Question

An electric field of 8.20 ✕ 105 V/m is desired between two parallel plates, each of area 25.0 cm2 and separated by 2.45 mm. There's no dielectric. What charge must be on each plate?

Answer 1

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

$E = \frac{\sigma }{\varepsilon _{0}}$

${\sigma }= E \times {\varepsilon _{0}}$

${\sigma }= 8.20 \times 10^{5} \times {8.854 \times 10^{-12}$

${\sigma }= 7.26 \times 10^{-6} C/m^{2}$

Charge, q = surface charge density x area

$q = 7.26 \times 10^{-6} \times 25 \times 10^{-4}$

q = 1.815 \times 10^{-8} C