Question

A capacitor of capacitance $10\mu \text{F}$ is connected to a battery of emf $4\text{V}$. Now this charged capacitor is disconnected from the battery and a dielectric slab of dielectric constant $4$ is inserted between the plates. Find the magnitude of the charge induced on the surface of the dielectric.

• A. $0$
• B. $40\mu \text{C}$
• C. $30\mu \text{C}$
• D. $20\mu \text{C}$

Answer 1

The correct option is C $30\mu \text{C}$

Given:
$C=10\mu \text{F};V=4\text{V};K=4$

Initial charge stored on the capacitor is given by

$Q=CV$

$\Rightarrow Q=10\times 4=40\mu \text{C}$

Now, when the dielectric is inserted, charge will be induced on the surface of the dielectric slab and that will be,

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

$\Rightarrow Q_{in}=40(1-\frac{1}{4})$

$\therefore Q_{in}=30\mu \text{C}$

Therefore, option $\left(C\right)$ is the right answer.

Why this question ? Concept: Induced charge on dielectric is given as, $Q_{in}=Q(1-\frac{1}{K})$ So, $Q^{{}^{\prime }}<Q$