Question

A parallel plate capacitor of capacitance c is connected to a battery and is charged to a potential difference v. Another capacitor of capacitance 2c is similarly charged to a potential difference 2v. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

Answer 1

Total energy of the system is 3cv²/2

Explanation:

Charge stored in the first capacitor = cv

Charge stored in the second capacitor = 2c x 2v = 4cv

When they are connected total charge of the system = 4cv - cv = 3cv

Because capacitance gets added in parallel connection,

Total capacitance of the parallel connection = c + 2c = 3c

we know that ,

charge = voltage x capacitance

q = cv

=> v = q/c

Hence voltage of the system = 3cv/3c = v

Now Energy of a capacitor is given by,

E = $\frac{1}{2} cv^2$

=> E = 1/2 x 3c x v² = 3cv²/2

Hence energy of the system is 3cv²/2

Answer 2

A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is then inserted in the space between the plates. Explain what changes, if any, occur in the values of: