Question

A parallel-plate capacitor has an area of $2.0c{m^2}$ and the plates are separated by $2.0mm$ . What is the capacitance?

Answer 1

Hint:As we know that, the capacitance of a parallel-plate capacitor is affected by the area of the plates, the distance between the plates, and the ability of the dielectric to support electrostatic forces, so we will go through the formula of capacitance in terms of area and the distance between plates.

Formula-used:- Capacitance of a parallel-plate capacitor: $C = {\varepsilon _ \circ }\dfrac{A}{d}$

Complete step by step answer:
Given that-
An area of a parallel-plate capacitor is $2.0c{m^2}$ or $2.0 \times {10^{ - 4}}{m^2}$ .
And the plates of the capacitor are separated by $2.0mm$ or $2.0 \times {10^{ - 3}}m$ .
So, as we know the formula to find the capacitance of a parallel-plate capacitor:
$\therefore C = {\varepsilon _ \circ }\dfrac{A}{d}$ …….(i)
where, $A$ is the area of each plate,
and $d$ is the distance which separate both the plates of the given capacitor,
${\varepsilon _ \circ }$ is a constant or vacuum permittivity, which is equals to $8.854 \times {10^{ - 12}}{C^2}.{N^{ - 1}}{m^{ - 2}}$ .
Now, we will put the given values of the facts in the eq(i) or the formula of capacitance:-
$
   \Rightarrow C = 8.854 \times {10^{ - 12}}.\dfrac{{2.0 \times {{10}^{ - 4}}}}{{2.0 \times {{10}^{ - 3}}}} \\
   \Rightarrow C = 8.854 \times {10^{ - 13}}F \\
 $
Hence, the required capacitance is $8.854 \times {10^{ - 13}}F$ .

Note:Increased capacitance can be achieved by either increasing plate area or decreasing plate gap. As a result, the capacitance of a parallel-plate capacitor can be enhanced by reducing the plate separation.