Consider the situation shown in the figure. The capacitor A has a charge $q$ on it whereas B is uncharged. The charge appearing on the capacitor B a long time after the switch is closed is [2001]

Question

Smart Achievers · Accepted Answer

(1)

When switch S is closed, there will be no shifting of negative charge from plate A to B as the charge −q is held by the charge +q, as unlike charges attract each other. Neither will there be any shifting of charge from B to A.

Hence, the charge on capacitor B = 0.

Solution

Q.
Consider the situation shown in the figure. The capacitor A has a charge $q$ on it whereas B is uncharged. The charge appearing on the capacitor B a long time after the switch is closed is [2001]

1 zero

2 $q / 2$

3 $q$

4 $2 q$

Ans.
(1)

When switch S is closed, there will be no shifting of negative charge from plate A to B as the charge −q is held by the charge +q, as unlike charges attract each other. Neither will there be any shifting of charge from B to A.

Hence, the charge on capacitor B = 0.

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Solution

Q. Consider the situation shown in the figure. The capacitor A has a charge q on it whereas B is uncharged. The charge appearing on the capacitor B a long time after the switch is closed is [2001]