JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Capacitance

A composite parallel plate capacitor is made up of two different dielectric materials with different thickness (

t_{1}

and

t_{2}

) as shown in figure. The two different dielectric material are separated by a conducting foil F. The voltage of the conducting foil is ______ V.

Answer: 60

Solution

The composite capacitor consists of two capacitors in series. The upper capacitor has dielectric constant

K_{1} = 3

and thickness

t_{1} = 1/2

mm, giving

C_{1} = \frac{A \cdot 3 ϵ _{0}}{1/2} = 6 A ϵ_{0}

. The lower capacitor has

K_{2} = 4

and thickness

t_{2} = 1

mm, giving

C_{2} = \frac{A \cdot 4 ϵ _{0}}{1} = 4 A ϵ_{0}

.

For series capacitors:

C_{e q} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{24 A ϵ _{0}}{10} = 2.4 A ϵ_{0}

.

Total charge:

q = C_{e q} \times 100 = 240 A ϵ_{0}

.

Voltage across

C_{2}

V_{2} = \frac{q}{C _{2}} = \frac{240 A ϵ _{0}}{4 A ϵ _{0}} = 60

V.

Since the foil is at the junction between the capacitors and the bottom plate is grounded, the foil voltage is 60 V.

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Integer answer