Question

A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation, the charge, potential, and capacitance respectively

• A. constant, decreases, decreases.
• B. increases, decreases, decreases.
• C. constant, decreases, increases.
• D. constant, increases, decreases.

Answer 1

The correct option is D

constant, increases, decreases.

Step 1: Given

A parallel plate capacitor is charged and then isolated.

Step 2: To find

The effect of increasing the plate separation on the charge, potential, and capacitance.

Step 3: Formula Used

The capacitance of a parallel plate capacitor is given by,

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

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

Step 4: Explanation

When the plate separation of a capacitor which is not connected to a battery increases, the charge on the capacitor remains the same as there is no other source of charge.

So, the charge remains constant.

The capacitance of a parallel plate capacitor is given by,

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

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

As the area and permittivity are constant, $C\propto \frac{1}{d}$

So, when the distance between the plates increases, capacitance decreases.

We know, charge on a capacitor is given by $Q=CV$

Here since $Q$ is constant,

$$\begin{gathered} V\propto \frac{1}{C} \\ \Rightarrow V\propto d(\because C\propto \frac{1}{d}) \end{gathered}$$

Therefore as the distance between the plates increases, voltage also increases.

Option C is correct.