The capacitance of a parallel plate capacitor with air as medium is 3 μ F. As a dielectric is introduced between the plates, the capacitance becomes 15 μ F. The permittivity of the medium in C2N-1 m-2 is

Question

The capacitance of a parallel plate capacitor with air as medium is 3 μ F. As a dielectric is introduced between the plates, the capacitance becomes 15 μ F. The permittivity of the medium in C2N-1 m-2 is (A) 8.15 × 10-11 (B) 0.44 × 10-10 (C) 15.2 × 10-12 (D) 1.6 × 10-14

Tardigrade Admin · Accepted Answer

The capacitance of air capacitor is given by

C_{0} = \frac{ε _{0} A}{d} = 3 μ F ....

(i)
When a dielectric of permittivity

ε_{r}

and dielectric constant

K

is introduced between the plates, then

C = \frac{K ε _{0} A}{d} = 15 μ F ...

(ii)
Dividing Eq. (ii) by Eg. (i), we get

\frac{C}{C _{0}} = \frac{\frac{K ε _{0} A}{d}}{\frac{ε _{0} A}{d}} = \frac{15}{3}

\Rightarrow K = 5

So, the permittivity of the medium

ε_{r} = ε_{0} K

= 8.85 \times 1 0^{- 12} \times 5

= 0.44 \times 1 0^{- 10}

Q. The capacitance of a parallel plate capacitor with air as medium is $3 μ F$ . As a dielectric is introduced between the plates, the capacitance becomes $15 μ F$ . The permittivity of the medium in $C^{2} N^{- 1} m^{- 2}$ is

$8.15 \times 1 0^{- 11}$

$0.44 \times 1 0^{- 10}$

$15.2 \times 1 0^{- 12}$

$1.6 \times 1 0^{- 14}$

Q. The capacitance of a parallel plate capacitor with air as medium is 3μF. As a dielectric is introduced between the plates, the capacitance becomes 15μF. The permittivity of the medium in C2N−1m−2 is

8.15×10−11

0.44×10−10

15.2×10−12

1.6×10−14

Q. The capacitance of a parallel plate capacitor with air as medium is $3 μ F$ . As a dielectric is introduced between the plates, the capacitance becomes $15 μ F$ . The permittivity of the medium in $C^{2} N^{- 1} m^{- 2}$ is

$8.15 \times 1 0^{- 11}$

$0.44 \times 1 0^{- 10}$

$15.2 \times 1 0^{- 12}$

$1.6 \times 1 0^{- 14}$