A parallel plate capacitor with area 200 ,cm^2 and separation bet…

MCQ

A parallel plate capacitor with area $200\,cm^2$ and separation between the plates $1.5\,cm$, is connected across a battery of $emf$ $V$. If the force of attraction between the plates is $25\times10^{-6}\,N$, the value of $V$ is approximately........$V$ $\left( {{\varepsilon _0} = 8.85 \times {{10}^{ - 12}}\,\frac{{{C^2}}}{{N{m^2}}}} \right)$

A
$150$
B
$100$
✓
$250$
D
$300$

✓

Answer

Correct option: C.

$250$

c
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, $F=$ $25 \times 10^{-6}\, \mathrm{N}$

$F=Q E$

$F=\frac{Q^{2}}{2 A \in_{0}} \quad$ ($E$ due to parallel plate

$=\frac{\sigma}{2 \in_{0}}=\frac{Q}{A 2 \in_{0}})$

But $Q=C V=\frac{\in_{0} A(V)}{d}$

$\therefore \,F = \,\frac{{({\in_0}A{V^2})}}{{{{\text{d}}^2} \times 2{\text{A}}{\in_0}}}$

$ = \frac{{{{\left( {{\in_0}A} \right)}^2} \times {V^2}}}{{{d^2} \times 2 \times \left( {A{\in_0}} \right)}} = \frac{{\left( {{\in_0}A} \right) \times {V^2}}}{{{d^2} \times 2}}$

or, $25 \times {10^{ - 6}} = \frac{{\left( {8.85 \times {{10}^{ - 12}}} \right) \times \left( {200 \times {{10}^{ - 4}}} \right) \times {V^2}}}{{2.25 \times {{10}^{ - 4}} \times 2}}$

$ \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}}}}$ $ \approx 250\,V$