Question

An air filled parallel plate capacitor is to be constructed which can store \[12\,\mu C\] of charge when operated at \[1200\,V\]. What can be the minimum plate area of the capacitor? The dielectric strength of air is \[3 \times {10^6}\,V{m^{ - 1}}\].
A. \[0.3\,{m^2}\]
B. \[0.45\,{m^2}\]
C. \[0.6\,{m^2}\]
D. \[0.75\,{m^2}\]

Answer 1

Hint: Use the formula for charge stored in a capacitor. Use the formula for the capacitance of a parallel plate capacitor to find the minimum plate area of the capacitor.The capacitor is a device in which electrical energy can be stored. It is an arrangement of two-conductors generally carrying charges of equal magnitudes and opposite sign and separated by an insulating medium.

Formula used:
The Formula for total charge in a capacitor is given by,
\[Q = CV\]
where, \[Q\] is the charge in the capacitor, \[C\] is the capacitance of the capacitor and\[V\] is the voltage applied across the capacitor.
Capacitance of parallel plate capacitor is given by,
\[C = \dfrac{{k{\varepsilon _0}A}}{d}\]
where, \[k\] is the dielectric constant, \[{\varepsilon _0}\] is the permittivity of the free space, \[A\]is the area of the parallel plates and\[d\] is the gap between the plates.
Electric field is given by,
\[E = \dfrac{V}{d}\]
where,\[V\] is the voltage applied and \[d\] is the distance between the terminals.

Complete step by step answer:
Here we have a parallel plate capacitor which can store \[Q = 12\mu C\] charge. Now, we know that capacitance of parallel plate capacitor is given by,
\[C = \dfrac{{k{\varepsilon _0}A}}{d}\]
Now, we also know that charge in a parallel plate capacitor can be written as,
\[Q = CV\]
where \[V\] is the voltage applied.
So, putting the value we have, \[Q = \dfrac{{k{\varepsilon _0}A}}{d}V\]
Now, we know that the strength of the electric field is given by, \[E = \dfrac{V}{d}\].
So, putting in the equation we have, \[Q = k{\varepsilon _0}AE\]
Here, we have, charge store \[Q = 12\mu C = 12 \times {10^{ - 6}}C\] strength of the field \[E = 3 \times {10^6}V{m^{ - 1}}\] dielectric constant of air is \[k = 1\], \[{\varepsilon _0} = 8.854 \times {10^{ - 12}}\].
So, putting the values we have,
\[12 \times {10^{ - 6}} = A \times 1 \times 8.854 \times {10^{ - 12}} \times 3 \times {10^6}\]
\[\Rightarrow A = \dfrac{{12}}{{3 \times 8.854}}\]
\[\therefore A = 0.45\]
So, the minimum plate area of the capacitor must be \[0.45{m^2}\] to operate at \[1200V\].

Hence, option B is the correct answer.

Note: The area of the parallel plate capacitor depends on the dielectric material used between the plates. If the material used between the plates is not air then the capacitance of the capacitor will be different for a fixed geometry and to operate in the \[1200V\] area of the plate will not be \[0.45{m^2}\]. It will be smaller than that. Since, the dielectric constant of any material is greater than vacuum or air.