Question

A parallel plate capacitor of capacitance $90\text{pF}$ is connected to a battery of $20\text{V}$. If a dielectric material of dielectric constant $K=5/3$ 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:
$C=90\text{pF};V=20\text{V};K=5/3$

Initial charge on the capacitor is given by,

$Q=CV..........\left(1\right)$

When dielectric is inserted, its capacitance becomes $KC$.

So, now charge on capacitor,

$Q^{\prime }=KCV$

So, charge induced in the dielectric is

$Q_{in}=Q^{\prime }-Q=CV(K-1)$

$Q_{in}=90\times {10}^{-12}\times 20(\frac{5}{3}-1)$

$\therefore Q_{in}=1200\times {10}^{-12}=1.2\text{nC}$

Therefore, option $\left(B\right)$ is correct.