Question

A capacitor has some dielectric between its plates, and the capacitor is connected to a $DC$ source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored, and the voltage will increase, decrease, or remain constant.

Answer 1

We know on removal of dielectric, capacitance will decrease by $\left(K\right)$

$C^{\prime }=\frac{C}{K}$

Energy stored in a capacitor

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

As charge remains same $U\alpha \frac{1}{C}$

Energy will increase.

Potential difference across the plate of capacitor

$V=\frac{Q}{C}$

$V\alpha \frac{1}{C}$

So, potential difference increases.

$V=Ed$

As $d$ remains constant, $E$ will also increase.

Final Answer: $C$ will decrease. Energy stored will increase. Electric field will increase. Charge stored will remain the same. $V$ will increase.