JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Capacitance

A parallel plate capacitor has plate area

100 m^{2}

and plate separation of 10 m. The space between the plates is filled up to a thickness 5 m with material of dielectric constant of 10. The resultant capacitance of the system is 'x' pF.
The value of

ε_{0} = 8.85 \times 1 0^{- 12} F . m^{- 1}

The value of 'x' to the nearest integer is ______

Answer: 161

Solution

A slab of thickness

t

and dielectric constant

K

inside a gap

d

is equivalent to an air gap of

d - t + \frac{t}{K}

, because the two regions are capacitors in series. Hence

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

With

A = 100 m^{2}

d = 10

t = 5

m and

K = 10

, the effective separation is

10 - 5 + \frac{5}{10} = 5.5

m, so

C = \frac{8.85 \times 1 0 ^{- 12} \times 100}{5.5} = 1.61 \times 1 0^{- 10} F

The resultant capacitance is

161

pF.

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Integer answer

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