A parallel plate capacitor stores a charge Q at voltage V. Suppose the area of the 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

Hint: In this question we have to find the quantity that will change if we double the area and the distance between the plates. In this question we will have to use the formula of capacitance of a parallel plate capacitor and we will find out how capacitance will get affected if we change the area and distance between the plates.

Complete answer:
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,
\[C = \dfrac{{{\varepsilon _ 0 }A}}{d}\]
Where,
${\varepsilon _ 0}$ 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,
\[\Rightarrow C = \dfrac{{{\varepsilon _ 0}2A}}{{2d}}\]
\[\Rightarrow C = \dfrac{{{\varepsilon _ 0}A}}{d}\]
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, $E= \dfrac{V}{d}$
We can say that Electric field changed.
Hence, Energy density, $U_d$ = $\dfrac{1}{2} \varepsilon _0 E^2$ 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.

Note: From above explanation we have seen that there will be no effect on capacitance of the capacitor if we double both the area and the distance between the plates of the capacitor. But if we double the area only and the distance between the plates of the capacitor remains the same. Then the capacitance of the capacitor will be doubled. Similarly, if we double the distance between the plates of the capacitor but the area of the plates remain same then the capacitor will be halved of its previous value.