Question

A parallel plate capacitor with a dielectric of dielectric constant $K$ between the plates has a capacitance $C$ and it is charged to $V\text{Volts}$. The dielectric slab is slowly removed from the capacitor and then re-inserted between its plates. The net work done by the system in this process is:

• A. $\frac{1}{2}(K-1)C{V}^2$
• B. $C{V}^2(K-1)K$
• C. $(K-1)C{V}^2$
• D. zero

Answer 1

The correct option is D zero

When dielectric is introduced between plates and again removed followed by re-insertion, then the capacitance and potential difference across the capacitor plate will be same as the initial value.

i.e., $C_i=C_f=C;V_i=V_f=V$

$\therefore U_f=U_i=U=\frac{1}{2}C{V}^2$

As we know that

$W_{ext}=U_f-U_i$

$W_{ext}=\frac{1}{2}C{V}^2-\frac{1}{2}C{V}^2=0$

Hence, option $\left(d\right)$ is correct.

Why this question? Tip: The process of insertion of dielectric, removal and re-insertion can be visualized as round trip phenomenon. Since the system involves presence of a conservative field (electrostatic field), therefore $W_{ext}$ = 0 in a round trip process.