Download Summary: The Electrical Poten- tal due to Parallel Lines of Charge

Physics 2460 Electricity and Magnetism I, Fall 2006, Supplement 1
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Summary: The Electrical Potental due to Parallel Lines of Charge
Suggested Reading:
Griffiths: Chapter 2, Sections 2.3 and 2.4, pages
77-96.
Wangsness: Chapters 5 Section 5.3, pages73079.
Physics 2460 Electricity and Magnetism I, Fall 2006, Supplement 1
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The Electrical Potential from Parallel Lines of Charge
Before looking at two parallel lines of charge, recall
that the electrical potential at a distance z from an
infinitely long line of charge, having charge density
λ is just
V (z) = −
λ
log z + C
2π0
where “C” is a constant (usually we will choose our
reference for such a case at z = 1 because then the
constant, which is the potential at the reference
point, is simply 0. If you refer to the notes for
Lecture 17 you will see that the constant term does
depend on the charge density λ, thus it will change
sign if λ changes sign.
IF we have a system of two very long wires having equal but opposite charge densities ±λ we can
use superposition to calculate the electric potential. We will place the lines of charge at x = ±a
and have them run parallel to the z-axis. In cylindrical coordinates the point P : [ρ, φ, z] is located
a distance ρ from the origin. The closest distance
from the positive line of charge is r+ and that from
the negative line of charge is r−.
Physics 2460 Electricity and Magnetism I, Fall 2006, Supplement 1
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The electric potential due to the positive line of
charge located at x = a is
2λ
log(r+) + C
4π0
while that due to the negative line of charge located
at x = −a is
2(−λ)
log(r−) − C
V−(P ) = −
4π0
V+(P ) = −
Physics 2460 Electricity and Magnetism I, Fall 2006, Supplement 1
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The total potential at P is then, by superposition,
VT (P ) = V+(P ) + V−(P )
2λ
[log(r−) − log(r+)]
=
4π0
2λ
r−
=
log
4π0
r
2+ r
λ
log −
=
4π0
r2+
Now, since ~r− = ρ~ + ~a and ~r+ = ρ~ − ~a where
~a = [a, 0, 0], the squared magnitudes of these two
vectors (in the plane at z = 0) are:
r2− = (~
ρ + ~a) · (~
ρ + ~a)
= [x + a, y, 0] · [x + a, y, 0],
= (x + a)2 + y 2
in Cartesian coordinates
r2+ = (~
ρ + ~a) · (~
ρ + ~a)
= [x − a, y, 0] · [x − a, y, 0],
= (x − a)2 + y 2
in Cartesian coordinates
so that the total potential at P is
2
2
2
r
λ
λ
(x + a) + y
V (ρ, φ, z) =
log −
=
log
4π0
r2+
4π0
(x − a)2 + y 2
If we take the ‘anti-log’ (exponential) of both
Physics 2460 Electricity and Magnetism I, Fall 2006, Supplement 1
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sides of this equation we obtain
(x + a)2 + y 2
= e4π0V (ρ,φ,z)/λ = e2η
2
2
(x − a) + y
where η = 2π0V (ρ, φ, z)/λ. Then, introducing
the hyperbolic trigonometric functions
2 sinh η = eη − e−η
and
2 cosh η = eη + e−η
where
eη = e2π0V (ρ,φ,z)/λ = cosh η + sinh η
and using
cosh2 η − sinh2 η = 1
we obtain
(x − a coth η)2 + y 2 =
a
sinh η
2
which is the equation of a circle centred at x =
a coth η and having a radius of a/ sinh η. (coth η =
cosh η/ sinh η.) We can plot these circles at fixed or
constant potential V (ρ, φ, z) to obtain the equipotential surfaces (which are right circular cylinders
parallel to the z-axis).
Physics 2460 Electricity and Magnetism I, Fall 2006, Supplement 1
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The blue circles define the surfaces of constant positive potential and the red circles the surfaces of
constant negative potential.
The black circles which intersect the equipotential surfaces show the electric field lines and are
obtained by taking the gradient of the potential.
Recall that the gradient of a scalar potential is the
electric field and that the gradient is normal to a
surface of constant potential.