Exam

JEE Main

Year

2018

Date

15 April

Shift

Evening Shift

Mode

Online

Subject

Physics

Chapter

Capacitance

JEE Main 2018 · 15 April · Evening Shift · Online · Physics · Capacitance

A parallel plate capacitor with area 200 ~cm^2 and separation...

A parallel plate capacitor with area $200 \mathrm{~cm}^2 and separation between the plates$1.5 \mathrm{~cm}$, is connected across a battery of emf V. If the force of attraction between the plates is$25 \times 10^{-6} \mathrm{~N}$, the value of$Vis approximately: \left.\left(\varepsilon_0=8.85 $\times$10^{-12} \frac{\mathrm{C}^2}{\text{ N.m }}\right)^2\right)

Step-by-step solution

Given area of Parallel plate capacitor, $A=200 \mathrm{ cm}^2$ Separation between the plates, $d=1.5 \mathrm{ cm}$ Force of attraction between the plates, $F=25 \times 10^{-6} \mathrm{ N}F=Q EF=\frac{Q^2}{2 A \epsilon_0} \quad$ (E due to parallel plate $\left.=\frac{\sigma}{2 \epsilon_0}=\frac{Q}{A 2 \epsilon_0}\right)$ But $Q=C V=\frac{\in_0 A(V)}{d}$
$\begin{gathered}\left.\therefore F=\frac{\left(\epsilon_0 A V^2\right)}{\mathrm{d}^2 \times 2 \mathrm{ A} \epsilon_0}\right)\\ =\frac{\left(\epsilon_0 A\right)^2 \times V^2}{d^2 \times 2 \times\left(A \epsilon_0\right)}=\frac{\left(\epsilon_0 A\right) \times V^2}{d^2 \times 2}\\ \text{ or, }25 \times10^{-6}=\frac{\left(8.85 \times 10^{-12}\right) \times\left(200 \times 10^{-4}\right) \times V^2}{2.25 \times 10^{-4} \times 2}\\ \Rightarrow V=\frac{25 \times 10^{-6} \times 2.25 \times 10^{-4} \times 2}{8.85 \times 10^{-12} \times 200 \times 10^{-4}} \approx250 \mathrm{ V} \end{gathered}$