A spherical conductor has a total surface area of 4 10⁻² m ^2 and a uniform surface charge density of 2 C m ⁻² . What is the magnitude of the electric field at a distance of 0.3 m from the centre of the sphere? ( 1 4 ₀ = 9 10^9 N m ^2 C ⁻² )

Step-by-step solution

First, find the total charge Q on the spherical conductor by multiplying the surface charge density by the surface area A : Q = A = ( 2 10⁻⁶ C m ⁻² ) (4 10⁻² m ^2 ) = 8 10⁻⁸ C The radius of the sphere R can be found from the surface area A = 4 R^2 : 4 R^2 = 4 10⁻² R = 0.1 m Since the given distance r = 0.3 m is greater than R , the point lies outside the sphere. The electric field is calculated assuming the entire charge is concentrated at the centre: E = 1 4 ₀ Q r^2 E = 9 10^9 8 10⁻⁸ (0.3)^2 E = 720 0.09 = 8000 N