An air-filled parallel plate capacitor has a capacitance of 1.3 pF.The separation of the plates is doubled, and wax is inserted between them. The new capacitance is 2.6pF. Find the dielectric constant of the wax.

a) Capacitance of air-filled plate capacitor,$C_0=1.3pF$

b) Separation between the plates is doubled.

c) New capacitance value,$C_K=2.6pF$

The addition of a dielectric substance increases the capacitance of a group of charged parallel plates. Since the dielectric lowers the effective electric field, the capacitance is inversely proportional to the electric field between the plates. We can use the formula of capacitance for a parallel plate capacitor with and without the dielectric. Taking the ratio, we can get the value of the dielectric constant of the wax.

Formula:

The capacitance value between two plates filled with a medium, $C=\frac{{\epsilon }_{0}KA}{d}$ …(i)

Here,${\epsilon }_0$ is permittivity of the free space, k is dielectric constant, A is area of cross section and is separation between the plates.

Taking the ratio of capacitance of a parallel plate capacitor with dielectric and without dielectric (that is for air, $\kappa =1$), we can the dielectric constant value of the wax as follows: (For new capacitance, separation doubles)

$$\begin{gathered} \frac{C_K}{C_0}=\frac{{\displaystyle \frac{{\epsilon }_{0}kA}{2d}}}{\frac{{\epsilon }_{0}A}{d}} \\ \frac{C_K}{C_0}=\frac{k}{2} \\ k=\frac{2C_K}{C_0} \\ =2\times \frac{2.6pF}{1.3pF} \\ =4 \end{gathered}$$

Hence, the value of the dielectric constant is 4 .