Question

Two parallel plate capacitors A and B have the same
separation $d=8.85 \times 10^{-4}m $ between the plates. The plate
areas of A and B are $0.04m^2 $ and $0.02m^2 $ respectively. A slab
of dielectric constant (relative permittivity) K = 9 has
dimensions such that it can exactly fill the space between the
plates of capacitor B.
(a) The dielectric slab is placed inside A as shown in figure
(i) A is then charged to a potential difference of 110 V.
Calculate the capacitance of A and the energy stored in it
(b) The battery is disconnected and then the dielectric slab is
removed from A. Find the work done by the external
agency in removing the slab from A.
(c) The same dielectric slab is now placed inside B, filling it
completely. The two capacitors A and B are then
connected as shown in figure (iii). Calculate the energy
stored in the system.

Answer 1

(a) Capacitor A is a combination of two capacitors $C_K \, and \, $
$C_0 $ in parallel. Hence,
$C_A=C_K+C_0= \frac {K \varepsilon_0A}{d}+ \frac {\varepsilon_0A}{d}=(K+1) \frac {\varepsilon_0A}{d} $
Here, $A=0.02 m^2. $ Substituting the values, we have
$C_A=(9+1) \frac {8.85 \times 10^{-12}(0.02)}{(8.85 \times 10^{-4})} $
$C_A=2.0 \times 10^{-9}F $
Energy stored in capacitor A, when connected with a
110 V battery is
$U_A= \frac {1}{2}C_AV^2= \frac {1}{2}(2 \times 10^{-9})(110)^2 $
$U_A=1.21 \times 10^{-5}J $
(b) Charge stored in the capacitor
$ q_A=C_AV=(2.0 \times 10^{-9})(110) $
$ q_A=2.2 \times 10^{-7}C $
Now, this charge remains constant even after battery is
disconnected. But when the slab is removed, capacitance
of A will get reduced. Let it be $C'_A $
$C'_A=\frac {\varepsilon_0(2A)}{d}= \frac {(8.85 \times 10^{-12})(0.04)}{8.85 \times 10^{-4}} $
$\, \, \, \, \, \, \, \, C'_A=0.4 \times 10^{-9}F $
Energy stored in this case would be
$U'_A= \frac {1(q_A)^2}{2 C'_A}= \frac {1(2.2 \times 10^{-7})^2}{2(0.4 \times 10^{-9})} $
$U'_A= 6.05 \times10^{-5}J>U_A $
Therefore, work done to remove the slab would be
$W=U'_A-U_A=(6.05 -1.21) \times 10^{-5} J $
or $ W=4.84 \times 10^{-5}J $
(c) Capacity of B when filled with dielectric is
$C_B= \frac {K\varepsilon_0A}{d}= \frac {(9)(8.85 \times 10^{-12})(0.02)}{(8.85 \times 10^{-4})} $
$C_B=1.8 \times 10^{-9}F $
These two capacitors are in parallel. Therefore, net
capacitance of the system is
$C=C'_A+C_B=(0.4 +1.8) \times10^{-9} F $
$\, \, \, \, \, \, \, \, \, \, C=2.2 \times 10^{-9} F $
Charge stored in the system is $q=q_A=2.2 \times 10^{-7} C $
Therefore, energy stored, $U= \frac {1}{2} \frac {q^2}{C} $
$U= \frac {1}{2} \frac {(2.2 \times 10^{-7})^2}{(2.2 \times 10^{-9})} \, or \, \, \, \, \, U=1.1 \times 10^{-5} J $