{"id":1883,"date":"2016-02-11T18:01:40","date_gmt":"2016-02-11T18:01:40","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1883"},"modified":"2024-08-19T14:06:10","modified_gmt":"2024-08-19T14:06:10","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-capacitores-2","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletrodinamica\/capacitores-condensadores\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-capacitores-2\/","title":{"rendered":"Capacitores – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

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

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C=Q\/U\u00a0 —\u00a0 como a capacit\u00e2ncia C \u00e9 constante para cada capacitor\u00a0 —\u00a0 se dobrar Q, U tamb\u00e9m dever\u00e1 dobrar\u00a0 —<\/b><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como as pilhas est\u00e3o associadas em s\u00e9rie\u00a0 —\u00a0 E=U=1,5 + 1,5\u00a0 —\u00a0 U=3,0V\u00a0 —\u00a0 C=Q\/U\u00a0 —\u00a0 6,0.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>=Q\/3,0\u00a0 —\u00a0 Q=18.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q=1,8.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C=Q\/U\u00a0 —\u00a0 10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>=Q\/10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Q=10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Q=10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>=0,1.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q=0,1\u03bcC\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Energia liberada pelo capacitor ao se descarregar\u00a0 —\u00a0 W=C.U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=22.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.(25)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 W=6,875J\u00a0 —\u00a0 essa energia foi utilizada para elevar o tijolo de massa 0,5kg a uma altura h e vale\u00a0 —\u00a0 W=mgh\u00a0 —\u00a0 6,875=0,5.10.h\u00a0 — \u00a0h=6,875\/5\u00a0 —\u00a0 h=1,375m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como a capacit\u00e2ncia C \u00e9 uma constante caracter\u00edstica de cada capacitor, as varia\u00e7\u00f5es entre Q e U s\u00e3o proporcionais\u00a0 —\u00a0 C=\u0394Q\/\u0394U\u00a0 —\u00a0 C=6.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\/(60 \u2013 50)\u00a0 —\u00a0 C=6.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>F\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>energia armazenada\u00a0 —\u00a0 W=C.U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=1,2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.(3.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 W=1,2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.9.10<\/b><\/span><\/span>6\u00a0<\/b><\/span><\/span><\/sup>\/2 —\u00a0 W=5,4J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A capacit\u00e2ncia \u00e9 constante e vale\u00a0 —\u00a0 C=Q\/U=60.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\/3=180.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\/9\u00a0 —\u00a0 C=20\u03bcF\u00a0 —\u00a0 quando U=20V\u00a0 —\u00a0 C=Q\/U\u00a0 —\u00a0 20.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0= Q\/20\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q=400\u03bcC<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Energia liberada pelo capacitor ao se descarregar\u00a0 —\u00a0 W=C.U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>.(16)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0\u00a0 —\u00a0 W=1,28J\u00a0 —\u00a0 essa energia \u00e9 utilizada para elevar o bloco de h\u00a0 —\u00a0 W=m.g.h\u00a0 —\u00a0 1,28=0,10.10.h\u00a0 —\u00a0 h=1,28m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

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

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O televisor deixa de funcionar no in\u00edcio dos 5 minutos quando a tens\u00e3o \u00e9 de 80V\u00a0 —\u00a0 a capacidade \u00e9 constante e vale\u00a0 —\u00a0 C=Q\/U=10\/1.000\u00a0 —\u00a0 C=10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>F\u00a0 —\u00a0 C=Q\/U\u00a0 —\u00a0 10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>=Q\/80\u00a0 —\u00a0 Q=80.<\/b><\/span><\/span><\/p>\n

10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Q=0,8C\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

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

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C= Q\/U\u00a0 —\u00a0 8,85.10<\/b><\/span><\/span>-12<\/b><\/span><\/span><\/sup>=Q\/10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Q=8,85.10<\/b><\/span><\/span>-10<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C= K\u03b5<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.A\/d\u00a0 —\u00a0 como o diel\u00e9trico \u00e9 o ar\u00a0 —\u00a0 K=1\u00a0 —\u00a0 A=200cm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=2.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>=2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 C= K\u03b5<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.A\/d —\u00a0 8,85.10<\/b><\/span><\/span>-12<\/b><\/span><\/span><\/sup>= 1.8,85.10<\/b><\/span><\/span>-12<\/b><\/span><\/span><\/sup>.2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\/d\u00a0 —\u00a0 d=2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>dobrando a \u00e1rea, a capacit\u00e2ncia C tamb\u00e9m dobra\u00a0 —\u00a0 dobrando d, a capacit\u00e2ncia C fica reduzida \u00e0 metade\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) C=Q\/U\u00a0 —\u00a0 40.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=Q\/40\u00a0 —\u00a0 Q=1.600.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q=1,6.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

b) W=C.U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=40.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.(40)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 W<\/b><\/span><\/span>=3,2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>J<\/b><\/span><\/span><\/p>\n

c) dobrando d, a capacit\u00e2ncia C cai pela metade\u00a0 —\u00a0 C=40\u03bcF\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>C=20\u03bcF<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) d=150m\u00a0 —\u00a0 E=3.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>V\/m\u00a0 —\u00a0 U=E.d=3.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>.150\u00a0 —\u00a0 U=450.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>V\u00a0 —\u00a0 C=Q\/U\u00a0 —\u00a0 1,6.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>=Q\/450.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Q=720.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q=7,2C
\n<\/b><\/span><\/span>b) energia fornecida pela descarga\u00a0 —\u00a0 W=C.U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=1,6.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>.(450.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 W=1,62.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0 P=W\/\u0394t=1,62.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>\/3.600\u00a0 —\u00a0 P=45.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>W\u00a0 —\u00a0 1 l\u00e2mpada – 60W\u00a0 —\u00a0 n l\u00e2mpadas \u2013 45.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>W\u00a0 —\u00a0 n=45.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\/60=0,75.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0n=7.500 l\u00e2mpadas<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a)\u00a0i=Q\/\u0394t\u00a0 —\u00a0 50.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=Q\/10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q=50C<\/b><\/span><\/span><\/p>\n

b)\u00a0C=\u00a0\u03b5<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.A\/d=9.10<\/b><\/span><\/span>-12<\/b><\/span><\/span><\/sup>.200.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\/2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>C=9,0.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>F<\/b><\/span><\/span><\/p>\n

c) C=Q\/U\u00a0 —\u00a0 9.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>=50\/U\u00a0 —\u00a0 U=50\/9.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>U=5,6.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>V<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/p>\n

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

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quanto mais diel\u00e9trico, maior ser\u00e1 C e consequentemente maior U e maior Q, pois como C \u00e9 constante, Q e U s\u00e3o diretamente proporcionais\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>1. Verdadeira\u00a0 —\u00a0 o potencial da placa A (positiva) \u00e9 maior que o da placa B (negativa).<\/b><\/span><\/span><\/p>\n

2. Falsa\u00a0 —\u00a0 o sentido do campo el\u00e9trico no interior das placas \u00e9 das cargas positivas para as negativas.<\/b><\/span><\/span><\/p>\n

3. Falsa\u00a0 —\u00a0 C=Q\/U\u00a0 —\u00a0 10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=Q\/100\u00a0 —\u00a0 Q=10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>C=100\u03bcC<\/b><\/span><\/span><\/p>\n

4. Verdadeira\u00a0 —\u00a0 cargas de mesmo sinal se repelem e cargas de sinais opostos se atraem.<\/b><\/span><\/span><\/p>\n

5. Verdadeira\u00a0 —\u00a0 a capacit\u00e2ncia \u00e9 inversamente proporciona \u00e0 dist\u00e2ncia entre as placas<\/b><\/span><\/span><\/p>\n

6. Verdadeira\u00a0 —\u00a0 essa \u00e9 a principal caracter\u00edstica dos capacitores<\/b><\/span><\/span><\/p>\n

R- 1.V; 2.F; 3.F; 4.V; 5.V; 6.V
\n<\/b><\/span><\/span><\/p>\n

20<\/b><\/span><\/span><\/span>–\u00a0<\/b><\/span><\/span><\/span>a) Como a capacit\u00e2ncia C \u00e9 diretamente proporcional \u00e0 constante diel\u00e9trica, C ficar\u00e1 tamb\u00e9m 5 vezes maior\u00a0 —\u00a0 C=Q\/U\u00a0 —\u00a0 como Q \u00e9 a mesma, para C ficar 5 vezes maior, U dever\u00e1 ficar 5 vezes menor\u00a0 —\u00a0 U=200\/5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>U=40V<\/b><\/span><\/span><\/p>\n

b) U=8.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>V\u00a0 —\u00a0 d=0,5mm=5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0 U=E.d=8.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 U=40.10<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>U=400V<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A ddp U \u00e9 a mesma, pois a fonte \u00e9 sempre a mesma\u00a0 —\u00a0 antes\u00a0 —\u00a0 E=U\/d\u00a0 —\u00a0 E=12\/d\u00a0 —\u00a0 depois\u00a0 —\u00a0 E\u2019=12\/2d\u00a0 —\u00a0 E\u2019=E\/2\u00a0 —\u00a0 o campo el\u00e9trico cai pela metade\u00a0 —\u00a0 C=\u03b5A\/d\u00a0 —\u00a0 Q.U=\u00a0\u03b5A\/d\u00a0 —\u00a0 Q=\u00a0\u03b5A\/d.U\u00a0 —\u00a0 observe nessa express\u00e3o que a \u00fanica grandeza vari\u00e1vel \u00e9 d e que \u00e9 inversamente proporcional a Q, ou seja, se d dobrar, Q ficar\u00e1 reduzida \u00e0 metade<\/b><\/span><\/span>\u00a0 —\u00a0 R- D<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A capacit\u00e2ncia C \u00e9 diretamente proporcional \u00e0 \u00e0 constante diel\u00e9trica K\u00a0 —\u00a0 como ela \u00e9 triplicada C fica 3 vezes maior\u00a0 —\u00a0 U permanece constante\u00a0 —\u00a0 W=C.U2\/2\u00a0 —\u00a0 assim a nova energia armazenada ficar\u00e1 3 vezes maior\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C=Q\/U\u00a0 —\u00a0\u00a0 8.10<\/b><\/span><\/span>-11<\/b><\/span><\/span><\/sup>=Q\/12\u00a0 —\u00a0 Q=96.10<\/b><\/span><\/span>-11<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 e \u2013 1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 n e \u2013 96.10<\/b><\/span><\/span>-11<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 n=6,0.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>\u00a0eletrons\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Veja sequ\u00eancia abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

27-<\/b><\/span><\/span><\/span> <\/b>Os dois \u00faltimos capacitores est\u00e3o em circuito aberto e n\u00e3o participam\u00a0 —\u00a0 observe a seq\u00fc\u00eancia abaixo\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a seq\u00fc\u00eancia abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

\"\"<\/p>\n

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

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Os capacitores C<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0e C<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0est\u00e3o ligados em paralelo e a \u00a0capacit\u00e2ncia equivalente dos dois\u00a0 \u00e9 C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>\u00a0= 2C. Portanto a diferen\u00e7a de potencial no capacitor C<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=C ser\u00e1 o dobro. Ficamos ent\u00e3o com 60 V no capacitor C<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- 60V<\/b><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O capacitor equivalente de \u00e1rea A e dist\u00e2ncia entre as placas d\u2019 que est\u00e1 em paralelo com os tr\u00eas capacitores C de \u00e1rea A e dist\u00e2ncia d vale C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=3C\u00a0 —\u00a0\u00a0\u03b5A\/d\u2019=3\u03b5A\/d\u00a0 —\u00a0 d\u2019=d\/3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a)Como est\u00e3o em paralelo\u00a0 —\u00a0 C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=1.400.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0+ 1.400.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=2,800.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>F\u00a0 —\u00a0 C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=Q\/U\u00a0 —\u00a0 2.800.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=Q\/170\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q=0,476C<\/b><\/span><\/span><\/p>\n

b) W=C.U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=2.800.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.(170)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 W=40.460.000.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W=40,46J<\/b><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e C<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0est\u00e3o em paralelo\u00a0 —\u00a0 C\u2019=2,0\u03bcF\u00a0 —\u00a0 C\u2019 est\u00e1 em s\u00e9rie com C<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=3\u03bcF\u00a0 —\u00a0 veja figura abaixo\u00a0 — <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

como C\u2019 est\u00e1 em s\u00e9rie com\u00a0 C<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>, eles possuem a mesma carga Q\u00a0 —\u00a0 C\u2019=Q\/U\u2019\u00a0 —\u00a0 U\u2019=Q\/2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 U\u2019\u2019=Q\/3.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 U\u2019 + U\u2019\u2019=120\u00a0 —\u00a0 Q\/2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0+<\/b><\/span><\/span><\/p>\n

Q\/3.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=120 (I)\u00a0 —\u00a0 C\u2019=Q\/U\u2019\u00a0 —\u00a0 Q=2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>U\u2019 (II)\u00a0 —\u00a0 (II) em (I)\u00a0\u00a0<\/b><\/span><\/span>—\u00a0\u00a0<\/b><\/span><\/span>2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>U\u20192.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0+\u00a0 2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>U\u2019=120\u00a0 —\u00a0 U\u2019 + 2\/3U\u2019=120\u00a0 —\u00a0 3U\u2019+ 2U\u2019=360\u00a0 —\u00a0 U\u2019=360\/5\u00a0 —\u00a0 U\u2019=72V\u00a0 —\u00a0 U\u2019\u2019=120 \u2013 72\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>U\u2019\u2019=48V\u00a0<\/b><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na seq\u00fc\u00eancia abaixo o c\u00e1lculo da capacit\u00e2ncia do capacitor equivalente\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

C\u2019=C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=4C\/11=4.11.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\/11\u00a0 —\u00a0 C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=4.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>F\u00a0 —\u00a0 C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=Q\/U\u00a0 —\u00a0 4.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=Q\/10\u00a0 —\u00a0 Q=4.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Todas est\u00e3o corretas\u00a0 —\u00a0 veja teoria\u00a0<\/b><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A capacit\u00e2ncia de um capacitor \u00e9 dada por C = q\/U de onde vem => U = q\/C<\/b><\/span><\/span><\/p>\n

No circuito em s\u00e9rie\u00a0 —\u00a0 \u03b5 = U1 + U2\u00a0 —\u00a0 \u03b5 = (q\/C1) + (q\/C2) => \u03b5\/q = (1\/C1) + (1\/C2)\u00a0\u00a0 —\u00a0\u00a0 \u03b5\/q=(C1 + C2)\/(C1 .C2)<\/b><\/span><\/span><\/p>\n

No circuito em paralelo\u00a0 —\u00a0 \u03b5 = U1\u00a0 —\u00a0 \u03b5 = q1\/C1 => q1 = \u03b5.C1\u00a0 —\u00a0 \u03b5 = U2\u00a0 —\u00a0 \u03b5 = q2\/C2 => q2 = \u03b5.C2\u00a0 —\u00a0 q1 + q2 = \u03b5.C1 + \u03b5.C2 = \u03b5.(C1 +C2) = 4q\u00a0\u00a0 —\u00a0 (C1 +C2) = 4q\/\u03b5 => C1 = 4q\/ \u03b5 – C2\u00a0 —\u00a0 ent\u00e3o\u00a0 —\u00a0 \u03b5\/q = (C1 + C2)\/(C1 .C2)\u00a0 —\u00a0 \u03b5\/q = (4q\/\u03b5)\/(C1 .C2) => C1 .C2 = 4q2\/\u03b5<\/b><\/span><\/span><\/p>\n

C1 .C2 = 4q2\/\u03b5\u00a0 —\u00a0 (4q\/\u03b5 – C2) .C2 = 4q2\/\u03b5\u00a0\u00a0 —\u00a0\u00a0 C22 – (4q\/\u03b5).C2 + (4q2\/\u03b52) = 0\u00a0\u00a0 —\u00a0\u00a0 C2 = 2q\/\u03b5\u00a0 —\u00a0 e ainda\u00a0 —\u00a0 C1 = 4q\/\u03b5\u00a0 – C2 = 4q\/\u03b5\u00a0 – 2q\/\u03b5\u00a0 —\u00a0\u00a0 C1 = 2q\/\u03b5\u00a0<\/b><\/span><\/span><\/p>\n

38-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub>\u00a0e C<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0est\u00e3o em paralelo\u00a0 —\u00a0 C\u2019=(2 + 3).10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 C\u2019=5.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>F\u00a0 —\u00a0 observe a figura abaixo e note que, como as capacit\u00e2ncias de C<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/p>\n

\"\"<\/p>\n

\u00a0e de C\u2019 s\u00e3o iguais, eles est\u00e3o sob mesma ddp U=5V\u00a0 —\u00a0 C\u2019=Q\/U\u00a0 —\u00a0 5.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=Q\/5\u00a0 —\u00a0 Q=25.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 C<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/U\u00a0 —\u00a0 2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/5\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=10.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=C<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.U=3.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.5\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=15.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=C<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>.U=5.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.5\u00a0 —\u00a0 Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=25.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 1\/C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=C\u2019 s\u00e9rie com C<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=5\/2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=2,5.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=Q\/U\u00a0 —\u00a0 2,5.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=Q\/5\u00a0 —\u00a0 Q=12,5.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Associados em s\u00e9rie\u00a0 —\u00a0 C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=2.3\/(2 + 3)\u03bcC\u00a0 —\u00a0 C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=1,2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=Q\/U\u00a0 —\u00a0 1,2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=Q\/100\u00a0 —\u00a0 Q=120.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q=<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup>120\u03bcC\u00a0 —\u00a0 como est\u00e3o em s\u00e9rie essa carga \u00e9 a mesma para os dois capacitores\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

40-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Depois que o capacitor est\u00e1 carregado, n\u00e3o passa mais corrente por ele e o circuito fica conforme esquema abaixo:\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R\u2019=U\/i\u00a0 —\u00a0 2=10\/i\u00a0 —\u00a0 i=5 A\u00a0 —\u00a0 c\u00e1lculo da ddp nos terminais do capacitor\u00a0 —\u00a0 R=U\/i\u00a0 —\u00a0 1-U\/5\u00a0 —\u00a0 U=5V\u00a0 —\u00a0 W=C.U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

W=4.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.(5)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0W=50\u03bcJ<\/b><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01- Falsa\u00a0 —\u00a0 o capacitor fica com ddp de 6V<\/b><\/span><\/span><\/p>\n

02- Falsa\u00a0 —\u00a0 a l\u00e2mpada ficar\u00e1 acesa at\u00e9 que o capacitor fique carregado<\/b><\/span><\/span><\/p>\n

03- Falsa\u00a0 —\u00a0 n\u00e3o h\u00e1 curto circuito, existe a l\u00e2mpada<\/b><\/span><\/span><\/p>\n

04- Falsa\u00a0 —\u00a0 C=Q\/U\u00a0 —\u00a0 2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=Q\/6\u00a0 —\u00a0 Q=12.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

05- Correta\u00a0 —\u00a0 W=C.U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.36\/2\u00a0 —\u00a0 W=3,6.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>J<\/b><\/span><\/span><\/p>\n

R- a correta \u00e9 a 05<\/b><\/span><\/span><\/p>\n

42-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Refazendo o circuito:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe que se trata de uma ponte de Wheatstone em equil\u00edbrio, pois o produto das resist\u00eancias opostas \u00e9 nulo e, nessas condi\u00e7\u00f5es o potencial em A \u00e9 igual ao potencial em B\u00a0 —\u00a0 U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

44-<\/b><\/span><\/span><\/span>\u00a0\u00a0 <\/b><\/span><\/span><\/span>I. Correta\u00a0 —\u00a0 o capacitor carregado interrompe a corrente no ramo do circuito onde est\u00e1 inserido o capacitor e o amper\u00edmetro.<\/b><\/span><\/span><\/p>\n

II. Correta\u00a0 —\u00a0 i=10\/5\u00a0 —\u00a0 i=2 A\u00a0 —\u00a0 R=U\/i\u00a0 —\u00a0 4=U\/2\u00a0 —\u00a0 U=8V\u00a0 —\u00a0 C=Q\/U\u00a0 —\u00a0 2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=Q\/8\u00a0 —\u00a0 Q=16\u03bcC.<\/b><\/span><\/span><\/p>\n

III. Falsa\u00a0 —\u00a0 \u00e9 8V\u00a0 —\u00a0 veja II.<\/b><\/span><\/span><\/p>\n

VI. Falsa\u00a0 —\u00a0 \u00e9 2 A\u00a0 —\u00a0 veja II.<\/b><\/span><\/span><\/p>\n

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

45-<\/b><\/span><\/span><\/span>\u00a0a) Chave aberta\u00a0 —\u00a0 observe na seq\u00fc\u00eancia abaixo que a tens\u00e3o nos terminais do capacitor com a chave aberta \u00e9 U=30V\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

C=Q\/U\u00a0 —\u00a0 3.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=Q\/30\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q=90\u03bcC<\/b><\/span><\/span><\/p>\n

b) Chave fechada\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 20=30\/i\u00a0 —\u00a0 i=1,5 A\u00a0 —\u00a0 o capacitor est\u00e1 no ramo onde tem o resistor de 10\u2126 e a tens\u00e3o em seus terminais vale\u00a0 —<\/b><\/span><\/span><\/p>\n

R=U\/i\u00a0 —\u00a0 10=U\/1,5\u00a0 —\u00a0 U=15V\u00a0 —\u00a0 C=Q\/U\u00a0 —\u00a0 3.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=Q\/15\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0Q=45\u03bcC<\/b><\/span><\/span><\/p>\n

46-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A corrente de curto circuito num gerador \u00e9 fornecida por i<\/b><\/span><\/span>cc<\/b><\/span><\/span><\/sub>=E\/r\u00a0 —\u00a0 6=E\/2\u00a0 —\u00a0 E=U=12V\u00a0 —\u00a0 C=Q\/U=2,4.10<\/b><\/span><\/span>-11<\/b><\/span><\/span><\/sup>\/12\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

C=2.10<\/b><\/span><\/span>-12<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 C=2pF\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

47-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Tirando o capacitor e aplicando a lei de Pouillet\u00a0 —\u00a0 200i + 300i -20 +10=0\u00a0 —\u00a0 i=1\/50 A\u00a0 —\u00a0 c\u00e1lculo da tens\u00e3o U nos terminais do capacitor\u00a0 —\u00a0 U=E \u2013 r.i=20 \u2013 300.1\/50\u00a0 —\u00a0 U=14V\u00a0 —\u00a0 Q=C.U=2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.14=28.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>C\u00a0 —<\/b><\/span><\/span>–\u00a0 Q=28\u03bcC<\/b><\/span><\/span>\u00a0 —\u00a0 P=R.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=500.(1\/50)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P=0,2W\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

48-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Como o capacitor perde 87,5% da sua carga em 3 ms, significa 3ms correspondem a 3\u03c4 (\u03c4 = R.C), ent\u00e3o \u03c4 <\/b><\/span><\/span>\uf020<\/b><\/span><\/span>= 1 ms:<\/b><\/span><\/span><\/p>\n

 <\/p>\n

\u03c4 = R . C\u00a0 —\u00a0 \u00a01 x 10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0= 50 . C >>> C = 2 x 10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0F\u00a0 —\u00a0 Q = C . U >>> Q = 2 x 20<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0x 5 x 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0= 0,1 C\u00a0 —\u00a0 Im = \u0394Q\/\u0394t = (0,875 x 0,1) \/ (3 x 10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>) =<\/b><\/span><\/span>\u00a029,2 A<\/b><\/span><\/span><\/p>\n

b) A energia armazenada no capacitor \u00e9 dada por E = (C . U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>) \/ 2<\/b><\/span><\/span><\/p>\n

E = [2 x 10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0x (5 x 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>)2] \/ 2\u00a0\u00a0 —\u00a0 \u00a0<\/b><\/span><\/span>E = 250 J<\/b><\/span><\/span><\/p>\n

 <\/p>\n

49-<\/b><\/span><\/span><\/span> a) C\u00e1lculo da capacit\u00e2ncia C da membrana de \u00e1rea 5,0.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>cm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>, se cada cm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0de \u00e1rea tem capacit\u00e2ncia de 0,8.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>F\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 cm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u2013 0,8.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>F\u00a0 —\u00a0 C=5.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>.0,8.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 C=4.10<\/b><\/span><\/span>-11<\/b><\/span><\/span><\/sup>\u00a0F\u00a0 —\u00a0 seja (n) a quantidade de \u00edons, cada um com carga e=1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>C, ent\u00e3o a quantidade de carga total (Q) que atravessar\u00e1 a membrana vale\u00a0 —\u00a0 Q=n.e=n.1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Q=1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>.n\u00a0 —\u00a0 defini\u00e7\u00e3o de capacit\u00e2ncia\u00a0 —\u00a0 C=Q\/U\u00a0 —\u00a0 4.10<\/b><\/span><\/span>-11<\/b><\/span><\/span><\/sup>\u00a0= 1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>.n\/80.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —<\/b><\/span><\/span>\u00a0n=2,0.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>\u00a0\u00edons<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

b) Pot\u00eancia el\u00e9trica entregue ao conjunto de\u00a0 \u00edons para despolarizar a membrana\u00a0 —\u00a0 P=i.U=(Q\/\u2206t).U=(n.\/\u2206t).e.U\u00a0 —\u00a0 P=5.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>.1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>.20.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>—\u00a0\u00a0<\/b><\/span><\/span>P=1,6.10<\/b><\/span><\/span>-12<\/b><\/span><\/span><\/sup>\u00a0W<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

50- <\/b><\/span><\/span><\/span>a) Cargas negativas (\u00edons negativos \u2013 Cl<\/b><\/span><\/span>–<\/b><\/span><\/span><\/sup>) deslocam-se para pontos de maior potencial, do interior para o exterior da c\u00e9lula\u00a0 —\u00a0 cargas positivas\u00a0 (\u00edons positivos \u2013 Ca<\/b><\/span><\/span>2+<\/b><\/span><\/span><\/sup>) deslocam-se para pontos de menor potencial, do exterior para o interior da c\u00e9lula.<\/b><\/span><\/span><\/p>\n

b) No interior da membrana (interior do capacitor), pelo gr\u00e1fico a ddp U=(64.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0\u2013 0)=64.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0 U=E.d\u00a0 — 64.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>=E.8.10<\/b><\/span><\/span>-9<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>E=8,0.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>V\/m ou N\/C<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

c) For\u00e7a el\u00e9trica sobre os \u00edons Cl<\/b><\/span><\/span>–<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>Cl<\/b><\/span><\/span><\/sub>=|q|.E=1.1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>.8.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0F<\/b><\/span><\/span>Cl<\/b><\/span><\/span><\/sub>=1,28.10<\/b><\/span><\/span>-12<\/b><\/span><\/span><\/sup>N\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 for\u00e7a el\u00e9trica sobre os \u00edons Ca<\/b><\/span><\/span>2+<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>Ca<\/b><\/span><\/span><\/sub>=|q|.E=2.1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>.8.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>Cl<\/b><\/span><\/span><\/sub>=2,56.10<\/b><\/span><\/span>-12<\/b><\/span><\/span><\/sup>N.<\/b><\/span><\/span><\/p>\n

d<\/b><\/span><\/span>)\u00a0<\/b><\/span><\/span>Pela defini\u00e7\u00e3o de capacit\u00e2ncia, sendo Q a carga el\u00e9trica na superf\u00edcie externa da membrana\u00a0 —\u00a0 C=Q\/U\u00a0 —\u00a0 12.10<\/b><\/span><\/span>-12<\/b><\/span><\/span><\/sup>=Q\/64.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>Q=7,68.10<\/b><\/span><\/span>-13<\/b><\/span><\/span><\/sup>C.<\/b><\/span><\/span><\/p>\n

51-<\/b><\/span><\/span><\/span> <\/b>Observe que, da maneira como eles est\u00e3o ligados voc\u00ea pode consider\u00e1-los como associados em s\u00e9rie\u00a0 —\u00a0 a express\u00e3o matem\u00e1tica da capacit\u00e2ncia de um capacitor \u00e9 C=\u03b5.S\/D\u00a0 —\u00a0 no caso, <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

C<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u03b5<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.S\/(d\/2) e C<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=\u03b5<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.S\/(d\/2)\u00a0 —\u00a0 na associa\u00e7\u00e3o s\u00e9rie de capacitores a capacidade do capacitor equivalente \u00e9 dada por\u00a0 —\u00a0 C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=produto\/soma=C<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.C<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/(C<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ C<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>= [\u03b5<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.S\/(d\/2). \u03b5<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.S\/(d\/2)]\/[\u03b5<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.S\/(d\/2) + \u03b5<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.S\/(d\/2)]\u00a0 —C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=2.\u03b5<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.\u03b5<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.S\/d(\u03b5<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ \u03b5<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>) \u00a0—\u00a0 Q=C<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>.U\u00a0 —\u00a0 Q=2.\u03b5<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.\u03b5<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.S\/d(\u03b5<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ \u03b5<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>).U\u00a0 — \u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Capacitores 01-\u00a0C=Q\/U\u00a0 —\u00a0 como a capacit\u00e2ncia C \u00e9 constante para cada capacitor\u00a0 —\u00a0 se dobrar Q, U tamb\u00e9m dever\u00e1 dobrar\u00a0 —\u00a0 R- C 02-\u00a0Como as pilhas est\u00e3o associadas em s\u00e9rie\u00a0 —\u00a0 E=U=1,5 + 1,5\u00a0 —\u00a0 U=3,0V\u00a0 —\u00a0 C=Q\/U\u00a0 —\u00a0 6,0.10-5=Q\/3,0\u00a0 —\u00a0 Q=18.10-5C\u00a0 —\u00a0 Q=1,8.10-4C\u00a0 —\u00a0\u00a0R- D 03-\u00a0C=Q\/U\u00a0 —\u00a0 10-3.10-3=Q\/102\u00a0 —\u00a0 Q=10-6.102\u00a0 —\u00a0 Q=10-4=10-1.10-3=0,1.10-3C\u00a0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1880,"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-1883","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1883","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=1883"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1883\/revisions"}],"predecessor-version":[{"id":10516,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1883\/revisions\/10516"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1880"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1883"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}