Question

A capacitor is charged using a battery which is then disconnected. A dielectric slab is introduced between the plates which results in
A) an increase in the potential difference across the plates and a reduction in the stored energy but no change in the charge on the plates.
B) a decrease in the potential difference across the plates and a reduction in the stored energy but no change in the charge on the plates.
C) a reduction of the charge on the plates and an increase in the potential difference across the plates.
D) an increase in stored energy but no change in the potential difference across the plates.

Answer 1

Hint: When a dielectric slab gets inserted between the capacitor plates, the capacitance of the capacitor will increase. Since the potential difference across the plates and the energy stored in the capacitor depends on the capacitance of the capacitor, these too will change.

Formulas used:
The capacitance of a capacitor with air between the plates is given by, $C={\displaystyle \frac{{\epsilon }_{0}A}{d}}$ where ${\epsilon }_0$ is the permittivity of free space, $A$ is the area of the plates and $d$ is the distance between the two plates.
The potential difference across the capacitor plates is given by, $V={\displaystyle \frac{Q}{C}}$ where $C$ is the capacitance and $Q$ is the charge on the plate.
The energy stored in the capacitor is given by, $U={\displaystyle \frac{Q^2}{2C}}$ where $Q$ is the charge on the plate and $C$ is the capacitance of the capacitor.

Complete step by step answer:
Step 1: Describing the problem at hand.
A capacitor was charged and disconnected. The charge on the plates remains as such. Then a dielectric slab was inserted between the plates. We have to determine how this action affects the potential difference across the capacitor plates and the energy stored in it.
Step 2: Expressing the relation for the capacitance to determine if it increases or decreases on the introduction of a dielectric slab.
The capacitance of a capacitor with air between the plates is given by,
$C={\displaystyle \frac{{\epsilon }_{0}A}{d}}$ where ${\epsilon }_0$ is the permittivity of free space, $A$ is the area of the plates and $d$ is the distance between the two plates.
Now on introducing a dielectric slab between the plates, the permittivity becomes $k{\epsilon }_0$ where $k$ is some constant.
So the capacitance then will be
$\Rightarrow C^{\prime }={\displaystyle \frac{k{\epsilon }_{0}A}{d}}$
$\Rightarrow C^{\prime }=kC$
i.e. the capacitance of the capacitor increases on inserting a dielectric slab between the plates.
Step 3: Using the obtained information about the increase in capacitance, determining if the potential difference across the plates decreases or increases on the introduction of a dielectric slab.
The potential difference across the capacitor plates will be $V={\displaystyle \frac{Q}{C^{\prime }}}$ where $C$’ is the new capacitance and $Q$ is the charge on the plate.
Since charge remains the same, the above relation suggests that $V\propto {\displaystyle \frac{1}{C^{\prime }}}$.
So as the capacitance increases the potential difference across the plates must decrease.
Step 4: Expressing the relation for the energy stored in the capacitor to see if it changes.
The energy stored in the capacitor will be $U={\displaystyle \frac{Q^2}{2C^{\prime }}}$ where $Q$ is the charge on the plate and $C$’ is the new capacitance of the capacitor.
From the above relation we have, $U\propto {\displaystyle \frac{1}{C^{\prime }}}$
We know the charge remains constant and the capacitance has increased. So the energy stored in the capacitor will decrease.

So we can conclude that as the dielectric gets inserted between the plates, the potential difference across the plates and the energy stored in the capacitor decreased while the charge remained constant. So the correct option is B.

Note:
When a capacitor is connected to a battery, the two plates of the capacitor will get charged depending on the polarity of the terminal of the battery to which each plate is connected. On disconnecting the battery, the charge formed on the plates will remain the same. The capacitor thus stores energy. Then on discharging the plates, the charge decreases.