What is the net electric flux through a closed surface surrounding an electric dipole ? Derive the expression for electric field intensity both inside and outside a uniformly charged spherical cell . What is the total charge enclosed by a closed surface , if the electric flux entering and leaving the surface are `20000 N// Cm^(2)` and `30000 N//Cm^(2)` respectively .
(Given `in_(0) = 8.85 xx 10^(-12) C^(2) N^(-1) m^(-2))`

Explanation

Electric Flux = `(q_("net"))/(in_(0)) = (0)/(in_(0)) = 0`
`therefore q_("net") = 0` for a dipole .
Field due to a uniformly charged thin spherical shell
Let `sigma ` be the uniform surface charge density of a thin spherical shell of radius R . The Gaussian surface will be a spherical surface centered at the centre of shell .

(i) At a point outside the shell (`r gt R`)
Since , E and S are in the same direction ,
`therefore phi_(E) = oint_(S) E * d S = (q)/(epsi_(0)) or E (4pi r^(2)) = (q)/(epsi_(0))`
`therefore E = (q)/(4pi epsi_(0) r^(2))`
Since `q = sigma xx 4 pi R^(2)`
`therefore E = (pi R^(2))/(epsi_(0) r^(2))`
Vectorially , `E = (sigma R^(2))/(epsi_0 r^(2)) r`
(ii) At a point inside the shell , `( r lt R`)

Here , the charge inside the Gaussian surface shell .
`therefore E(4 pi r^(2)) = 0`
`therefore " " E = 0 `
The variation of electric field intensity E with distance from the centre of a uniformly charged spherical shell is shown in figure .

Now `phi_("net") = phi_("in") - phi_("out") = 20000 - 30000 = - 10000 (N)/(cm^(2))`
So , `q_("net") = epsi_(0) xx phi_("net")`
`= 8.85 xx 10^(-12) xx - 10^(4)`
`= -8.85 xx 10^(-8) C`
Negative sign shows a negative excess change is present .