Question

A 2.50 F capacitor is charged to 857 V and a 6.80F capacitor is charged to 652 V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each and the charge on each? [Hint: charge is conserved.]

Answer 1

Given ,
C = 2.50 µF
C = 2.50×10^-6 F
V = 857 V
Charge. In first capacitor,(Q) = C*V
Q =2.50×10^-6 F *857 V
Q = 2,142.5 *10^-6 coulmb

For second capacitor
C' = 6.80 µF
C' = 6.80×10^-6 F
V' = 652 V
Q' = 6.80×10^-6 F * 652 V
Q' = 4,433.6 *10^-6 coulmb

The negative plate are connected with eachother, So they are connected in parallel
Resultant capacitance C'' = C + C'
C'' =2.50×10^-6 F + 6.80×10^-6 F
C'' = 9.3 ×10^-6 F

Total charge Q'' = Q + Q'
Q'' = 2,142.5 *10^-6coulmb + 4,433.6*10^-6 coulmb
Q'' = 6,576.1*10^-6 coulmb

Total potential difference V'' = Q'' / C''
V'' = 6,576.1 *10^-6coulmb / 9.3 ×10^-6 F
V'' = 707.12 V

Charge on first capacitor q = CV''
q = 2.50×10^-6 F * 707.12 V
q = 1,767.8*10^-6 C

Charge on second capacitor q'' = C'V''
q'' = 6.80×10^-6 F * 707.12 V
q'' = 4,808.416*10^-6 C

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