Question

Following operations can be performed on a capacitor.
X - connect the capacitor to a battery of emf E.
Y - disconnect the battery.
Z - reconnect the battery with polarity reversed.
W - insert a dielectric slab in the capacitor.
Then which of the following is/are true

• 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 XYW
• 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

Answer: (B), (C), (D)

The correct options are
B The charge appearing on the capacitor is greater after the action XWY than after the action XYW
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 $WX$

(a)


Stored electrical energy remains the same (equal to $\frac{1}{2}C{E}^2$) through operations X, Y and Z. But there is thermal energy developed due to flow of charge during operation Z.

(b)


Charge on capacitor is greater after operation XWY than after operation XYW.


(c)

Since electrical energy stored is $\frac{q^2}{2C}$, the energy stored after operation WXY is greater than the energy stored after operation XYW.

(d)

Potential difference across the plates is the same in both cases (operation XW and WX). So field in the capacitor is also the same (electric field is given by $\frac{E}{d}$ where $E$ is the potential difference across the plates and $d$ is the distance between the plates.