Charge on spherical conductors

amazing problem, thanks

Consider two equal yet opposite charges placed on either side of the sphere which produce an electric charge at the centre , as the distance between the charges tends to infinity, they tend to produce a near horizontal electric field at the centre

Let us make some quick observations - The potential due to the external system is 0 at the centre, hence we will take the origin as our refference point for the potential due to this field (not necessary ) - The sphere being conducting will gather surface charge over it so as to maintain its equipotential surface

Now from image theory we know that the Induced dipole will have each charge as

$\frac {QR}{L}$ at a distance $\frac {R^2}{L}$ from centre each

and the dipole moment ?

their product times 2

= $\frac {QR^3}{L^2}$

now the electric field at the centre (or around the centre) by external charges is simply $E_o = \frac {kQ}{L^2}$

using both equations we claim

$\frac {ER^3}{k}$ = dipole moment induced

now, comes the part which i think is most awesome

, The potential at any point in space outside or just on sphere is the sum of the potential of image charges and the sum of the potential due to external charges with appropriate refference points,

We have to link the potential with surface charge density

First , Let us find the potential

it is

$\frac {kpcos(\theta)}{r^2} for \quad the \quad dipole (w.r.t \infty \quad and \quad -rEcos(\theta$ for external field (w.r.t 0 ) )

which can be easily written down as

(the minus signs comes from the opposite orientation of dipoles)

$E\left( \frac { { r }^{ 3 } }{ R^2 } cos(\theta )-cos(\theta ) \right)$

now we use gauss law at the surface to claim that

${ E }_{ just\quad out }-{ E }_{ just\quad in }=\frac { \sigma }{ \varepsilon } \approx \quad \frac { \delta V }{ \delta r } { | }_{ at\quad r=R }=\quad -3Ecos(\theta)$

(Because electric field just inside is 0)

(E represents the normal component of field, the one that contributes to flux)

now we simply integrate it all over one surface using

$q=\int { \sigma dA }$ over the needed area

to get answer as 3

(i think every one can do that part, so i skipped)

Quite a good problem.What one could think of this is analysing a similar situation in which the field is constant everywhere.So we consider a charge distribution on the surface of a sphere given by $sigma=kcos@$ .This charge distribution produces a constant field everywhere inside given by $E=k/3£$ (This could be analysed with the case of 2 overlapping oppositely charged clouds just to completely overlap).So note that the net field inside must be zero so that the sphere produces a constant field everywhere.So we have a exactly same situation.The distribution of charge on the sphere is $sigma=kcos@$ where @ is the angle made with the X axis and $E=k/3£$ So now its just a matter of integration......Or you may refer to IRODOV problem 3.17......The integral gives $Q=3ER^(2)/4k$ as the required answer....