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 ....... $\times 10^{-4}\, m ^{2}$
AIIMS 2019, Medium
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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.
$9 \times 10^{-9}=\frac{\left(8.85 \times 10^{-12}\right) A (2.4)}{10^{-6}}$
$A=\frac{\left(9 \times 10^{-15}\right)}{\left(8.85 \times 2.4 \times 10^{-12}\right)}$
$=4.2 \times 10^{-4} m ^{2}$
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