Question
Tue May 01, 2012 By: Shubhendu Mishra
 

HOW TO DERIVE AN EXPRESSION FOR ELECTRIC FIELD INTENSITY AT A POINT OUTSIDE A CHARGED CONDUCTING SPHERE

Expert Reply
Thu May 10, 2012
Let ? be the uniform surface charge density of a thin spherical shell of
radius R (Fig. 1.31). The situation has obvious spherical symmetry. The
field at any point P, outside or inside, can depend only on r (the radial
distance from the centre of the shell to the point) and must be radial (i.e.,
along the radius vector).
(i) Field outside the shell: Consider a point P outside the
shell with radius vector r. To calculate E at P, we take the
Gaussian surface to be a sphere of radius r and with centre
O, passing through P. All points on this sphere are equivalent
relative to the given charged configuration. (That is what we
mean by spherical symmetry.) The electric field at each point
of the Gaussian surface, therefore, has the same magnitude
E and is along the radius vector at each point. Thus, E and
?S at every point are parallel and the flux through each
element is E ?S. Summing over all ?S, the flux through the
Gaussian surface is E × 4 ? r 2. The charge enclosed is
? × 4 ? R2. By Gauss’s law
E × 4 ? r 2 =
?/?0  4? R2
where q = 4 ? R2 ? is the total charge on the spherical shell.
electric field vector  , E=(q/4??0 r2) r        r is unit vector along the radial dir.
 
The electric field is directed outward if q > 0 and inward if
q < 0. This, however, is exactly the field produced by a charge
q placed at the centre O. Thus for points outside the shell, the field due
to a uniformly charged shell is as if the entire charge of the shell is
concentrated at its centre.
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