Question

A capacitor is filled with an insulator and a certain potential difference is applied to its plates. The energy stored in the capacitor is U. Now the capacitor is disconnected from the source and the insulator is pulled out of the capacitor. The work performed against the forces of the electric field in pulling out the insulator is 4U. Then the dielectric constant of the insulator is:
A.4.
B.8.
C.5.
D.3.

Answer 1

Hint: A capacitor is defined as an electrical component which stores energy electrostatically in an electric field. Potential difference is defined as the energy which is released in transfer of unit charge from one point to another point.
Formula used:
The formula of the energy stored in the insulator is given by,
$U = \dfrac{{{q^2}}}{{2C}}$
Where energy is U the charge is q and the capacitance is C.
The formula of the capacitance is given by,
$C = \dfrac{{A \cdot {\varepsilon _o}}}{d} \cdot k$
Where capacitance is C the area of the insulator is A the dielectric constant is k the distance between the plates is d and k is permittivity of dielectric constant.

Complete answer:
It is given in the problem that a capacitor is filled with an insulator and a certain potential difference is applied to its plates the energy stored in the capacitor is U when the capacitor is disconnected from the source and the insulator is pulled out of the capacitor the work performed against the forces of the electric field in pulling out the insulator is 4U and we need to find the value of dielectric constant.
The capacitor originally had the energy U and the work required to pull the insulator out is 4U so the final energy is equal to 5U.
$ \Rightarrow {U_f} = 4U + U$
$ \Rightarrow {U_f} = 5U$………eq. (1)
The capacitance of the insulator with an insulator,
$ \Rightarrow C = \dfrac{{A{\varepsilon _o}}}{d} \cdot k$………eq. (2)
The energy of the insulator is given by,
$ \Rightarrow U = \dfrac{{{q^2}}}{{2C}}$………eq. (3)
When we remove the insulator from the capacitor then the capacitance gets ${C_o}$ which will be to,
$ \Rightarrow {C_o} = \dfrac{{A \cdot {\varepsilon _o}}}{d}$………eq. (4)
And the stored energy will be,
$ \Rightarrow {U_f} = \dfrac{{{q^2}}}{{2{C_o}}}$………eq. (5)
Since,
$ \Rightarrow {U_f} = 5U$
Replacing the value of $U$and ${U_f}$ in the above equation from the equation (3) and equation (5) we get
$ \Rightarrow {U_f} = 5U$
$ \Rightarrow \left( {\dfrac{{{q^2}}}{{2{C_o}}}} \right) = 5\left( {\dfrac{{{q^2}}}{{2C}}} \right)$
$ \Rightarrow \left( {\dfrac{1}{{{C_o}}}} \right) = \left( {\dfrac{5}{C}} \right)$
$ \Rightarrow \dfrac{C}{{{C_o}}} = 5$
Replacing the value of $C$ and ${C_o}$ in the above relation from equation (2) and equation (4) we get.
$ \Rightarrow \dfrac{C}{{{C_o}}} = 5$
$ \Rightarrow \dfrac{{\left( {\dfrac{{A{\varepsilon _o}}}{d} \cdot k} \right)}}{{\left( {\dfrac{{A \cdot {\varepsilon _o}}}{d}} \right)}} = 5$
$ \Rightarrow k = 5$
The dielectric constant value is equal to $k = 5$.

The correct answer for this problem is option C.

Note:
It is advisable for the students to remember and learn the formula of the energy stored in the capacitor also the formula of the capacitance for a dielectric constant. The work done to pull out the insulator is 4U and the energy stored is already U so the total energy becomes 5U. The energy stored in the insulator was U and the work done to remove the insulator is 4U and therefore the final energy on the capacitor will be 5U and U students should not make a mistake here. Also in the formula of the capacitor after the removal of the insulator the term k disappears as the insulator is removed and there is nothing in between the plates of the capacitors.