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Chapter 3: Q3.36P (page 160)
Show that the electric field of a (perfect) dipole (Eq. 3.103) can be written in the coordinate-free form
Answer
The given relation is proved.
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In Section 3.1.4, I proved that the electrostatic potential at any point
in a charge-free region is equal to its average value over any spherical surface
(radius R )centered at .Here's an alternative argument that does not rely on Coulomb's law, only on Laplace's equation. We might as well set the origin at P .Let be the average; first show that
(note that the in da cancels the out front, so the only dependence on R
is in itself). Now use the divergence theorem, and conclude that if Vsatisfies
Laplace's equation, then,.
Find the potential in the infinite slot of Ex. 3.3 if the boundary at x = 0 consists of two metal strips: one, from y = 0 to y = a/2, is held at a constant Potential , and the other, from y = a/2 to y = a , is at potential .
(a) Using the law of cosines, show that Eq. 3.17 can be written as follows:
Whereand are the usual spherical polar coordinates, with the axis along the
line through . In this form, it is obvious thaton the sphere, localid="1657372270600" .
(a) Find the induced surface charge on the sphere, as a function of . Integrate this to get the total induced charge . (What should it be?)
(b) Calculate the energy of this configuration.
Use Green's reciprocity theorem (Prob. 3.50) to solve the following
two problems. [Hint:for distribution 1, use the actual situation; for distribution 2,
removeq,and set one of the conductors at potential .]
(a) Both plates of a parallel-plate capacitor are grounded, and a point charge qis
placed between them at a distance xfrom plate 1. The plate separation is d. Find the induced charge on each plate. [Answer: ;]
(b) Two concentric spherical conducting shells (radii aand b)are grounded, and a point charge is placed between them (at radius r). Find the induced charge on each sphere.
Here's an alternative derivation of Eq. 3.10 (the surface charge density
induced on a grounded conducted plane by a point charge qa distance dabove
the plane). This approach (which generalizes to many other problems) does not
rely on the method of images. The total field is due in part to q,and in part to the
induced surface charge. Write down the zcomponents of these fields-in terms of
qand the as-yet-unknown -just below the surface. The sum must be zero,
of course, because this is inside a conductor. Use that to determine .
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