Question

Following operations can be performed on a capacitor:
X − connect the capacitor to a battery of emf ε.
Y − disconnect the battery.
Z − reconnect the battery with polarity reversed.
W − insert a dielectric slab in the capacitor.
(a) In XYZ (perform X, then Y, then Z) the stored electric energy remains unchanged and no thermal energy is developed.
(b) The charge appearing on the capacitor is greater after the action XWY than after the action XYZ.
(c) The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.
(d) The electric field in the capacitor after the action XW is the same as that after WX.

Answer 1

(b) The charge appearing on the capacitor is greater after the action XWY than after the action XYZ.
(c) The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.
(d) The electric field in the capacitor after the action XW is the same as that after WX.

Justification of option (b)

If the potential is held constant, that is, the battery remains attached to the circuit, then the charge on the capacitor increases by a factor of K on inserting a dielectric of a dielectric constant K between the plates of the capacitor.

Mathematically,
q = Kq_​0
Here, q₀ and q are the charges without dielectric and with dielectric, respectively.
The amount of charge stored does not depend upon the polarity of the plates.
Thus, the charge appearing on the capacitor is greater after the action XWY than after the action XYZ.

Justification of option (c)

Since the battery is disconnected before inserting a dielectric, the amount of charge remains constant, that is, q = q_​0, because after the battery is disconnected, the capacitor gets no source to store charge from. In other words, the capacitor is now an isolated system where the amount of charge is conserved and so is the energy U as $\frac{1}{2}$q $\epsilon$. Hence, inserting a dielectric after disconnecting the battery will not bring any change in the amount of charge stored in the capacitor. So, the energy stored in the capacitor will also not change after the action XYW.
However, during the action WXY, the amount of charge that will get stored in the capacitor will get increased by a factor of K, as the battery is disconnected after inserting a dielectric between the plates of the capacitor and the energy stored will also get multiplied by a factor of K.
Thus, the electric energy stored in the capacitor is greater after the action WXY than after the action XYW.

Justification of option (d)

The electric field between the plates E depends on the potential across the capacitor and the distance d between the plates of the capacitor.
Mathematically,
E = $\frac{\epsilon }{d}$
In either case, that is, during actions XW and WX, the potential remains the same, that is, $\epsilon$. Thus, the electric field E remains the same.

Denial of option (a)

During the action XYZ, the battery has to do extra work equivalent to $\frac{1}{2}$CV² to change the polarity of the plates of the capacitor. In other words, the total work to be done by the battery will be $\frac{1}{2}$CV² + $\frac{1}{2}$CV². This extra work done will be dissipated as heat energy. Thus, thermal energy is developed. However, the stored electric energy remains unchanged, that is, $\frac{1}{2}$CV².