Question

A parallel plate capacitor stores a charge Q at a voltage V. Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change?

a) Capacitance
b) Charge
c) Voltage
d) Energy density​

Answer 1

Answer:

energy density is answer

Explanation:

Given, the charge stored by the capacitor is Q and voltage is V. Let the capacitance of the capacitor is C.

The capacitance of a parallel plate capacitor is given by,

C=ε0Ad

Where,

ε0 is the permittivity of free space,

d is the distance between the plates of the capacitor,

A is the area of the plates

If we double the area and distance between the plates then the capacitance will become like the equation given below,

⇒C=ε02A2d

⇒C=ε0Ad

Above equation shows that there will be no effect on the capacitance of the parallel plate capacitor.

If C is constant and Q is constant, V will also be constant. By the formula, E=Vd

We can say that Electric field changed.

Hence, Energy density, Ud = 12ε0E2 will change.

Result- Hence from above explanation, we have seen that there will be no effect on the capacitance if we double the area and the distance between the plates of the capacitor.

So, options D is correct.