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44 COULOMB'S LAW

Example

I f we express all positions in rectangular coordinates, we can easily write down an
explicit form for (2-10). Using (1-13) and (1-14), and noting that the various charges are
designated by the subscripts / rather than by primes, we find that (2-10) becomes

f qq, [ ( * - * , ) * + (y- y t ) 9 . + { ' - * , ) * ] . .

In a sense, (2-12) provides a simple recipe for solving problems, since once the values of
all the charges and their positions in rectangular coordinates are given, all that remains
is to substitute these numbers into (2-12) and to simplify the result as much as possible.

4 CONTINUOUS DISTRIBUTIONS OF CHARGE

We often encounter situations in which the other charges are so close together
compared to the other distances of interest that we can regard them as being
continuously distributed, much as we can treat a glass of water, on a laboratory scale,
as a continuous distribution of mass by neglecting its molecular structure. We can deal
with such a case by considering a region of the charge distribution that is so small that
the charge within it can be written as dq' and treated as a point charge; this is
illustrated in Figure 2-4. We can still use (2-10), but now the sum wil l become an
integral over the complete charge distribution so that

a , dq'R

F < = i / " k < 2 - 1 3 >
where (2-2) continues to be applicable.

I f the charges are distributed throughout a volume, we can introduce a v o l u m e
charge density p, which is defined as the charge per unit volume and hence wil l be
measured in coulombs/(meter) 3. (We wil l write this charge density as p c h in the
infrequent cases in which it might be confused with the p of cylindrical coordinates.)
Then the charge contained in a small source volume dr' wi l l be given by

dq' = p ( r ' ) dr' (2-14)

dq-

Figure 2-4. Charge element of a continu-
ous distribution.

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