Chapter 26, Problem 1OQ

(i)

To determine

The change in the capacitance of the capacitor.

Answer to Problem 1OQ

Option (a) increase.

Explanation of Solution

Write the expression for capacitance of a capacitor.

    $C=k{C}_0$

Here, $C$ is the capacitance of a capacitor, $C_0$ is the initial capacitance of a capacitor and $k$ is the dielectric constant.

The above equation signifies that the capacitance of the capacitor increases when a dielectric is introduced between the plates.

Conclusion:

Therefore, option (a) increase; is correct.

(ii)

To determine

The change in the charge of the capacitor.

Answer to Problem 1OQ

Option (a) increase.

Explanation of Solution

Write the expression for charge in the capacitor.

    $q=k{q}_0$

Here, $q$ is the charge in the capacitor and $q_0$ is the initial charge on the capacitor.

The above equation signifies that the charge of the capacitor increases when a dielectric is introduced between the plates.

Conclusion:

Therefore, option (a) increase; is correct.

(iii)

To determine

The change in the potential difference of the capacitor.

Answer to Problem 1OQ

Option (c) stay the same.

Explanation of Solution

The capacitor remains connected to the battery while the dielectric is introduced between the plates, due to which the potential difference in the capacitor remains the same.

Conclusion:

Therefore, option (c) stay the same; is correct.

(iv)

To determine

The change in energy stored in the capacitor.

Answer to Problem 1OQ

Option (a) increase.

Explanation of Solution

Write the expression for energy stored in a capacitor.

    $E=\frac{q^2}{2C}$

Substitute $k{q}_0$ for $q$ and $k{C}_0$ for $C$ in the above equation to find $E$.

    $\begin{array}{l}E=\frac{{\left(k{q}_0\right)}^2}{2k{C}_0}\\ E=\frac{k^2{q}_0}{2k{C}_0}\\ E=k\frac{q_0}{2C_0}\end{array}$

The above equation signifies that the energy stored in the capacitor increases when a dielectric is introduced between the plates.

Conclusion:

Therefore, option (a) increase; is correct.