Question

A capacitor stores 50 µC charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of 100 µC flows through the battery. Find the dielectric constant of the material inserted.

Answer 1

Given:

Initial charge on the capacitor = 50 µC

Now, let the dielectric constant of the material inserted be k.

As 100 µC of extra charge flows through the battery, the net charge on the capacitor becomes

50 + 100 = 150 µC

Now,
$$\begin{gathered} C_1=\frac{q_1}{V}=\frac{{\in }_{0}A}{d}...\left(i\right) \\ {C}_2=\frac{q_2}{V}=\frac{{\in }_{0}Ak}{d}...\left(ii\right) \end{gathered}$$
On dividing (i) by (ii), we get
$$\begin{gathered} \frac{C_1}{C_2}=\frac{q_1}{q_2}=\frac{1}{k} \\ \Rightarrow \frac{50}{150}=\frac{1}{k} \\ \Rightarrow k=3 \end{gathered}$$

Thus, the dielectric constant of the given material is 3.