Prova 2 - Fisica 3

SEX (I'm A... EEEEE Unholy LOVE AGAIN Babyt LADy Kal Candidate me Bye BLUE Ai NARANJNGAZ Where Doyou Gor Today? SolEIL Publicitettlare D," itenhoe mowat MUN Part selcthes PROVA 2 1 Questão 1 — CAMPO DE FORÇA F → = α − y x ^ + x y ^ x 2 + y 2 a). calcular o trabalho realizados no caminho fechado C C = C 1 ∪ C 2 ∪ C 3 ∪ C 4 C 1 , C 3 d l → = x ^ d x C 2 , C 4 d l → = y ^ d y W = ∮ C F → . d l → = ∫ C 1 F → . d l → + ∫ C 2 F → . d l → + ∫ C 3 F → . d l → + ∫ C 4 F → . d l → = α ∫ a − a − y x ^ + x y ^ x 2 + y 2 | 0 x ^ d x = y = a + α ∫ a a − y x ^ + x y ^ x 2 + y 2 | x = a 0 y ^ d y + α ∫ a − a − y x ^ + x y ^ x 2 + y 2 | y = − a 0 x ^ d x + α ∫ a − a − y x ^ + x y ^ x 2 + y 2 | x = − a 0 x ^ d y = α ∫ − a a − a d x x 2 + a 2 + ∫ a − a d y y 2 + a 2 + + ∫ a − a a d x x 2 + a 2 + ∫ a − a d y y 2 + a 2 = = 4 α ∫ − a a d u u 2 + a 2 = 8 α a ∫ 0 a d u u 2 + a 2 = .. u = a tan β d u = a d β u 2 + a = a tan 2 β + a 2 = a 2 ( 1 + sin 2 β ) cos 2 β = a 2 ( cos 2 β + sin 2 β ) cos 2 β = a 2 cos 2 β = α a cos β β = arc β ( 0 ) = 0 β ( a ) = arc tan ( a ) = π a . . . . = 8 α ∫ 0 π / 4 d β cos α x β = = 8 α . π / 4 os α a = = 8 α ( π / 4 − 0 ) = 2 π α . W = 2 π α Questao 2 σ = - \frac{Q}{4 \pi R^2} Encontrar potencial eletrico desta configuracao de carga. Determinamos campo eletrico em tudo espaco. \vec{E} = E(r) \hat{r} d\vec{a} = \hat{r} r^2 dΩ \oint_S \vec{E} \cdot d\vec{a} = \frac{1}{ε}Q_{int}(r) = 4\pi r^2 E(r) = \frac{1}{ε}Q_{int}(r) E(r) = \frac{1}{4 \pi ε_ο} \frac{Q_{int}(r)}{r^2} = k \frac{Q_{int}(r)}{r^2} → r < R \quad Q_{int}(r) = Q → r > R \quad Q_{int}(r) = Q + 4\pi R^2 σ = 0 ⇒ E(r) = k \frac{Q}{r^2} \quad r < R E(r) = 0 \quad r > R Potencial eletrico V(r) = - \int_r^∞ \vec{E} \cdot d\vec{l'} = - \int_r^∞ E(r') dr' → r < R V(r) = - \int_r^R \frac{Q}{r'^2} dr' + \int_R^∞ 0 dr' = = - kQ \left[-\frac{1}{r'} \right]_r^R = kQ \left(\frac{1}{r} - \frac{1}{R} \right) → r > R V(r) = - \int_r^∞ 0 dr = 0 V(r) − \frac{kQ}{R} R Questao 3 a b c R_2 R_1 ε R_a d → t < 0 \quad V_a > V_b = V_c = V_d → t > 0 \quad V_a = V_b > V_c > V_d Equacoes das malhas. \frac{V_a - V_d + V_d - V_c - V_a}{R_a I_1} = 0 \quad \Rightarrow I_1 = \frac{ε}{R_a} −ε \frac{V_b - V_c + V_c - V_a + V_a - V_b}{R_2 I_2} = 0 \frac{q(t)}{C} \frac{-R_a I_1}{C} R_2 I_2 + \frac{q(t)}{C} = R_a I_1, \quad I_2 = \frac{dq}{dε} R_2 \frac{dq}{dt} + \frac{q(t)}{C} = ε \quad / \frac{1}{R_2} dy/dt + (1/R2C)y = ε/R2 -> determina a carga na placa do capacitor -> soluções q(t) = Ae^(-t/R2C) + εC q(0) = 0 =>, A + εC = 0 q(t) = εC[1 - e^(-t/R2C)] -> corrente I2(t) = dy/dt I2(t) = (ε/R2)e^(-t/R2C) -> No tempo t1: q(t1) = (2/3)εC 1/3 εC = εC[1 - e^(-t1/R2C)] e^(-t1/R2C) = 2/3 => t1 = R2C ln(3/2) -> Desligamos o chave em t1 t > t1 q(t) = (εC/3) - q' q' - carga que saiu da placa I(t) = dq'/dt Vc > Vb > Vd Vc - Vb + Vb - Vd + Vd - Vb/I R2 = 0 q(t)/C (R1 + R2)/R dq'/dt - (εC/3 - q')1/C = 0 R dq'/dt + q'/C = ε/3 => dq'/dt + q'/(RC) = ε/(3R) q'(t) = Ae^(-t/RC) + (εC/3) q'(t1) = 0 => q(t1) = (εC/3) Ae^(-t1/RC) + (εC/3) = 0 A = -(εC/3)e^(-t1/RC) q'(t) = (εC/3)[1 - e^(-t-t1/RC)] carga no placa d(t) = (εC/3) - q'(t) = (εC/3)e^(-t/RC) -> corrente I = dq'/dt = (ε/3R)e^(-t-t1/RC) I = (ε/3R)e^(-t-t1/RC)