Question

Calculate the area of the plates of a one farad parallel plate capacitor if separation between plates is $1mm$ and plates are in vacuum:
(A)\[18 \times {10^8}{m^2}\]
(B)\[0.3 \times {10^8}{m^2}\]
(C)\[1.3 \times {10^8}{m^2}\]
(D)\[1.13 \times {10^8}{m^2}\]

Answer 1

Hint: Parallel plate capacitor setup made up of two conductor plates connected by external potential difference to store charge by establishing electric field between plates. The capacitance of parallel plate capacitor is given by $C = \dfrac{{A{\varepsilon _0}}}{d}$ .

Formula used:
Capacitance of parallel plate is given by following formula:
$C = \dfrac{{k{\varepsilon _0}A}}{d}$where, $A$is area of plates of parallel plate capacitor, $k$ is permittivity of dielectric material${\varepsilon _0}$is permittivity of vacuum and $d$is separation between the plates.

Complete step by step answer:
Capacitance of capacitor is capacity to store charge. Every capacitor has capacitance according to its formation using different types of electrodes, placing them in different positions and using various types of dielectric material between them.
Capacitance of capacitor is given as,
$C = \dfrac{Q}{V}$
where, Q is electric charge, V is potential difference across plates and C is capacitance.
Capacitors have many types based on electrode position for example: parallel plate capacitor, based on use of dielectric materials, charge carrying capacity for example: fixed capacitor and variable capacitor.
In parallel plate capacitor electrodes are placed parallel to each other and connected to the battery to generate electric fields between them to store charge. Capacity of it can be enhanced by using dielectric material between plates. Capacitance of parallel plate is given by following formula:
$C = \dfrac{{k{\varepsilon _0}A}}{d}$where, $A$is area of plates of parallel plate capacitor, $k$ is permittivity of dielectric material${\varepsilon _0}$is permittivity of vacuum $(8.854 \times {10^{ - 12}}F{m^{ - 1}})$ and $d$is separation between the plates.
To calculate the area of the plates of a one farad parallel plate capacitor, we have separation between plates is $1mm$ and ${\varepsilon _0} = 8.854 \times {10^{ - 12}}F{m^{ - 1}}$.
As there is vacuum between two plates $k = 1$ .
So, $C = \dfrac{{{\varepsilon _0}A}}{d}$
\[A = \dfrac{{dC}}{{{\varepsilon _0}}}\]
By putting given values we get,
\[A = \dfrac{{1 \times {{10}^{ - 3}}m}}{{8.854 \times {{10}^{ - 12}}F{m^{ - 1}}}} \\
\therefore A= 1.13 \times {10^8}{m^2}\]

Hence, the correct answer is option (D) \[1.13 \times {10^8}{m^2}\].

Note: From this question we can think about how big a size of capacitor is required to have capacitance one farad. This size can be reduced by placing dielectric material between plates, but then also to get one farad capacitance is not achievable. So, generally capacitors in the market are of capacitance micro to Pico farad.