A) The capacitance decreases
B) The charge on the capacitor increases
C) The voltage across the plates increases
D) The electrostatic energy stored in the capacitor increases
Correct Answer: B
Solution :
As the distance between the plates of capacitor is increased Capacitance C decreases as \[C=\frac{A{{\varepsilon }_{0}}}{d}\] or \[C\propto \frac{1}{d}\] (ii) Charge on the capacitor remains constant. (iii) Potential difference increases as Q = C V (iv) Electrostatic energy increases as \[U=\frac{1}{2}QV\] and for constant Q, \[U\propto V\]
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