Page 336 - Intro to Tensor Calculus

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The differential equation for the equipotential curves is obtained by taking the negative reciprocal of
the slope of the field lines. This gives

kq(x−a) kq(x+a)
dy r 3 2 − r 3 1
= .
dx kqy kqy
3
3 −
r 1 r 2
This result can be written in the form

(x + a)dx + ydy (x − a)dx + ydy
− 3 + 3 =0
r r
1 2
which simplifies to the easily integrable form

dr 1 dr 2
− 2 + 2 =0
r 1 r 2
in terms of the new variables r 1 and r 2 . An integration produces the equipotential curves
1 1
− =C 2
r 1 r 2
1 1
or p − p =C 2 .
2
2
(x + a) + y 2 (x − a) + y 2
The potential function for this problem can be interpreted as a superposition of the potential functions
kq kq
V 1 = − and V 2 = associated with the isolated point charges at the points (−a, 0) and (a, 0).
r 1 r 2
Observe that the electric lines of force move from positive charges to negative charges and they do not
cross one another. Where field lines are close together the field is strong and where the lines are far apart
the field is weak. If the field lines are almost parallel and equidistant from one another the field is said to be
~
uniform. The arrows on the field lines show the direction of the electric field E. If one moves along a field
line in the direction of the arrows the electric potential is decreasing and they cross the equipotential curves
~
at right angles. Also, when the electric field is conservative we will have ∇× E =0.
In three dimensions the situation is analogous to what has been done in two dimensions. If the electric
~
~
field is E = E(x, y, z)= P(x, y, z) b e 1 + Q(x, y, z) b e 2 + R(x, y, z) b e 3 and ~r = x b e 1 + y b e 2 + z b e 3 is the position
~
vector to a variable point (x, y, z) on a field line, then at this point d~ and E must be colinear so that
r
~
r
d~ = KE for some constant K. Equating like coefficients gives the system of equations
dx dy dz
= = = K. (2.6.19)
P(x, y, z) Q(x, y, z) R(x, y, z)
From this system of equations one must try to obtain two independent integrals, call them u 1 (x, y, z)= c 1
and u 2 (x, y, z)= c 2 . These integrals represent one-parameter families of surfaces. When any two of these
~
surfaces intersect, the result is a curve which represents a field line associated with the vector field E. These
type of field lines in three dimensions are more difficult to illustrate.
~
The electric flux φ E of an electric field E over a surface S is defined as the summation of the normal
~
component of E over the surface and is represented
ZZ Nm 2
~
φ E = E · ˆ n dσ with units of (2.6.20)
S C