(a) Using Gauss's law, derive an expression for the electric field intensity at any point outside a uniformly charged thin spherical shell of radius R and charge density `sigma C//m^(2)`. Draw the field lines when the charge density of the sphere is (i) positive, (ii) negative.
(b) A uniformly charged conducting sphere of 2.5 m in diameter has a surface charge density of 100 `mu C//m^(2)`. Calculate the
(i) charge on the sphere
(ii) total electric flux passing through the sphere.

Explanation

(a) Electric field intensity at any point outside a uniformly charged spherical shell :
Consider a thin spherical shell of radius R with centre O. Let charge +q is uniformly distributed over the surface of the shell.
Let P be any point on the sphere `S_(1)` with centre O and radius r.
According to Gauss's law

`underset(s)oint vec(E).vec(ds)=(q)/(in_(0)) rArr underset(E) oint vec(E).vec(n) ds=(q)/(in_(0))`
`:. E ointds=(q)/(in_(0))rArr E. 4pi r^(2)=(q)/(in_(0)) " " :. E=(1)/(4pi in_(0)).(q)/(r^(2))`
At any point on the surface of the shell, r=R
`:. e=(1)/(4 pi in_(0)).(q)/(R^(2))`
If `sigma` is charge density `:. q = 4 pi R^(2)sigma`
`:.E=(1)/(4 pi in_(0)) . ( 4 pi R^(2)sigma)/(R^(2))` Hence, `E=(sigma)/(in_(0))`
Electric field lines when the charged density of the sphere:
(i) Positive (ii) Negative

(b) Here diameter = 2.5 m `:. (2.5)/(2)=1.25 m`.
Charge density `sigma = 100 mu c // m^(2)=100 xx 10^(-6)=10^(-4)c//m^(2)`
`(i) q=4 pi R^(2) sigma = 4xx3.14 (1.25)^(2) xx 10^(-4)=19.625xx10^(-4)=1.96xx10^(-3) C`.
(ii) Total electrifix `phi_(E)=(q)/(in_(0))`
` :. phi_(E)=(1.96xx10^(-13))/(8.85xx10^(-12))=0.221xx10^(9)=2.2xx10^(8) Nm^(2)//C`.