Answer:
q = 1.815 \times 10^{-8} C
Charge on one plate is positive in nature and on the other plate it is negative in nature.
Explanation:
E = 8.20 x 10^5 V/m, A = 25 cm^2, d = 22.45 mm
According to the Gauss's theorem in electrostatics
The electric field between the two plates
[tex]E = \frac{\sigma }{\varepsilon _{0}}[/tex]
[tex]{\sigma }= E \times {\varepsilon _{0}}[/tex]
[tex]{\sigma }= 8.20 \times 10^{5} \times {8.854 \times 10^{-12}[/tex]
[tex]{\sigma }= 7.26 \times 10^{-6} C/m^{2}[/tex]
Charge, q = surface charge density x area
[tex]q = 7.26 \times 10^{-6} \times 25 \times 10^{-4}[/tex]
q = 1.815 \times 10^{-8} C
