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

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.

Step-by-step 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 = (\sigma )/(\varepsilon _(0))`

`{\sigma }= E * {\varepsilon _(0)}`

`{\sigma }= 8.20 * 10^(5) * {8.854 * 10^(-12)`

`{\sigma }= 7.26 * 10^(-6) C/m^(2)`

Charge, q = surface charge density x area

`q = 7.26 * 10^(-6) * 25 * 10^(-4)`

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