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

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 x 10^3 nF = 4 micro F ; V = 14 V

a)Q = CV

Q = 4 x 10^-6 x 14 = 56 x 10^-6 C = 56 micro-C

Hence, Q = 56 uC

b)U = 1/2 C V^2

U = 0.5 x 4 x 10^-6 x 14^2 = 392 x 10^-6 J

Hence, U = 392 x 10^-6 J = 3.92 x 10^-4 J

c)charge remains conserved.

we know that the same charge flows through the capactors connected in series.

Q1 = Q2 = Q3 = 56 uC

V1 = Q1/C1 = 56 x 10^-6/4 x 10^-6 = 14 V

V2 = Q2/C2 = 56 x 10^-6/100 x 10^-9 = 560 V

V3 = Q3/C3 = 56 x 10^-6/200 x 10^-9 = 280 V

Hence, V1 = 14 V ; V2 = 560 V and V3 = 280 V

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