JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Capacitance

A parallel plate capacitor is formed by two plates each of area

30 π cm^{2}

separated by 1 mm. A material of dielectric strength

3.6 \times 1 0^{7} Vm^{- 1}

is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is

7 \times 1 0^{- 6} C

, the value of dielectric constant of the material is :

{

Use :

\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} Nm^{2} C^{- 2}}

Answer: (D)

Solution

At dielectric breakdown, the electric field reaches its maximum value

E = 3.6 \times 1 0^{7}

V/m.

For a parallel plate capacitor:

E = \frac{σ}{K ϵ _{0}} = \frac{Q}{K A ϵ _{0}}

Rearranging:

K = \frac{Q}{A ϵ _{0} E}

Substituting values:

K = \frac{7 \times 1 0 ^{- 6}}{30 π \times 1 0 ^{- 4} \times \frac{1}{4 π \times 9 \times 1 0 ^{9}} \times 3.6 \times 1 0 ^{7}}

K = \frac{7 \times 1 0 ^{- 6} \times 4 π \times 9 \times 1 0 ^{9}}{30 π \times 1 0 ^{- 4} \times 3.6 \times 1 0 ^{7}} = \frac{7 \times 36}{30 \times 3.6} = \frac{252}{108} = 2.33

The answer is (D) 2.33.

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Single correct option