A capacitor of capacitance 9 nF having dielectric slab of _r=2.4 , dielectric strength 20 MV m^ -1 , and potential difference =20 ~V . Calculate area of plates.

Correct answer: B. 4.2 × 10^ -4 ~m^2 Here, C=9 nF, _r=2.4, V=20 volt Dielectric strength =20 MV m^ -1 Let separation between plates =d aligned & E=(V)/(d) 20 × 10^6=(20)/(d) & d=10^ -6 ~m aligned Now, C=( _0 A _r)/(d) aligned & 9 × 10^ -9 = 8.85 × 10^ -12 × A × 2.4 10^ -6 & A= 9 × 10^ -15 8.85 × 2.4 × 10^ -12 =4.2 × 10^ -4 ~m^2 aligned

AIIMS 201925 May

Exam: AIIMS
Year: 2019
Date: 25 May
Subject: Physics
Chapter: Capacitance

AIIMS 2019 · 25 May · Physics · Capacitance

A capacitor of capacitance 9 nF having dielectric slab of _r=2.4 ,...

A capacitor of capacitance

9 \mathrm{nF}

having dielectric slab of

\varepsilon_r=2.4

, dielectric strength

20 \mathrm{MV} \mathrm{m}^{-1}

, and potential difference

=20 \mathrm{~V}

. Calculate area of plates.

Options

A $2.1 \times 10^{-4} \mathrm{~m}^2$
B $4.2 \times 10^{-4} \mathrm{~m}^2$
C $1.4 \times 10^{-4} \mathrm{~m}^2$
D $2.4 \times 10^{-4} \mathrm{~m}^2$

Correct answer

B. $4.2 \times 10^{-4} \mathrm{~m}^2$

Step-by-step solution

Here,

C=9 \mathrm{nF}, \varepsilon_r=2.4, V=20

volt Dielectric strength

=20 \mathrm{MV} \mathrm{m}^{-1}

Let separation between plates

=d

\begin{aligned} & E=\frac{V}{d} \Rightarrow 20 \times 10^6=\frac{20}{d} \\ & d=10^{-6} \mathrm{~m}\end{aligned}

Now, $C=\frac{\varepsilon_0 A \varepsilon_r}{d}

\begin{aligned} & 9 \times 10^{-9}=\frac{8.85 \times 10^{-12} \times A \times 2.4}{10^{-6}} \\ & A=\frac{9 \times 10^{-15}}{8.85 \times 2.4 \times 10^{-12}}=4.2 \times 10^{-4} \mathrm{~m}^2\end{aligned}$$$

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