Question

The surface density of charge on the surface of a charged conductor in the air is 0.885 `(muC)/m^2`. The outward force per unit area of charged conductor is ______

`(ε_0 = 8.85 xx 10^-12 C^2/(N - m^2))`

Options

• `5 xx 10^-2 N/m^2`

• `5 xx 10^-3 N/m^2`

• `8.85 xx 10^-2 N/m^2`

• `4.425 xx 10^-2 N/m^2`

Answer 1

The surface density of charge on the surface of a charged conductor in the air is 0.885 `(muC)/m^2`. The outward force per unit area of charged conductor is `underline(4.425 xx 10^-2 N/m^2)`.

Explanation:

`sigma = 0.885 (muC)/(m^2) = 0.885 xx 10^-6` c/m²

F = `sigma^2/(2epsilon_0) = (0.885 xx 0.885 xx 10^-12)/(2 xx 8.85 xx 10^-12) = 0.04425 = 4.425 xx 10^-2 N/m^2`