Question

A capacitor of capacitance $C$ is charged fully by connecting a battery of emf $E$. It is then disconnected from the battery. If the separation between its plates is doubled then how will the following change?
(i) Charge stored by the capacitor
(ii) Field strength between the plates
(iii) Energy stored by the capacitor
Justify your answer for each.

Answer 1

A capacitor is a device that stores large amounts of charge which is retained within the plates of the capacitor in the form of energy. The law of conservation of charges is applied in-order to find out if the changes made have any effect on the amount of charges stored by the capacitor. The equation for the electric field is applied to determine whether the field that is set up is dependent on the distance between the plates. The formula for the total energy stored by the capacitor is applied to obtain the factor by which the energy varies.

Complete step by step answer:
The capacitance of a parallel plate capacitor when there is a uniform electric field set-up between the plates of the capacitor is given by an equation:
${C_0} = \dfrac{{A{\varepsilon _0}}}{d}$ -----($1$)
Where, ${C_0}$ stands for original capacitance before the changes are made.
The question says that the distance between the plates are doubled and hence the relation obtained in equation ($1$) becomes:
$C = \dfrac{{A{\varepsilon _0}}}{{2d}}$ ------($2$)
Where, $C$ is the new capacitance that is obtained.

We know that a capacitor's role is to store large amounts of charge which means that the potential difference between its plates is also high. The main purpose or the main advantage of a capacitor is its ability to hold the energy produced due to the build-up of these charges. This is done by holding the energy in the electric field induced between its plates even after disconnecting the supply voltage.

(i) We can see from the relation obtained in equation ($1$) that the charge on the capacitor does not even depend on the distance $d$ between the plates. The explanation is as follows:
From the equation it is clearly seen that capacitance is inversely related to the distance between its plates. When the capacitor is charged and then disconnected from the voltage supply as given in the question the charge of the capacitor remains unchanged when there is a variation in the distance between its plates.

This is due to the concept of law of conservation of charges. It states that the net charge of the system must be constant. The equation for the charge on the capacitor is given by:
$Q = CV$
When the distance between the plates are increased or doubled in this case, then, the value of the capacitance reduces as these quantities are inversely proportional to each other (as per equation ($1$)). Due to this the potential difference between the plates increases in-order to compensate or balance out the decreased amount of capacitance. Hence the charge of the capacitor remains unchanged when the distance $d$ is doubled.

(ii) To find out if strength of the electric field changes with a change in $d$ we must first look if it directly depends on the separation distance. This is done by determining an equation for the electric field induced in between the plates. This equation is as follows:
${E_0} = \dfrac{\sigma }{{{\varepsilon _0}}}$ ------($3$)
This equation is said to be the net electric field between the two capacitor plates as electric field is set-up only in the inner regions, that is, the region between the plates and this is set-up due to the two charged plates of the capacitor (which are opposite in polarity and hence the electric field is induced from the positive to the negative plate).

It can be clearly seen that the electric field is independent of the distance between the plates of the capacitor from equation ($3$) and hence the electric field remains the same as well when there is a change in the $d$ value.

(iii) A capacitor charges up through charging cycles and hence there is some amount of work done in charging the initially uncharged plates of the capacitor and this is stored in the form of some potential energy. This charging is done when electric charges transfer from one plate to another. This potential energy of the capacitor is given by the equation (in terms of charge and capacitance):
${U_0} = \dfrac{{{Q^2}}}{{2C}}$ ----($4$)
Let this be the initial potential energy of the capacitor. We are aware that the capacitance is dependent on the distance of separation of the plates. We now determine how the capacitance changes with a change in distance by substituting the value of ($1$) in ($2$). From equations ($1$) and ($2$) we can see that:
$C = \dfrac{{{C_0}}}{2}$ ----($5$)

Hence we can see that the new value of capacitance reduces by half of the original capacitance. We next substitute this equation ($5$) in ($4$) to find out how the energy is affected.
$\Rightarrow U = \dfrac{{{Q^2}}}{{2\left[ {\dfrac{{{C_0}}}{2}} \right]}}$
We now do a little manipulation to write the new potential energy in terms of the general formula for potential energy and we get:
$\Rightarrow U = 2\left[ {\dfrac{{{Q^2}}}{{2C}}} \right]$
$\therefore U = 2{U_0}$
Hence it is clear from the above equation that potential energy also gets doubled as it is also inversely related to the capacitance which in-turn is inversely proportional to the distance between the plates.

Additional information: The capacitance of a capacitor is defined as the charge required to be supplied to either of the conductor plates of the capacitors in-order to increase the potential difference between them by a unit amount. The electrical capacitance is the measure of its ability to store energy.

This potential energy is stored when work is done in transferring charges between the plates or in other words is defined as the work done in charging the capacitor. The factors on which a parallel plate capacitor is dependent on is the area of the plates, the permittivity of the medium and the distance between the plates. For plates with finite area the field lines bend at the edges and this effect is known as fringing of the field.

Note: The electric field at the outer regions, that is, above and below the surface of the plates of the capacitor are not considered as it is zero due to the fact that the electric field due to the two plates cancel out each as they are the same in magnitude. In reality, the capacitors do not actually store charge. They transfer the charges and try to balance out the quantity of charges in both the plates’ in-order to conserve charge ensuring that the net charge of the two plates is always neutral or zero.