A 4.00 - 103-nF parallel plate capacitor is connected to a 14.0-V battery and ch

Question

A 4.00 - 103-nF parallel plate capacitor is connected to a 14.0-V battery and charged. a) What is the charge Q on the positive plate of the capacitor? b) What is the electric potential energy stored in the capacitor? The 4.00 middot 103-nF capacitor is then disconnected from the 14.0- V battery and used to charge two uncharged capacitors, a 100.-nF capacitor, a 200.-nF capacitor, connected in series. c) After charging, what is the potential difference across each of the three capacitors? How much work would be done by an electric field in moving a alpha particle from a point at a potential of + 180.0 V to a point at a potential of-60.0 V? (alpha particle electric charge is 2e).

Explanation / Answer

7) C = 4*10^3 nF , V = 14 V

(a) Q = VC

Q = 14*4*10^3*10^-9

Q = 5.6*10^-5 C

(b) U = 0.5cv^2

U = 0.5*4*10^3*10^-9*14*14

U = 0.0004 J

(c) C1 = 4*10^3*10^-9 F, C2 = 100 nF, C3 = 200 nF

in series combination 1/C = 1/4000 +1/100 +1/200

C = 65.6 nF

Q = VC = 14*65.6

Q = 918.4 nC

in seriec combiantion charge is constant

Q1 = Q2 = Q3 = 918.4 nC

V1 = 918.4/4000 = 0.023 V

V2 = 918.4/100 = 9.18 V

V3 = 918.4/200 = 4.6 V

8) V1 = 180 V, V2 = -60 V

q = 2 e

W = q(V1 - V2)

W = 2*1.6*10^-19(180+60)

W = 7.68*10^-17 J

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