JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Capacitance

A parallel plate capacitor with plate area A and plate separation

d = 2 m

has a capacitance of

4 μ F

. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant

K = 3

(as shown in figure) will be :

Answer: (C)

Solution

The original capacitance is

C_{original} = \frac{A ε _{0}}{d} = 4 μ F

. When half the space is filled with a dielectric of constant

K = 3

, the system can be viewed as two capacitors in series:

C_{1} = \frac{A ε _{0}}{d /2} = 2 C

(air half) and

C_{2} = \frac{K A ε _{0}}{d /2} = 2 K C = 6 C

(dielectric half). The equivalent capacitance is

C_{new} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{2 C \cdot 6 C}{2 C + 6 C} = \frac{3 C}{2} = \frac{3}{2} \times 4 = 6 μ F

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Single correct option