Example 9-10.
Consider the special case of a single point charge qlocated at the origin. The
electric field due to this point charge is
E=E(x, y, z ) =^1
QF =^1
4 π
0q
r^2ˆer (9 .138)where ˆeris a unit vector in spherical coordinates and ris the distance of a test charge
Qfrom the origin to the position (x, y, z). This vector field produces field lines and
the strength of the vector field is proportional to the flux across some surface placed
within the electric field. In the case where a sphere of radius rand centered at the
origin is placed within the electric field, then the flux is calculated from the surface
integral
Flux =
∫∫SE ·dS=∫∫SE·ˆendS (9 .139)where ˆen = ˆer in spherical coordinates. Also the element of area dS in spherical
coordinates is given by dS=r^2 sin θ dθdφ ˆerso that equation (9.139) can be expressed
as
Flux =
∫∫SE·dS=∫ 2 π0[∫π01
4 π
0q
r^2r^2 sin θ dθ]
dφ = q(^0)
(9 .140)
This result states that the flux is a constant no matter what size sphere is placed
about the point charge. If the sphere were made of rubber and could be deformed
into some other simple closed surface, the number of field lines passing through the
new surface would also be the same constant as above. This is because the dot
product E ·ˆen selects an element of area perpendicular to the field lines and the
flux is proportional to the number of these lines. Note that if the point charge were
outside the closed surface, then the flux would be zero, since field lines entering the
surface at one point must exist at some other point and then the sum of the flux
would be zero.
One can say that if there were n-charges q 1 , q 2 ,... , qninside a simple closed surface
and Eiwas the electric field associated with the ith charge, then E =
∑ni=1Ei would
represent the total electric field and the flux across any simple closed surface due to
this total electric field would be
∫∫SE·dS=∑ni=1
∫∫SEi·dS
=∑ni=1qi
0 (9 .141)