Question

A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following will remain the same?

• A. The electric field in the capacitor
• B. The charge on the capacitor
• C. The potential difference between the plates
• D. The stored energy in the capacitor

Answer 1

The correct option is B The charge on the capacitor

Since capacitor is isolated. It means battery is not connected. So there is no way for the charge to inflow or outflow.
$\therefore Q=\text{constant}$

And we know that insertion of dielectric slab in PPC increases the capacitance of PPC by $K$ times.

$\therefore C^{\prime }=KC$

From formula,

$Q=CV$

$\because C\uparrow \therefore V\downarrow$

And, energy stored in capacitor

$U=\frac{Q^2}{2C}$

Since, $Q$ is constant and $C$ increases, so stored energy will decrease.

So, finally, we can say that only charge on the capacitor remains constant.

Therefore, option $\left(B\right)$ is correct.