encostando-os haver\u00e1 passagem de cargas el\u00e9tricas entre eles at\u00e9 que seja atingido o equil\u00edbrio<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

eletrost\u00e1tico quando atingem o\u00a0mesmo potencial V\u00a0 —\u00a0 sejam Q\u2019<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0e Q\u2019<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0as novas cargas ap\u00f3s o contato. T\u00eam-se duas condi\u00e7\u00f5es\u00a0 —\u00a0 A soma alg\u00e9brica das cargas antes do contato \u00e9 a mesma que a soma alg\u00e9brica das cargas depois do contato\u00a0 —\u00a0\u00a0Q<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ Q<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=Q\u2019A\u00a0+ Q\u2019<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 os potenciais s\u00e3o os mesmos ap\u00f3s o contato\u00a0 — V=KQ\u2019A\/RA\u00a0 —\u00a0 V=KQ\u2019<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 KQ\u2019<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0= KQ\u2019<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0Q\u2019<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/Q\u2019<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 observe que, se R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Q\u2019<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=Q\u2019<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=(Q<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ Q<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>)\/2.<\/b><\/span><\/span><\/span><\/p>\n

Observe na express\u00e3o Q\u2019<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/Q\u2019<\/b><\/span><\/span><\/span>B\u00a0<\/b><\/span><\/span><\/sub><\/span>= R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0que, como a soma das cargas el\u00e9tricas antes e depois do contato \u00e9 a mesma, a rela\u00e7\u00e3o Q\u2019<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/Q\u2019<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 constante\u00a0 —\u00a0 assim, se R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>>R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>, obrigatoriamente\u00a0Q\u2019<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0> Q\u2019<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D.<\/b><\/span><\/span><\/span><\/p>\n

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

Superf\u00edcies equipotenciais \u2013 trabalho da for\u00e7a eletrost\u00e1tica<\/b><\/span><\/span><\/span><\/p>\n

50-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>No capacitor plano considerado, a diferen\u00e7a de potencial entre as placas \u00e9 de U=1V e a dist\u00e2ncia entre as placas vale d=1\u03bcm=1.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 num capacitor plano \u00e9 v\u00e1lida a rela\u00e7\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

U=E.d\u00a0 —\u00a0 1=E.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E=1\/10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E=10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>V\/m\u00a0 —\u00a0 E=1MV\/m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A.<\/b><\/span><\/span><\/span><\/p>\n

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

51-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O trabalho (W) realizado pela for\u00e7a el\u00e9trica no transporte dessa carga de q=10C da nuvem para o solo \u00e9 igual \u00e0 energia el\u00e9trica (E) liberada pelo raio nessa transfer\u00eancia e, ela ocorre devido \u00e0 diferen\u00e7a de potencial U=100.106V=10.107V entre\u00a0 a nuvem e o solo\u00a0\u00a0 —\u00a0 E=W=q.U=10×10.107=100.107J\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1J \u2013 3.10-7\u00a0kWh\u00a0 —\u00a0 100x107J \u2013 W\u00a0 —\u00a0 W=E=3.10-7.100.107=300kWh\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Resolu\u00e7\u00e3o Comentada \u2013 2014- 2013 ELETROST\u00c1TICA 01- Lei de Coulomb: Cargas el\u00e9tricas de mesmo sinal se repelem e de sinais opostos se atraem — analise atentamente a figura abaixo: Observe que pela lei de Coulomb F=KQq\/d2\u00a0que — F4A\u00a0= F2A\u00a0e que F1A\u00a0= F3A\u00a0— adicionando vetorialmente essas for\u00e7as voc\u00ea obt\u00e9m a for\u00e7a resultante\u00a0\u00a0pedida, cuja dire\u00e7\u00e3o e sentido est\u00e3o indicados na figura —<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3323,"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-3325","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3325","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=3325"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3325\/revisions"}],"predecessor-version":[{"id":11077,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3325\/revisions\/11077"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3323"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3325"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}