Question

A 4.0 uF parallel plate capacitor is connected to a 5 V
battery. The work done by an external agent to slowly
increase the separation between the plates to double
of its original value is​

Answer 1

$\bold{ \red{ \huge{ \underline{Answer:-}}}}$

➤ The energy initially stored by the capacitor = 1/2×C×V² or Q/2*C

➤ When the plate separation is doubled, the charge remains constant and the capacitance is halved.

➤ The work done in separating the plates is the difference in energy stored =

$\frac{Q^2}{2} \times 0.5 \times C - \frac{Q^2}{2} \times C$

➤ This represents double the initial energy less the initial energy.

The work done in adding this energy is =

$\frac{Q^2}{2} \times C$

➤ Therefore,

Final Answer =

$= > \frac{(2 \times 10 - 4) {}^{2}}{(2 \times 10 \times 10 { }^{ - 6} )} \\ = > \bold{ \boxed{2 \: mJ}}$

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Hope, it helps.