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) $692pJ$

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
$U = \dfrac{1}{2}C{V^2}$
Here C is the capacitance of the conductor and its formula is
$C = {\varepsilon _0}\dfrac{A}{d}$
Where ${\varepsilon _0}$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
$\Rightarrow {U_i} = \dfrac{1}{2}C{V^2}$---(i)
Substituting the value of C and V in equation (i),
$\Rightarrow {U_i} = \dfrac{1}{2} \times 12 \times {(10)^2}$
$\Rightarrow {U_i} = \dfrac{{1200}}{2}$
$\Rightarrow {U_i} = 600pJ$
Step III:
When the dielectric slab with constant K is inserted between the slab, then the capacitance will be
${C^{'}} = K\dfrac{{{\varepsilon _0}A}}{d}$
Or ${C^{'}} = KC$
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
$\Rightarrow {U_f} = \dfrac{1}{2}\dfrac{{{q^2}}}{{{C^{'}}}}$
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,
$\Rightarrow {U_f} = \dfrac{1}{2}.\dfrac{{120 \times 120 \times {{10}^{ - 24}}}}{{6.5 \times 12 \times {{10}^{ - 12}}}}$
$\Rightarrow {U_f} = 92pJ$
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.