Question

A parallel plate condenser with a dielectric constant $K$ between the plates has a capacity $C$ and is charged to a potential $V$ volt. The dielectric slab is slowly removed from, between the plates and then reinserted. The net work done by the system in this process is

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

Answer 1

Answer: (A)

Energy stored in parallel plate capacitor of Capacitance $C$ and Potential $V$ is
$U=\frac{1}{2}C{V}^2$
When the dielectric slab is removed from, between the capacitor energy stored
$U=\frac{1}{2}{C}^{\prime }{V}^2$
$C$ is the capacitance in the presence of dielectric
$C^{\prime }$ is the capacitance in the absence of dielectric
and we have $C=K{C}^{\prime }$ where $K$ is the dielectric constant
When the dielectric slab is reinserted the potential energy becomes
U = $\frac{1}{2}C{V}^2$
Hence the net work done in the process is zero