Question

A parallel-plate air capacitor of capacitance $C_0$ is connected to a cell of emf $V$ and then disconnected from it. A dielectric constant $K$, which can just fill the air gap of capacitor, is now inserted in it. Which of the following is incorrect?

• A. The potential difference between the plates decreases $K$ times.
• B. The energy stored in the capacitor decreases $K$ times.
• C. The change in energy is $\frac{1}{2}{C}_0{\epsilon }^2(K-1)$.
• D. The change in energy is $\frac{1}{2}{C}_0{\epsilon }^2(1-\frac{1}{k})$.

Answer 1

The correct option is A The potential difference between the plates decreases $K$ times.

As the capacitor is disconnected from cell so charge will remain constant .

here $Q=C_0ϵ$

Before inserting dielectric potential difference between the plates is $V_0=\frac{Qd}{A{ϵ}_0}$

after inserting, potential $V=\frac{Qd}{Ak{ϵ}_0}=\frac{V_0}{k}$

before inserting dielectric the energy is $U_0=\frac{Q^2}{2C_0}$

after inserting, the energy is $U=\frac{Q^2}{2k{C}_0}=\frac{U_0}{k}$

Change in energy $=U_0-U=\frac{Q^2}{2C_0}-\frac{Q^2}{2k{C}_0}=\frac{Q^2}{2C_0}(1-\frac{1}{k})=\frac{(C_0ϵ{)}^2}{2C_0}(1-\frac{1}{k})=\frac{1}{2}{C}_0{ϵ}^2(1-\frac{1}{k})$