Question

On increasing the plate separation of charged condenser its energy:
A. Remains unchanged
B. Decreases
C. Increases
D. None of these.

Answer 1

Hint: here, we will use the formula of the energy of a charged condenser to find the relation between the energy and the distance between the plates. This energy will be dependent on the capacitance and the potential between the plates.

Complete step by step answer:
We will calculate the energy in the charged condenser by using the following formula
$$U = \dfrac{1}{2}C{V^2}$$
Here, $C$ is the capacitance between the parallel plates and $V$ is the potential between the plates.
Now, capacitance is defined as the ability of an object to store an electric charge. It is also defined as the ratio of the change in an electric field to the change in the electric potential of a system. Therefore, the capacitance between the parallel plates is given by
$C = \dfrac{{A{\varepsilon _0}}}{d}$
Here, $A$ is the area of plates, ${\varepsilon _0}$ is the dielectric constant in free space, and $d$ is the distance between the two plates.
Now, the potential is defined as the ability of a system to do some work. The Potential between the two plates is given by
$V = \dfrac{{Qd}}{{A{\varepsilon _0}}}$
Here, $Q$ is the charge on the plates, $d$ is the distance between the plates, $A$ is the area of plates, and ${\varepsilon _0}$ is the dielectric constant in free space.
Now, putting the values of $C$ and $V$ in the formula of energy $U$ , we get
$U = \dfrac{1}{2}C{V^2}$
$ \Rightarrow \,U = \dfrac{1}{2} \times \dfrac{{A{\varepsilon _0}}}{d} \times {\left( {\dfrac{{Qd}}{{A{\varepsilon _0}}}} \right)^2}$
$ \therefore \,U = \dfrac{1}{2} \times \dfrac{{{Q^2}d}}{{A{\varepsilon _0}}}$
Now, from the above relation, we get
$U \propto \,d$
Therefore, the energy of the charged condenser will increase with an increase in the separation of the plates.

So, the correct answer is “Option C”.

Note:
As the plates are carrying opposite charges, therefore, the plates will attract each other.
Hence, to separate the plates the mechanical work requires should be more.
This mechanical work is in the form of energy.