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Question

See the diagram. Area of each plate is 2.0m2 and d=2×10×103m. A charge of 8.85×108C is given to Q. Then the potential of Q becomes:
573457.JPG
  1. 10V
  2. 13V
  3. 8.825V
  4. 66.7V

A
8.825V
B
13V
C
10V
D
66.7V
Solution
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Given circuit can be visualized as two parallel capacitors C1 and C2. Let charge on the two capacitors be Q1 and Q2.

Given:
Q1+Q2=8.85×108C
C1=εoAd=2εo20×103=100εo
C2=εoA2d=2εo2×20×103=50εo
Net capacitance is C=C1+C2=150εo

From definition of capacitance,
V=QC=8.85×108150εo
V104εo150εo=66.7 V

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573457.JPG
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