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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?

medium

Electrostatic Potential and Capacitance

2015

physics

The energy stored in the capacitor decreases $K$ times.

The change in energy stored is $\frac{1}{2} CV^2 \left(\frac{1}{K} - 1 \right).$

The charge on the capacitor is not conserved.

The potential difference between the plates decreases $K$ times.

Explanation

To solve this problem, let's analyze the situation step-by-step. $\\$ Initially, the capacitor is charged by a cell of emf $V. \\$ The initial charge on the capacitor is given by: $\\ Q = CV \\$ The initial energy stored in the capacitor is: $\\ U_{initial} = \frac{1}{2} CV^2 \\$

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

Explanation

Our AI powered practice platform can help you achieve your doctor dream.

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?