Question

An air capacitor of capacitance $10\mu \text{F}$ is connected to a battery of $10\text{V}$. Now the space between its plates is completely filled with an insulator $(K=2)$. The magnitude of induced charge on the dielectric is

• A. $100\mu \text{C}$
• B. $50\mu \text{C}$
• C. $25\mu \text{C}$
• D. $200\mu \text{C}$

Answer 1

The correct option is A $100\mu \text{C}$

Given:
$C=10\mu \text{F}=10\times {10}^{-6}\text{F}$
$V=10\text{V};k=2$

Initial charge on capacitor will be

$Q=CV=10\mu \text{F}\times 10\text{V}=100\mu \text{C}$
Now, as battery is connected and dielectric of $k=2$ is introduced in between the plates, the capacitance will increased to $kC$, thus, the final charge on the plates will be
$Q_f=kCV=2\times 10\times 10=200\mu C$

Induced charge is given by the formula,

$Q_{in}=Q_f(1-\frac{1}{K})$

$\Rightarrow Q_{in}=200(1-\frac{1}{2})=100\mu \text{C}$

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

Key concept- Induced charge on the surface of dielectric inside a capacitor.