A parallel plate capacitor has plate area 100 m2 and plate separation… | JEE Main 2021 (18 Mar, Shift 1)

Q74JEE Main 2021 · 18 MAR · SHIFT 1Easy

Dielectrics

A parallel plate capacitor has plate area 100 m2 and plate separation of 10 m. The space between the plates is filled up to a thickness 5 m with a material of dielectric constant of 10. The resultant capacitance of the system is 'x' pF.

The value of $\varepsilon$ 0 = 8.85 $\times$ 10 $-$ 12 F.m ${}^{-1}$ .

The value of 'x' to the nearest integer is _____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 161

$C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$ with $A = 100\ \text{m}^2$ , $d = 10$ m, $t = 5$ m, $K = 10$ .
$d - t + \frac{t}{K} = 5 + 0.5 = 5.5$ m.
$C = \frac{8.85 \times 10^{-12} \times 100}{5.5} = 1.609 \times 10^{-10}$ F $\approx 161$ pF.
So $x = 161$ .

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