Question

On increasing the plate separation of a charged condenser, the energy

• A. increases
• B. decreases
• C. remains unchanged
• D. becomes zero

Answer 1

Answer: (A)

increases

Answer 2

The energy which is stored in the condenser is given by
$E=\frac{1}{2} \cdot \frac{q^{2}}{C}$ ...(i)
where $q$ is charge and $C$ the capacitance.
Also, $C=\frac{\varepsilon_{0} A}{d}$ ...(ii)
From Eqs. (i) and (ii), we get
$\Rightarrow E =\frac{1}{2} \cdot \frac{q^{2} d}{\varepsilon_{0} A}$
$\Rightarrow E \propto d$
When plate separation $d$ is increased energy increases.