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

Answer 1

Answer: (A)

zero

Energy of capacitor removal of slab $Ui=\frac{1}{2}C{V}^2$

Energy of capacitor offer removal of slab $V_f=\frac{1}{2}C{V}^2$

$W=-\Delta V$

$W=0\left(A\right)$