Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?

The potential difference between the plates decreases K times

The energy stored in the capacitor decreases K times

The change in energy stored is ½CV²(1/K - 1)

The charge on the capacitor is not conserved

Step-by-Step Solution

When the capacitor is disconnected from the battery, the charge ( $Q$ ) on its plates remains isolated and thus conserved ( $Q$ = constant). Therefore, the statement 'The charge on the capacitor is not conserved' is incorrect.

Let's analyze the other statements:

Capacitance: When a dielectric of constant $K$ is inserted, the new capacitance becomes $C' = KC$ .
Potential Difference: Since $Q$ is constant, $V' = Q/C' = Q/(KC) = V/K$ . The potential decreases $K$ times (Correct).
Energy Stored: The initial energy is $U_i = Q^2/2C$ . The final energy is $U_f = Q^2/2C' = Q^2/2KC = U_i/K$ . The energy decreases $K$ times (Correct).
Change in Energy: $\Delta U = U_f - U_i = \frac{1}{2} \frac{Q^2}{KC} - \frac{1}{2} \frac{Q^2}{C} = \frac{Q^2}{2C} (\frac{1}{K} - 1)$ . Since $Q=CV$ , initial energy $\frac{Q^2}{2C} = \frac{1}{2}CV^2$ . Thus, $\Delta U = \frac{1}{2}CV^2 (\frac{1}{K} - 1)$ (Correct).

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