Capacitors C1 = 3450 nF and C2 = 700 nF are initially connected in parallel and

Question

Capacitors C1 = 3450 nF and C2 = 700 nF are initially connected in parallel and hold a combined charge of Qtoti = 783 nC. Find the charge on each capacitor Q1i , Q2i and the total energy Utoti stored in the system. The capacitors are then disconnected and reconnected, still in parallel, but with the terminals of one capacitor reversed, so that the positively charged side of C1 is now connected to the negatively charged side of C2 and vice versa. Find the combined charge Qtotf and total energy stored in the final state. (Remember, the total charge one the entire capacitor is zero, but Q is the magnitude of the charge on one side.) Answers: Q1i = 651 nC, Q2i = 132 nC, Utoti = 73.9 nJ, Qtotf = 519 nC, Utotf = 32.4 nJ Solutions needed

Explanation / Answer

when they are connected in parallel

potential difference across them must be equal

Q1/C1 = Q2/C2

Q1 = 69Q2/14

Q1+Q2 is given = 783 nC

Q2(83/14) = 783 nC

Q2 = 132.07 nC

Q1 = 650.9 aprox = 651 nC

after they are connected differently

let charge amount x has transferred from C1 to C2 such that charges on them becomes equal

what is initially a positive terminal of C2 will become negative and vice versa

so chargeo n C2 will be x-Q2

Q1-x , Q2 + x

(Q1-x)/C1 = (x-Q2)/C2

(Q1C2 + Q2C1) / (C1+C2) = x

x = 219.5 nC

Q1' = 431.46

Q2' = 87.5

Qtotaf = 519 nC

I hope this is helpful

You can calculate energy from Q^2 / 2C equation for both and sum them up.

Let me know if you need any clarification

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