Answer:
[tex]A = 0.2875 m^2[/tex]
Explanation:
As we know that
[tex]Q = 4.6 \mu C[/tex]
E = 1.8 kV/mm
now we know that electric field between the plated of capacitor is given as
[tex]E = \frac{\sigma}{\epsilon_0}[/tex]
now we will have
[tex]1.8 \times 10^6 = \frac{\sigma}{\epsilon_0}[/tex]
[tex]\sigma = (1.8 \times 10^6)(8.85 \times 10^{-12})[/tex]
[tex]\sigma = 1.6 \times 10^{-5} C/m^2[/tex]
now we have
[tex]\sigma = \frac{Q}{A}[/tex]
now we have area of the plates of capacitor
[tex]A = \frac{Q}{\sigma}[/tex]
[tex]A = \frac{4.6 \times 10^{-6}}{1.6 \times 10^{-5}}[/tex]
[tex]A = 0.2875 m^2[/tex]
