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

Since both capacitors have the same plate area A and plate separation d, their capacitance depends only on the dielectric constant K.

Let the capacitance of X (with air, K = 1) be C_X = C    ...(i)

Let the capacitance of X (with dielectric, K = 4) be C_Y = 4C    ...(ii)

a. For two capacitors connected in series, the equivalent capacitance C_eq is calculated using the formula:

`1/C_"eq" = 1/C_X + 1/C_Y`

Substituting the given equivalent capacitance (4 µF) and the relationship between C_X and C_Y:

`1/4 = 1/C + 1/(4 C)`

`1/4 = (4 + 1)/(4 C)`

`1/4 = 5/(4 C)`

4C = 5 × 4

4C = 20

C = `20/4`

C = 5 µF

Thus,

C_X = 5 µF    ...[From equation (i)]

C_Y = 4 × 5    ...[From equation (ii)]

C_Y = 20 µF

b. In a series combination, the charge Q on each capacitor is the same.

Total charge (Q) = C_eq × V_total

= 4 µF × 6 V

= 24 µC

Potential difference across X: 

V_X = `Q/V_X`

= `24/5`

= 4.8 V

Potential difference across Y:

V_Y = `Q/V_Y`

= `24/20`

= 1.2 V