Question

A parallel plate capacitor having capacitance $12\text{pF}$ is charged by a battery to a potential difference of $10\text{V}$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant $6.5$ is placed between the plates. The work done by the capacitor on the slab is:

• A. $600\text{pJ}$
• B. $560\text{pJ}$
• C. $692\text{pJ}$
• D. $508\text{pJ}$

Answer 1

The correct option is D $508\text{pJ}$

Energy stored in a charged capacitor is given by

$U=\frac{1}{2}C{V}^2=\frac{Q^2}{2C}$

Here, $C=12\times {10}^{-12}\text{F};V=10\text{Volts}$

$\Rightarrow U_i=\frac{1}{2}\times 12\times {10}^{-12}\times \left(100\right)$

$\therefore U_i=6\times {10}^{-10}\text{J}$

After insertion of slab, final energy of capacitor

$U_f=\frac{U}{K}=\frac{6\times {10}^{-10}}{6.5}$

The energy dissipated during the process is

$\Delta U=U_i-U_f$

$\Rightarrow (\Delta U{)}_{Lost}=(1-\frac{1}{6.5})\times 6\times {10}^{-10}$

$\therefore (\Delta U{)}_{Lost}=508\text{pJ}$

As we know that
work done on the slab = Energy dissipated

$\therefore$ Work done on the slab $=508\text{pJ}$.

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

Why this question ? Tip: When a dielectric is inserted inside a capacitor, the charge $\uparrow$ & final energy of the capacitor decreases. The work done on slab by capacitor dissipates in form of heat produced.