Question

An air filled parallel plate capacitor has a capacitance$1pF$. The separation between the plates is doubled and wax is inserted between the plates, then the capacitance becomes $2pF$. What is the dielectric constant of the wax?
A. 8
B. 2
C. 4
D. 10

Answer 1

Using the constant of parallel plate capacitor where capacitance of a capacitor depends on Area of the plate and the distance between the plates.

Formula Used: $C=\frac{K{{\varepsilon }_{0}}A}{d}$
Where;
C is the capacitance of the capacitor
K is the dielectric constant of the capacitor
A is the area of the plates
d is the distance between the two plates

Complete step by step solution:
Capacitance of the capacitor having air medium between the plates
$C=\frac{{{\varepsilon }_{0}}A}{d}.......\,\,(1)$
When the distance between the plates is doubled and the medium is filled with wax
${{C}^{'}}=\frac{K{{\varepsilon }_{0}}A}{{{d}^{'}}}$
Where;
${{d}^{'}}=2d$
Thus, we get ${{C}^{'}}=\frac{K{{\varepsilon }_{0}}A}{2d}.......\,\,(2)$
Dividing equation (1) and (2), we get
$\frac{C}{{{C}^{'}}}=\frac{2}{K}$
Here, $C=1pF\,\,and\,\,{{C}^{'}}=2pF$
Therefore,

& \frac{1}{2}=\frac{2}{K} \\\ & Or\,\,K=4 \\\ \end{aligned}$$ **The correct answer is (c).** **Additional Information:** The capacitor is a type of device which stores energy for a small interval of time. Parallel plate capacitor is one of the types of capacitor which has arrangement of electrodes and insulating material. It’s capacitance depends on the medium filled between the plates of the capacitor, the area of the plates and the distance between the plates. We know there is a certain amount of charge on the plates. If we supply more charge, the potential increases and it leads to leakage of charge. **Note:** If we change the medium between the plates keeping the distance between the plates and area of the plates constant, the new capacitance is K time the capacitance of the capacitor in the air medium( Where K is the dielectric constant of the capacitor).