Question

A dielectric is inserted into a capacitor while keeping the charge constant. Identify what happens to the potential difference and the stored energy?
A. The potential difference decreases and the stored energy increases
B. Both the potential difference and stored energy increase
C. The potential difference decreases and the stored energy decreases
D. Both the potential difference and stored energy decrease
E. Both the potential difference and stored energy remain the same

Answer 1

Hint: First determine how the introduction of the dielectric would affect the capacitance by using the fact that the dielectric constant is the ratio of the permittivity in that medium to permittivity of free space, and the capacitance is directly proportional to the permittivity in free space. Then, we know that capacitance can be defined as the ratio of charge and potential difference. Use this to determine how the potential difference would change.
Consequently, extrapolate this behaviour of change in potential difference to the energy stored in the capacitor (since the energy stored is nothing but the electric potential energy) to finally reach the appropriate conclusion.

Formula used:
Potential difference across a parallel plate capacitor: $V=\dfrac{Q}{C}$
Energy stored in a capacitor $U = \dfrac{Q^2}{2C}$

Complete step-by-step answer:
The capacitance of a parallel plate capacitor is given by: $C = \dfrac{A\epsilon_0}{d}$, where A is the area between the two plates, $\epsilon_0$ is the electrical permittivity of free space, and d is the distance between the two plates.
When a dielectric is introduced between the plates of the capacitor, the permittivity of the area between the capacitor plate to electric field changes as:
 $\epsilon = k\epsilon_0$, where k is the dielectric constant which always has a value k>1.
We know that a dielectric is an insulating material that inherits the capability to store charge when placed between two parallel conducting plates of the capacitor. The dielectric material decided how effectively a capacitor stores charge. Higher k value implies that the capacitor can store more charge.
Therefore, the capacitance is now given as:
$C^{\prime} = \dfrac{Ak\epsilon_0}{d} = kC$
Now, capacitance is also defined as the ratio of the electric charge to the potential difference between the parallel capacitor plates. This is given as:
$C=\dfrac{Q}{V} \Rightarrow V= \dfrac{Q}{C}$
When the dielectric is introduced, the new potential difference becomes:
$V^{\prime} = \dfrac{Q}{C^{\prime}} = \dfrac{Q}{kC} = \dfrac{V}{k}$
Thus, we see that with the introduction of the dielectric, the potential difference decreases.
Now, the energy stored in a capacitor is given as
$U = \dfrac{1}{2}CV^2 = \dfrac{1}{2}C.\dfrac{Q^2}{C^2} = \dfrac{Q^2}{2C}$
When the dielectric is introduced, the energy stores becomes:
$U^{\prime} = \dfrac{Q^2}{2C^{\prime}} = \dfrac{Q^2}{2kC} = \dfrac{U}{k}$
This means that the energy stored in the capacitor also decreases.
Thus, the correct option would be: D. Both the potential difference and stored energy decrease.

So, the correct answer is “Option D”.

Note: Remember that a dielectric is able to store electrical charge by the process of polarization. This occurs by the production of electric dipoles inside the dielectric by the application of an external electric field or a potential difference.
Dielectrics can also be of two types: active and passive.
Those materials that polarize easily for the storage of electrical energy are known as active dielectrics (example: piezoelectric), whereas the dielectrics that restrict the storage of electrical energy are passive dielectrics (example: glass).