Question

Two parallel plate capacitors X and Y are connected in series to a 6 V battery. They have the same plate area and same plate separation but capacitor X has air between its plates, whereas capacitor Y contains a material of dielectric constant 4.

1. Calculate the capacitances of X and Y, if the equivalent capacitance of the combination of X and Y is 4 µF.
2. Calculate the potential difference across the plates of X and Y.

Answer 1

a. For a parallel plate capacitor:

C = `(epsilon A)/d`

If the dielectric constant K is introduced:

C' = KC

For capacitors in series:

`1/C_"eq" = 1/C_1 + 1/C_2`

Relation between capacitances:

Since they have the same geometry,

Capacitor X (air): C_X = C    ...(i)

Capacitor Y (dielectric K = 4): C_Y = 4C    ...(ii)

`1/C_"eq" = 1/C + 1/(4C)`

`1/4 = 5/(4C)`    ...[C_eq = 4 µF]

C = 5 µF

C_X = C    ...[From equation (i)]

∴ C_X = 5 µF

C_Y = 4C    ...[From equation (ii)]

= 4 × 5 µF

= 20 µF

b. In a series circuit, the charge on each capacitor is the same, and the voltage is inversely proportional to the capacitance.

Total voltage (V) = 6 V

Using VX : VY:

`1/C_X : 1/C_Y`

= `1/5 : 1/20`

= 4 : 1

V_X = `4/5 xx 6`

= 4.8 V

V_Y = `1/5 xx 6`

= 1.2 V