Question

A capacitor stores $50\mu \text{C}$ charge when connected across battery. When the gap between the plates is filled with a dielectric, a charge of $100\mu \text{C}$ flows through the battery. Find the dielectric constant of the material inserted.

• A. $3$
• B. $6$
• C. $7$
• D. $5$

Answer 1

The correct option is A $3$

Initial charge stored $Q_1=50\mu \text{C}$.

Let, the dielectric constant of the material inserted is $K$.

Given, extra charge flown through battery is $100\mu \text{C}$.

Therefore total, charge stored in capacitor $Q_2=100+50=150\mu \text{C}$.

Now in initial case.
$C_1=\frac{A{\epsilon }_0}{d}$

$\Rightarrow \frac{Q_1}{V}=\frac{A{\epsilon }_0}{d}$

$\Rightarrow Q_1=\frac{VA{\epsilon }_0}{d}...\left(i\right)$

In the second case:

Capacitance, $C_2=\frac{KA{\epsilon }_0}{d}$

Charge stored,

$Q_2=\frac{VKA{\epsilon }_0}{d}...\left(ii\right)$

Dividing (i) & (ii) we get,

$\frac{Q_1}{Q_2}=\frac{1}{K}$

$\Rightarrow \frac{50}{150}=\frac{1}{K}$

$\therefore K=3$

Hence, option (a) is the correct answer.