A) decreases
B) does not change
C) becomes zero
D) increases
Correct Answer: D
Solution :
Charge remains constant after charging. If the battery is removed after charging, then the charge stored in the capacitor remains constant. q = constant New capacitance \[C'=\frac{{{\varepsilon }_{0}}A}{d'}\] As \[d'>d.\]Hence, C' < C Hence, potential difference between the plates \[V'=\frac{q}{C'}\] \[V'\alpha =\frac{1}{C'}\] As capacitance decreases, so potential difference increases.
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