A capacitor of capacitance 9 n F having dielectric slab of varepsilon_{ r }=2.4 dielectric strength

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2. Electric Potential and Capacitance

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A capacitor of capacitance $9 n F$ having dielectric slab of $\varepsilon_{ r }=2.4$ dielectric strength $20\, MV / m$ and $P.D. =20 \,V$ then area of plates is ....... $\times 10^{-4}\, m ^{2}$

A

$2.1$

B

$4.2$

C

$1.4$

D

$2.4$

(AIIMS-2019)

Solution

Let the separation between plants is $d.$ Therefore,

$E =\frac{ v }{ d }$

Substitute the values.

$E =\frac{ v }{ d }$

$20 \times 10^{6}=\frac{20}{ d }$

$d =10^{-6} m$

The expression of capacitance is given by,

$C =\frac{\varepsilon_{0} A \varepsilon_{ r }}{ d }$

Substitute the values.

$9 \times 10^{-9}=\frac{\left(8.85 \times 10^{-12}\right) A (2.4)}{10^{-6}}$

$A=\frac{\left(9 \times 10^{-15}\right)}{\left(8.85 \times 2.4 \times 10^{-12}\right)}$

$=4.2 \times 10^{-4} m ^{2}$

Standard 12

Physics

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