A) decreases
B) does not change
C) becomes zero
D) increases
Correct Answer: D
Solution :
Key Idea: Charge remains constant after charging. If the battery is removed after charging then the charge stored in the capacitor remains constant. q = constant Change in capacitance \[C'=\frac{{{\varepsilon }_{0}}\,A}{d'}\] \[Asd'>d\] hence, \[C'<C\] Hence, potential difference between the plates or \[V'\,\,\propto \,\,\frac{1}{C'}\] As capacitance decreases, so potential difference increases. Note: If the battery remains connected, the charge stores increases. Also the potential difference V becomes constant.You need to login to perform this action.
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