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A dielectric is inserted into a capacitor while the charge on it is kept constant. 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 increases.
C. The potential difference increases and the stored energy decreases.
D. Both the potential difference and stored energy decreases.
E. Both the potential difference and stored energy remains the same.

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Last updated date: 23rd May 2024
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Answer
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Hint: To solve this question, we have to remember that the potential difference between the plates of the capacitor when a dielectric slab of thickness t is introduced between the two plates of the capacitor is, $V = \dfrac{q}{{{\varepsilon _0}A}}\left( {d - t + \dfrac{t}{K}} \right)$, where A is the area of each plate, d is the separation between them, K is the dielectric constant and q is the charge on the capacitor.

Complete answer:
As we know that,
A capacitor is the arrangement of two or more conducting plates which are separated by air or any medium (dielectric or conducting). It is used to store charge or energy.
The potential difference between the plates, when the plates are separated by air is given by,
$ \Rightarrow V = \dfrac{{qd}}{{A{\varepsilon _0}}}$ …….. (i)
where A is the area of each plate, d is the separation between them and q is the charge on the capacitor.
When a dielectric medium with dielectric constant K, is inserted between the plates and the charge on it kept as constant, then the potential difference between the plates will become
$ \Rightarrow V' = \dfrac{q}{{{\varepsilon _0}A}}\left( {d - t + \dfrac{t}{K}} \right)$ ……. (ii)
Here t is the thickness of the dielectric slab, (tFrom equation (ii), we can see that the factor $\left( {d - t + \dfrac{t}{K}} \right)$ is small as compared to d in equation (i).
This will result in a decrease in potential difference.
Now, we know that
The energy stored in a capacitor is given by,
$ \Rightarrow U = \dfrac{1}{2}C{V^2}$, where C is the capacitance and V is the potential difference between the plates.
Here,
Energy store, U is directly proportional to V and in our case, V decreases,
Thus, the energy stored, U will also decrease.
Hence, when a dielectric is inserted into a capacitor while the charge on it is kept constant, then both the potential difference between the plated and the energy stored will decrease.

So, the correct answer is “Option D”.

Note:
When a dielectric slab is inserted between the plates of the capacitor, which is kept connected to the battery, i.e. the charge on it increases, then the capacitance (C) increases, potential difference (V) between the plates remains unchanged and the energy stored in the capacitor increases.
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