Question

A parallel plate capacitor filled with air is connected to a battery. A dielectric material of dielectrical constant is K=4 is added between the two plates while the capacitor remains connected to the battery. Determine how the following five quantities are affected by the insertion of the dielectric

a) capacitance

b) voltage across the capacitor's plates

c) charge on the plates of the capacitor

d) energy stored in the capacitor.

If a quantity changes specifiy whether it increases or decreases and bby what factor. Justify and show all work. Thank you.

Answer 1

Answer:

Let us go to the basics first.

(a)Let Capacitance = C (without dielectric)

On insertion of dielectric, new Capacitance C’ = KC = 4C (Answer a)

Thus, capacitance increases by a factor of 4.   (Answer a)

(b)Let voltage = V (without dielectric)

C = Q / V ……………Eqn.1

On insertion of dielectric, new Voltage = V’

We know that charge remains unaffected (i.e., constant) on addition of dielectric,

So C’ = Q / V’

=>4C = Q / V’

=>4(Q / V) = Q / V’ (From Eqn.1)

=>4 / V = 1 / V’

=>V’ = V/4 (Answer b)

Thus, voltage decreases and becomes 1/4^th.   (Answer b)

(C)As stated above, charge remains unaffected (i.e., constant) on addition of dielectric. (Answer c)

(d)Energy E (initial) = (1/2)CV²

E’ = (1/2)C’V’²

=>E’ = (1/2)(4C)(V/4)²

=>E’ = (1/2)(CV²/4)

Thus, E’ = E / 4   (Answer d)

Thus, energy decreases and becomes 1/4^th.   (Answer d)

Thanks!!!