Four identical capacitors are connected in series with a 10 V battery as shown in the figure. Potentials at A and B are

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10 V, 0 V

7.5 V, -2.5 V

5 V, -V

7.5V, 2.5 V

answer is B.

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Detailed Solution

Let q = charge on each capacitorqC+qC+qC+qC=10 ⇒ qC=104 ⇒ qC=2.5 ∴ VA−VN=3qC =3×2.5=7.5 V VA−0=7.5V ⇒ VA=7.5V again VN−VB=qC ⇒ 0−VB=2.5V ⇒ VB=−2.5V

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