Gauss.s Theorem : It states that total electric flux through a closed surface is equal to `(1)/(epsilon_(0))` times the magnitude of charge enclosed by it. Electric field due to uniformly charged spherical shell : Consider a spherical shell of radius R. .O. is the centre of the spherical shell ad + Q charge be uniformly distributed over the surface of the shell.
Let `sigma` = Surface charge density
`therefore` Total charge on the shell = `Q = sigma (4 pi R^(2))`
(i) `vec(E )` at a point outside the shell :
Let P be a point outside the shell at a distance r (OP = r), where electric field intensity is to be determined. Draw a spherical Gaussian surface of radius r with centre at O. Consider a small area elements dS on this Gaussian surface
By symmetry consideration, the magnitude of `vec(E)` is same at all points on this surface and direction is radially outward.
Now flux through area dS is given by
`d phi_(E) = vec(E).vec(dS) = vec(E).hat(n) dS`
where `hat(n)` is a unit vector normal to area dS
or `d phi_(E) = E (1) = ds cos 0^(@)`
= E dS `" " [therefore hat(E) || hat(n), therefore theta = theta]`
Total flux through whole surface
`phi_(E) = oint E dS`
Using Gauss.s theorem,
`phi_(E) = (Q)/(epsilon_(0))`
From eqns, (2) and (3)
`E oint dS = (Q)/(epsilon_(0))`
`[ therefore "Area of Gaussian spherical surface" = 4 pi r^(2)]`
`E (4 pi r^(2)) = (Q)/(epsilon_(0))`
`rArr E = (Q)/(4 pi r^(2) epsilon_(0)) = (1)/(4 pi epsilon_(0)) (Q)/(r^(2))`
which is also the expression for electric field due to a point charge It means that for a point lying outside the uniformaly charged spherical shell, the whole charged spherical shell behaves like a point charge.
Using eqn. (1) in eqn. (4), we have
`E = (1)/(4 pi epsilon_(0)) (sigma(4 pi R^(2)))/(r^(2))`
`E = (sigma R^(2))/(epsilon_(0) r^(2))`
(ii) `vec(E)` at a point on the surface of shell (r = R) :
From eqn. (4), we get
`E = (Q)/(4 pi epsilon_(0)R^(2))`
But `Q = (4 pi R^(2)) sigma`
`therefore E = ((4 piR^(2)))/(4 pi epsilon_(0)R^(2)) = (sigma)/(epsilon_(0)) or E = (sigma)/(epsilon_(0))`
`vec(E)` at a point inside the shell :
It the point P lies inside the shell at a distance `r lt R`, then radius of Gaussian surface will be less than the radus of shell. Therefore, in thi case, Gaussian surface encloses no charge, i.e., q= 0 Using Gauss.s theorem,
`oint vec(E) . vec(dS) = (q)/(epsion_(0)) = 0`
`therefore E = 0`
i.e., electric field intensity at all points inside a uniformly charged shell is zero.
The variation of `vec(E)` with distance from the centre of uniformly charged spherical shell is shown in fig. given below.
