Question

If we increase 'd' of a parallel condenser to '2d' and fill wax to the whole empty space between its two plate, then capacitance increase from $1\;{\rm{pF}}$ to $2\;{\rm{pF}}$. What is the dielectric constant of wax?
A. 2
B. 4
C. 6
D. 8

Answer 1

The above problem can be resolved using the concept and applications of capacitors. The capacitors are the major energy restoring electrical devices, that have numerous applications both in the electrical as well as the mechanical analysis. Then, the mathematical formula for the capacitance at initial is used. If we are filling the space by way, then there should be some change in capacitance, that can be utilised to calculate the value of dielectric constant.

Complete step by step answer:
Given:
The initial value of capacitance is, ${C_1} = 1\;{\rm{pF}}$.
The final value of capacitance is, ${C_2} = 2\;{\rm{pF}}$.
The initial capacitance of condenser is,
${C_1} = \dfrac{{A{\varepsilon _0}}}{d}................................................\left( 1 \right)$
Here, A is the area of the condenser plates, ${\varepsilon _0}$ is the permittivity of free space and d is the separation within the condensers.
After filling the empty space with wax. The final value of capacitance is,
${C_2} = \dfrac{{KA{\varepsilon _0}}}{{2d}}......................................\left( 2 \right)$
Here, K is the dielectric constant of the wax.
Taking the ratio of equation 1 and 2 as,

\dfrac{{{C_1}}}{{{C_2}}} = \dfrac{{A{\varepsilon _0}/d}}{{KA{\varepsilon _0}/2d}}\\\ \dfrac{{{C_1}}}{{{C_2}}} = \dfrac{2}{K} \end{array}$$ Solve by substituting the values as, $$\begin{array}{l} \dfrac{{{C_1}}}{{{C_2}}} = \dfrac{2}{K}\\\ \dfrac{{1\;{\rm{pF}}}}{{2\;{\rm{pF}}}} = \dfrac{2}{K}\\\ K = 4 \end{array}$$ Therefore, the dielectric constant of wax is 4 and option (B) is correct. **Note:** Try to understand the meaning of capacitor, and along with knowledge of the fact that, how capacitance of the capacitor, varies with separation and the value of dielectric constant. There is a direct relation for the capacitance and the dielectric constant of the medium. Moreover, this relation is also useful to determine the area of plates, taken into consideration.