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 = $\dfrac{{{\varepsilon _o}A}}{d}$
Therefore \[Q = \dfrac{{{\varepsilon _o}A\left( V \right)}}{d}\]
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.