The surface charge density of a conducting sphere is `8.85xx10^(-10)C//m^(2)` and the electric field intensity at a distance of 4m from the centre of the sphere is `10^(-2)V//m`. The radius of the sphere, assuming the sphere to be in vacuum is

Options

A. 3 cm
B. 4 mm
C. 4 cm
D. 4 km

Answer

C

Explanation

`E=(sigmaR^(2))/(epsi_(0)kr^(2))`
`therefore" "R^(2)=(Eepsi_(0)kr^(2))/sigma=(10^(-2)xx8.85xx10^(-12)xx1xx16)/(8.85xx10^(-10))`
R = 4 cm