Surface density of charge on a sphere of radius R in terms of electric intensity E at a distance in free space is
`(epsi_(0)` = permittivity of free space)

Options

A. `epsi_(0) E((R)/(r))^(2)`
B. `(epsi_(0) ER)/(r^(2))`
C. `epsi_(0) E((r)/(R))^(2)`
D. `(epsi_(0) Er)/(R^(2))`

Answer

C

Explanation

According to Gauss's law ,
Total flux through Gaussain surface
`phi = oint E * ds oint_(s) E ds = E * 4 pi r^(2)`
If the charge enclosed by Gaussian surface is q .
according to Gauss's theorem
`E * 4 pi r^(2) = (q)/(epsi_(0)) implies E = (1)/(4 pi epsi_(0)) (q)/(r^(2)) " " ... (i)`
If `sigma ` is uniform surface charge density of spherical shell , then
`q = 4 pi R^(2) sigma " " ... (ii)`
Substituting Eq.(ii) in Eq.(i) . we get
` E = (sigma)/(epsi_(0)) (R^(2))/(r^(2)) therefore sigma = ( epsi_0 Er^(2))/(R^(2))`