Question

The separation between the plates of a charged parallel-plate capacitor is increased. Which of the following quantities will change?

• A. Charge on the capacitor
• B. Potential difference across the capacitor
• C. Energy density between the plates
• D. None of these

Answer 1

The correct options are
B Potential difference across the capacitor

Since the charge always remains conserved in al isolated system, it will remain the same

Now,

$V=\frac{Qd}{{\epsilon }_{0}A}$

Here, Q, A, and d are the charge, area, and distance between the plates respectively.

Thus as d increases, V increases.

Energy is given by

$E=\frac{qV}{2}$

So.it will also increase with increase in the value of the potential.

Energy density u, that is, energy stored per unit volume in the electric field is given by

$u=\frac{1}{2}{\epsilon }_0{E}^2$

So, u will remain constant with increase in distance between the plates.