A capacitor havin

Given

Capacitance =100 μF

Initial battery voltage used = 24V

Second voltage used = 12V

a) Charges on the capacitor before and after the reconnection.

Before reconnection, the battery used is 24V, hence

We know charge present on a capacitor is given by

q = CV 1)

Where q = charge

C=capacitance

V=voltage

Substituting in 1)

Similarly, after connection of 12V battery –

When C = 100μf

V = 12V

Using equation 1)

b) Charge flown through the 12V battery.

C = 100, V = 12V

From 1),

Substituting the values,

c) Work is done by the battery, and its magnitude is as follows

We know

Where V = applied voltage across the capacitor

W = work done and

q = charge on the surface of the parallel plate capacitor

Which gives,

is the amount of work done on the battery.

d) Decrease in electrostatic field energy

Electrostatic field energy stored is given by –

, c = capacitance

v= voltage across capacitor

Initially, electrostatic field energy stored is given by -

Final Electrostatic field energy

Decrease in Electrostatic field energy

=

=

=

Substituting the values

Capacitance =100 μF

Initial battery voltage used = 24V

Second voltage used = 12V

=

e) Heat developed during the flow of charge after reconnection

After reconnection

C = 100μc, V = 12v

We know the energy stored, E in capacitor is given by

Where c is the capacitance and v is the applied voltage

Substituting values –

This is the amount of energy developed as heat when the charge flows through the capacitor.