Question

A parallel-plate capacitor is connected to a battery, which maintains a potential difference $V_0$ between its plates. A slab of dielectric constant $K$ is slowly inserted between the plates, completely filling the space between them. If $Q_0$ is the charge on the plates before the dielectric is inserted, how much work is done by mechanical forces on the slab in the whole process ?

• A. $K{Q}_0{V}_0$
• B. $\frac{1}{2}{Q}_0{V}_0(K-1)$
• C. $-\frac{K{Q}_0{V}_0}{2}$
• D. $-\frac{1}{2}{Q}_0{V}_0(K-1)$

Answer 1

The correct option is D $-\frac{1}{2}{Q}_0{V}_0(K-1)$


Suppose the capacitance before the dielectric is inserted is $C_0$

The initial Charge, $Q_0=C_0{V}_0$

Now when the dielectric is inserted, charge
$Q^{\prime }=K{C}_0{V}_0$

As $K>1\Rightarrow Q^{\prime }>Q_0$

As we know that work done by battery will be

$W=QV$

$\therefore W_{battery}=(Q^{\prime }-Q_0)V_0=(K-1)C_0{V}_0^2$

$\Rightarrow W_{battery}=(K-1)Q_0{V}_0$

Using conservation of energy, we have

$W_{mech}+W_{batt}=\Delta U$

No kinetic energy present because slab inserted slowly.

$\Rightarrow W_{mech}+(K-1)Q_0{V}_0=\frac{1}{2}K{C}_0{V}_0^2-\frac{1}{2}{C}_0{V}_0^2$

$\Rightarrow W_{mech}=\frac{1}{2}{C}_0{V}_0^2(K-1)-(K-1)Q_0{V}_0$

$\Rightarrow W_{mech}=\frac{1}{2}{Q}_0{V}_0(K-1)-Q_0{V}_0(K-1)$

$\Rightarrow W_{mech}=-\frac{1}{2}{Q}_0{V}_0(K-1)$

Observe that Electric force does positive work and in turn the applied mechanical forces do negative work on the dielectric slab.

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

Key concept- Work done by external forces on the dielectric when it is inserted in a parallel plate capacitor with battery connected.