Question

The separation between the plates of a charged isolated parallel plate capacitor is increased. Which of the following quantities will change (seperation is very small in comparision to the area of plates)?

• A. Charge on the capacitor
• B. Potential difference across the capacitor
• C. Electric field between the plates
• D. Capacitance of the capacitor

Answer 1

Answer: (B), (D)

Capacitance of the capacitor

Since the capacitor is isolated , no charge can enter or leave the plates and therefore total charge won't change.

So, option (a) is incorrect.

Capacitance of a parallel plate capacitor is given by $$C=\frac{{\epsilon }_{0}A}{d}$$Due to an increase in separation between the plates, the capacitance will decrease.

From $Q=CV$, evidently $V$ will increase.
So, option (b) and (d) are correct.

Electric field due parallel plate capacitor is given by, $$E=\frac{Q}{A{\epsilon }_0}$$ As $Q$ and $A$ are constant, so $E$ remains the same.

So, option (c) is incorrect.

Hence, options (b) and (c) are the correct alternatives.