A parallel plate capacitor has plate area 50 cm 2 and plate separation 3 mm . The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 4 . The capacita

Question

A parallel plate capacitor has plate area 50 cm 2 and plate separation 3 mm . The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 4 . The capacitance becomes ( ϵ 0 ​ = permittivity of free space)

TestHub · Accepted Answer

The capacitance of a parallel plate capacitor partially filled with a dielectric is given by:

$C = \frac{ϵ _{0} A}{d - t + \frac{t}{K}}$

Given:
$A = 50 cm^{2} = 50 \times 1 0^{- 4} m^{2}$
$d = 3 mm = 3 \times 1 0^{- 3} m$
$t = 1 mm = 1 \times 1 0^{- 3} m$
$K = 4$

Substituting the values:

$C = \frac{ϵ _{0} \times 50 \times 1 0 ^{- 4}}{3 \times 1 0 ^{- 3} - 1 \times 1 0 ^{- 3} + \frac{1 \times 1 0 ^{- 3}}{4}}$

$C = \frac{ϵ _{0} \times 50 \times 1 0 ^{- 4}}{( 2 + \frac{1}{4} ) \times 1 0 ^{- 3}}$

$C = \frac{ϵ _{0} \times 50 \times 1 0 ^{- 4}}{\frac{9}{4} \times 1 0 ^{- 3}}$

$C = \frac{200 \times 1 0 ^{- 4}}{9 \times 1 0 ^{- 3}} ϵ_{0}$

$C = \frac{20 ϵ _{0}}{9}$

Answer: $\frac{20 ϵ _{0}}{9}$

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