Question

A parallel plate capacitor having capacitance 12pF is charged by a battery to a potential difference of 10V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates the work done by the capacitor on the slab is
A) $508pJ$
B) $560pJ$
C) $600pJ$
D) $692pJ$

Answer 1

Hint: A capacitor is a device that stores energy in the form of an electric field. A capacitor consists of two metallic plates that are separated with the help of an insulator. The energy that is stored by a capacitor is known as its capacitance.

Complete step by step answer:
Step I:
The energy stored in the capacitor is written by the equation
$U={\displaystyle \frac{1}{2}}C{V}^2$
Here C is the capacitance of the conductor and its formula is
$C={\epsilon }_0{\displaystyle \frac{A}{d}}$
Where ${\epsilon }_0$is the permittivity of the dielectric and its value is $6.5$
A is the area and d is the distance between the plates
Step II:
The energy stored by the capacitor before inserting the slab is given by
$\Rightarrow U_i={\displaystyle \frac{1}{2}}C{V}^2$---(i)
Substituting the value of C and V in equation (i),
$\Rightarrow U_i={\displaystyle \frac{1}{2}}\times 12\times (10{)}^2$
$\Rightarrow U_i={\displaystyle \frac{1200}{2}}$
$\Rightarrow U_i=600pJ$
Step III:
When the dielectric slab with constant K is inserted between the slab, then the capacitance will be
$C^{{}^{\prime }}=K{\displaystyle \frac{{\epsilon }_{0}A}{d}}$
Or $C^{{}^{\prime }}=KC$
Step IV:
When the battery is disconnected, the charge remains the same in case of a capacitor. Therefore, final energy stored by the capacitor is given by
$\Rightarrow U_f={\displaystyle \frac{1}{2}}{\displaystyle \frac{q^2}{C^{{}^{\prime }}}}$
Step V:
Since $q=CV$
$\Rightarrow q=12\times 10$
$\Rightarrow q=120pF$
$1pF={10}^{-12}F$
$\Rightarrow 120pF=120\times {10}^{-12}F$
Step VI:
Substituting all the values in and solving,
$\Rightarrow U_f={\displaystyle \frac{1}{2}}.{\displaystyle \frac{120\times 120\times {10}^{-24}}{6.5\times 12\times {10}^{-12}}}$
$\Rightarrow U_f=92pJ$
Step VII:
Work done by the capacitor is given by
$W+U_f=U_i$
$\Rightarrow W=U_i-U_f$
On substituting the corresponding values, we get
$\Rightarrow W=600-92$
On simplifications,
$\Rightarrow W=508pJ$

$\therefore$ The work done by the capacitor is $508pJ$. Hence, Option A is the right answer.

Note:
It is possible to store more energy in the capacitor. The energy stored can be increased if more plates are connected together in place of a single plate. This will increase the surface area of the capacitor. Also, the distance between the plates should be reduced in order to increase the capacitance.