Download Electrostatic energy of charges in front of conducting planes

Electrostatic energy of charges in front of conducting planes
Consider the following configurations with point charges
in the presence of conductors. Using the method of images, find for each case the solution for the electrostatic
potential by specifying the position and value of the image charges, and the potential energy of the system.
a) A charge q is at distance a from an infinite conducting
plane.
b) Two opposite charges +q and −q are at mutual distance d and at the same distance a from an infinite conducting plane.
c) A charge q is at distances a and b from two mutually
perpendicular, infinite conducting planes, respectively.
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a)
a
q
a
a
+q
d
x
0
b)
)
q
−q
b
Solution
The image charges configurations for the
a)
three cases are as in the figure. In all
−q
−q
a +q
cases, the forces on the charges are the
same as if the image charges were rex
al, because the force only depends on
0
)
the local electric field at the charge lo+q
−q
a +q
cation, and the electric field (outside the
conductors) is the same as if all charges
b
were real.
a) In this case, the potential problem is
solved by placing an image charge q ′ =
+q
−q
−q in the position symmetrical to the
real charge with respect to the plane.
To evaluate the potential energy we suppose to displace the
charge q along the x axis, from the initial position x = a to
(−q)
x = +∞. Correspondingly, the image charge moves from x = −a
2
2
to x = −∞. The force on the real charge q is Fx = −k0 q /(2x)
(directed towards the plane) and the work done by Fx to move q
to x = +∞ is
Z
Z ∞
q2
q 2 ∞ dx
=
−k
.
Fx dx = −k0
W =
0
4 a x2
4a
d
b)
+q
a
d
−q
Fx
q
a
x
(1)
There is no additional work due to move the image charge to x = −∞ since what actually move are
surface charges on the conductor, but since the latter is kept at a zero potential, no work is done.
Thus, the potential energy is U = W = −k0 q 2 /4a, that is half the potential energy of the system
of two real charges q and −q at a distance 2a, Ureal = −k0 q 2 /2a. The 1/2 reduction may be simply
understood because for a symmetrical displacement of real charges the work would be given by
Z −∞
Z +∞
F+q→−q dx ,
(2)
F−q→+q dx +
Wreal =
+a
where F−q→+q = −F+q→−q = Fx . Thus
Z
Z +∞
Fx dx −
Wreal =
+a
−a
−∞
Fx dx = 2
−a
Z
+∞
Fx dx = 2W ,
(3)
+a
thus Ureal = 2W = 2U .
Another way to get the above result is to evaluate and compare the electrostatic energy of the
system for the “real” case (i.e. if all charges were real) and the image charge case. In both cases,
because of the cylindrical symmetry the electrostatic field is a function of x and r = (y 2 + z 2 )1/2 only,
i.e. E = E(x, r), and In the first case (all real charges) we would have
Z
Z ∞
Z ∞
Z ∞ Z ∞
2
es
3
2
2πrdrǫ0 E2 (x, r)/2 ,
(4)
dx
2πrdrǫ0 E (x, r)/2 = 2
dx
Ureal = d xǫ0 E /2 =
−∞
0
0
2
0
because due to specular symmetry with respect to the x = 0 plane E2 is an even function in x, i.e.
E2 (x, r) = E2 (−x, r). In the case of the charge in front of the conducting plane we have
Z ∞ Z ∞
es
U =
dx
2πrdrǫ0 E2 (x, r)/2 ,
(5)
0
0
because E = 0 for x < 0 (i.e. inside the conductor), while the field is the same as the “real” case
es
for x > 0, thus U es = Ureal
/2. The electrostatic energy includes both the interaction energy and the
“self-energy” of the charges, i.e. for the “real” system
es
int
Ureal
= Us,+q + Us,−q + Ureal
= 2Us + k0
q2
,
2a
(6)
where we used the fact that Us,+q = Us,−q ≡ Us . In the “image” case, we have U es = Us + U int since
es
int
there is only one real charge, thus from U es = Ureal
/2 we also obtain U int = Ureal
/2.
(Actually the self-energy Us → ∞ if the charges are exactly point-like, but this issue is not really
relevant here. In any case the divergence may be cured by assuming an arbitrarily small, but non-zero
radius for the charges.)
b) The above argument on the electrostatic energy tells us that also in this case the potential energy
is half of the “real” case: U = Ureal /2 where
q2
q2
q2
Ureal = k0 −2 − 2 + 2 2
.
(7)
d
2a
(d + 4a2 )1/2
The factors of 2 appear because we have two identical contributions from pairs of opposite charges
interacting at a distance d, two contributions from opposite charges at a distance 2a, and two
contributions from equal charges at a distance (d2 + 4a2 )1/2 .
Let us get to the same conclusion by thinking again to the work done by interaction forces when
the charges are displaced to infinity. Assume that all charges were real and suppose to displace them
symmetrically up to infinity. The total work done is the sum of the line integral of all the force
terms between each pair of charges. For each pair, the two force terms exerted by one charge on
the other are equal in modulus and opposite in direction, so that the line integral of each of the
two force terms along the path (because of symmetry) is equal and gives the same contribution to
the total work. Now let us turn to the system of real charges with their images. The interaction
between the two real charges yields a k0 q 2 /d contribution to the total work as in the “real” case.
For what concerns the interaction of each real charge with an image, the contribution is one half the
corresponding term in the “real” case, because it is only the force on the real charge that actually
does work. Finally, with respect to the real case the contribution corresponding to the interaction
between image charges must be discarded, because there is no such interaction anymore. Thus, we
write the potential energy as
2
q
1
1 q2
1
q2
U = k0 − − 2 + 2 2
= Ureal .
(8)
2
1/2
d
2 2a 2 (d + 4a )
2
c) The argument on the electrostatic energy now tells us that in this case the potential energy is one
quarter of the “real” case, U = Ureal /4, because the integral of the energy density now extends only
over one quarter of the whole space.
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Alternatively, we may again considering the six terms contributing to the potential energy in
the “real” case, which are identical to those of case b) (for d = 2b), and use the above discussed
rule of thumb: to obtain the energy for the equivalent system of real and image charges keep the
whole contribution of “real-real” pairs, keep one half of the “real-image” interactions, and discard
the “image-image” interaction. This yields
1 q2 1 q2
q2
1
1
U = k0 −
=
−
+
Ureal ,
(9)
2 2b 2 2a 2 (4b2 + 4a2 )1/2
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where Ureal is given by Eq.(7) for d = 2b.
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