Question

A parallel plate capacitor with air between the plates has a capacitance of $8pF(1pF={10}^{-12}F)$. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$?

Answer 1

Step 1: Given data

Initial capacitance, $C=8pF$

Step 2: To find

The capacitance when distance between the plates is reduced by half and a dielectric of dielectric constant $6$ is filled between the plates.

Step 3: Formula used

Capacitance, $C=\frac{{\epsilon }_{0}A}{d}$

When a dielectric is present between the plates, the capacitance, $C=\frac{K{\epsilon }_{0}A}{d}$

where $K$ is the dielectric constant, ${\epsilon }_0$ is the permittivity of free space, $A$ is the area of the plate and $d$ is the distance between the plates.

Step 4: Calculation

The initial capacitance,

$$\begin{gathered} C=\frac{{\epsilon }_{0}A}{d}=8\times {10}^{-12} \\ \frac{{\epsilon }_{0}A}{d}=8\times {10}^{-12} \end{gathered}$$

Capacitance after the distance between the plates is reduced to half and a dielectric is filled between the plates with dielectric constant, $K=6$

$$\begin{gathered} C\text{'}=\frac{K{\epsilon }_{0}A}{d\text{'}} \\ C\text{'}=\frac{6\times {\epsilon }_{0}A}{\left({\displaystyle \frac{d}{2}}\right)}\left(\because d\text{'}=6,K=6\right) \\ C\text{'}=2\times 6\times \frac{{\epsilon }_{0}A}{d} \\ C\text{'}=2\times 6\times C\left(\because C=\frac{{\epsilon }_{0}A}{d}\right) \\ C\text{'}=2\times 6\times 8\times {10}^{-12} \\ C\text{'}=96\times {10}^{-12} \\ C\text{'}=96pF \end{gathered}$$

Therefore, the new capacitance is $96pF$.