First slide

Capacitance

Question

In the arrangement will be the potential difference between the terminals A and B after the battery is removed and the switch S is closed ?

Moderate

A

3.33 volt
B

4 volt
C

5 volt
D

10 volt

Solution

Total charge stored in the system is Q = $10 \times \frac{3 \times (2 + 4)}{3 + (2 + 4)} μC = 20 μC$
after the battery is removed and S is closed, total charge in the system between A and B, is Q₁ = Q = 20µC
$∴ V_{AB} = \frac{Q'}{2 + 4} volt = \frac{20}{6} volt = 3.33 volt$

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