Question

A parallel plate capacitor with area 200 $c{m^2}$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 $25 \times {10^{ - 6}}$N, the value of V is approximately: $\left( {{\varepsilon _o} = 8.85 \times {{10}^{ - 12}}\dfrac{{{C^2}}}{{N.{m^2}}}} \right)$
(A). 150V
(B). 100V
(C). 250V
(D). 300V

Answer 1

Hint: To attempt this question one must have prior knowledge about the capacitors, use the formula of electric field between two parallel plates of capacitor i.e. $E = \dfrac{\sigma }{{2{\varepsilon _o}}}$ to find the relation of force and potential difference across Capacitors plates (V), use these details to get closer towards the solution to the problem.

Complete step-by-step answer:
According to the given information the area of parallel capacitors is given 200 $c{m^2}$, distance between the parallel plates is 1.5 cm and the force of attraction generating between the parallel plates of capacitors is $25 \times {10^{ - 6}}$N
Since we know that the electric field generated between the parallel plates of the capacitor is calculated by formula $E = \dfrac{\sigma }{{2{\varepsilon _o}}}$ here $\sigma$ is the surface charge density here $\sigma = \dfrac{Q}{A}$ and ${\varepsilon _o}$is the absolute permittivity of material used
So $E = \dfrac{\sigma }{{2{\varepsilon _o}}}$ also can be written as $E = \dfrac{Q}{{2A{\varepsilon _o}}}$ (equation 1)
We know that formula to calculate the force between the parallel plates of capacitors is $F = \dfrac{1}{2}QE$
So the net force between the plates = force by +ve plate + force by –ve plate
Thus ${F_{net}} = {F_ + } + {F_ - }$
Substituting the given values in the above equation, we get
\Rightarrow $$${F_{net}} = \dfrac{1}{2}QE + \dfrac{1}{2}QE$$ \Rightarrow {F_{net}} = QE$$ Substituting the value of E form equation 1 $${F_{net}} = Q \times \dfrac{Q}{{2A{\varepsilon _o}}}$$ $ \Rightarrow {F_{net}} = \dfrac{{{Q^2}}}{{2A{\varepsilon _o}}}Taking this equation as equation 2 Since we know that Q = CV here C is the capacitance of the capacitor and V is the potential of capacitor also since C = $\dfrac{{{\varepsilon _o}A}}{d}$ ThereforeQ = \dfrac{{{\varepsilon o}A\left( V \right)}}{d}$Substituting the value of Q in equation 2 We get${F{net}} = \dfrac{{{{\left( {{\varepsilon _o}AV} \right)}^2}}}{{{d^2} \times 2A{\varepsilon _o}}} $ \Rightarrow $$${F_{net}} = \dfrac{{{{\left( {{\varepsilon _o}A} \right)}^2} \times {V^2}}}{{{d^2} \times 2\left( {A{\varepsilon _o}} \right)}}
\Rightarrow $$${F_{net}} = \dfrac{{\left( {{\varepsilon _o}A} \right) \times {V^2}}}{{{d^2} \times 2}}$$ Now substituting the given values in the above equation to find the value of V $$25 \times {10^{ - 6}} = \dfrac{{\left( {8.85 \times {{10}^{ - 12}}} \right) \times \left( {200 \times {{10}^{ - 4}}} \right) \times {V^2}}}{{{{\left( {1.5 \times {{10}^{ - 2}}} \right)}^2} \times 2}}$$ \Rightarrow {V^2} = \dfrac{{25 \times {{10}^{ - 6}} \times 2.25 \times {{10}^{ - 4}} \times 2}}{{\left( {8.85 \times {{10}^{ - 12}}} \right) \times \left( {200 \times {{10}^{ - 4}}} \right)}}$$ $ \Rightarrow V = \sqrt {0.063559 \times {{10}^6}} $$
$\Rightarrow$V = 252.09 V $\approx$ 250 V
So the value of V is 250 V
Hence option C is the correct option.

Note: In the above question we came across the term parallel plate capacitor which can be defined as the pair of 2 parallel electrodes which helps to store some finite amount of energy before the breakdown of dielectric occurs here when a pair of parallel plates are connected with a battery which transfer the energy to plates which stores the energy thus the generation of electric field takes place between the parallel plates of capacitors.