Q130 / 150 · MHT CET 2026 (19 April Shift 2) · Physics
A parallel plate air capacitor has capacity 'C' farad, potential 'V' volt and energy 'E' joule. When the gap between the plates is completely filled with dielectric
- AV increases, E decreases.
- BV decreases, E increases.
- Cboth V and E increase.
- Correct answerDboth V and E decrease.
Solution
Since the capacitor is isolated, the charge $Q$ on the plates remains constant.
The initial capacitance is $C$, potential is $V = \dfrac{Q}{C}$, and energy is $E = \dfrac{Q^2}{2C}$.
When a dielectric of dielectric constant $K > 1$ is introduced, the new capacitance becomes $C' = KC$.
The new potential is $V' = \dfrac{Q}{C'} = \dfrac{Q}{KC} = \dfrac{V}{K}$.
Since $K > 1$, $V' < V$, so the potential decreases.
The new energy is $E' = \dfrac{Q^2}{2C'} = \dfrac{Q^2}{2KC} = \dfrac{E}{K}$.
Since $K > 1$, $E' < E$, so the energy also decreases.
Therefore, both $V$ and $E$ decrease.
Answer: both V and E decrease.
The initial capacitance is $C$, potential is $V = \dfrac{Q}{C}$, and energy is $E = \dfrac{Q^2}{2C}$.
When a dielectric of dielectric constant $K > 1$ is introduced, the new capacitance becomes $C' = KC$.
The new potential is $V' = \dfrac{Q}{C'} = \dfrac{Q}{KC} = \dfrac{V}{K}$.
Since $K > 1$, $V' < V$, so the potential decreases.
The new energy is $E' = \dfrac{Q^2}{2C'} = \dfrac{Q^2}{2KC} = \dfrac{E}{K}$.
Since $K > 1$, $E' < E$, so the energy also decreases.
Therefore, both $V$ and $E$ decrease.
Answer: both V and E decrease.
Hence, the correct answer is option: 'D'.