Question

A parallel plate capacitor of capacitance $90\text{pF}$ is connected to a battery of emf $20\text{V}$. If a dielectric material of dielectric constant $K=\frac{5}{3}$ is inserted between the plates, the magnitude of induced charge will be:

• A. $0.9\text{nC}$
• B. $1.2\text{nC}$
• C. $0.3\text{nC}$
• D. $2.4\text{nC}$

Answer 1

The correct option is B $1.2\text{nC}$

Given, before inserting the dielectric:

Initial capacitance, $C_0=90\text{pF}$,
Potential difference across the plates, $E=20\text{V}$,
Dielectric constant, $K=\frac{5}{3}$,

After inserting dielectric:

Capacitance, $C_1=K{C}_0$
$\Rightarrow C_1=\frac{5}{3}\times 90=150\text{pF}$

Therefore charge supplied by battery;
$q=C_{1}E$
$\Rightarrow q=150\times 20=3000\text{pC}$
$\Rightarrow q=3000\times {10}^{-12}\text{C}\text{or}3\text{nC}$

The magnitude of induced charge is given by the formula,$$q^{\prime }=q(1-\frac{1}{K})$$$\Rightarrow q^{\prime }=3[1-\frac{1}{\left(\frac{5}{3}\right)}]=3\times \left[\frac{2}{5}\right]$

$\therefore q^{\prime }=\frac{6}{5}=1.2\text{nC}$

Hence, option (b) is the correct answer.