A parallel plate capacitor is charged fully by using a battery. Then, without disconnecting the battery, the plates are moved further apart. Then,
KVPY 2009, Medium
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(d)
As plates are moved far away further, capacity of capacitor $\left(C=\frac{\varepsilon_0 A}{d}\right)$ decreases.
As battery remains connected during this activity, potential difference between plates remains same.
So, the charge on plates $(Q=C V)$ decreases.
Also, energy of capacitor $\left(U=\frac{1}{2} C V^2\right)$ decreases as capacitance decreases. So, correct option is $( d )$.
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