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 flow of charge through the cell after insertion of dielectric will be

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

Answer 1

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

Given:
$C=90\text{pF}$

Now charge for $V=20\text{volt}$

$Q=CV$

$Q=90\times 20\text{pC}.....\left(1\right)$

When the dielectric medium is present then capacitance is,

$C^{\prime }=KC$

$C^{\prime }=\frac{5}{3}\times 90\text{pF}$

$\therefore$ Charge $Q^{\prime }=C^{\prime }V=\frac{5}{3}\times 20\times 90\text{pC}....\left(2\right)$

Now the induced charge is,

$Q^{\prime }-Q=(\frac{5}{3}-1)\times 20\times 90\times {10}^{-12}\text{C}$

$=1200\times {10}^{-12}\text{C}=1.2\text{nC}$