Example 3.2. A point charge q is situated a distance a from the center of agrounded conducting sphere of radius R (Fig. 3.12). Find the potential outsidethe sphere.V=0FIGURE 312FIGURE 3.13SolutionExamine the completely different configuration, consisting of the point charge qtogether with another point charge--.(3.15)placed a distance(3.16)to the right of the center of the sphere (Fig. 3.13). No conductor, now-just thetwo point charges. The potential of this configuration isVr) =(3.17)where s and e' are the distances from q and q', respectively. Now, it happens (seeProb. 3.8) that this potential vanishes at all points on the sphere, and therefore fitsthe boundary conditions for our original problem, in the exterior region."Conclusion: Eq. 3.17 is the potential of a point charge near a grounded con-ducting sphere. (Notice that b is less than R, so the "image" charge q' is safelyinside the sphere-you cannot put image charges in the region where you are cal-culating V; that would change p, and you'd be solving Poisson's equation with"Tis solution is due to William Thomson (later Loed Kelvin), who published ik in 1848, when hewas just 24. It was apparently inspired by a theorem of Apollonius (200 BC) that says the locus ofpoints with a fed ratio ef distances from two given points is a spbere. See .C. Maswell "Treatise onElectricity and Magnetism, Vel. 1," Dover, New York, p. 245. I thank Gabriel Karl for this interestinghistory.3.2 The Method of Images129the wrong source.) In particular, the force of attraction between the charge andthe sphere isq*Ra4xeo (a - Ra1(3.18)4T€0 (a - b) Problem 3.9 In Ex. 3.2 we assumed that the conducting sphere was grounded(V = 0). But with the addition of a second image charge, the same basic modelwill handle the case of a sphere at any potential Vo (relative, of course, to infin-ity). What charge should you use, and where should you put it? Find the force ofattraction between a point charge q and a neutral conducting sphere.Chapter 3 Potentials!Problem 3.10 A uniform line charge à is placed on an infinite straight wire, a dis-tance d above a grounded conducting plane. (Let's say the wire runs parallel to thex-axis and directly above it, and the conducting plane is the xy plane.)(a) Find the potential in the region above the plane. [Hint: Refer to Prob. 2.52.](b) Find the charge density o induced on the conducting plane.

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Problem 3.9 In Ex. 3.2 we assumed that the conducting sphere was grounded (V = 0). But with the addition of a second image charge, the same basic model will handle the case of a sphere at any potential Vo (relative, of course, to infin- ity). What charge should you use, and where should you put it? Find the force of attraction between a point charge q and a neutral conducting sphere.

Problem 3.9 In Ex. 3.2 we assumed that the conducting sphere was grounded (V = 0). But with the addition of a second image charge, the same basic model will handle the case of a sphere at any potential Vo (relative, of course, to infin- ity). What charge should you use, and where should you put it? Find the force of attraction between a point charge q and a neutral conducting sphere.

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Electric field

A physical field present around an electrically charged particle that exerts a force on all other charged particles is called an electric field. The charged particles present in the field either repel or attract. From electric charges or time-varying magnetic fields, the electric field originates. One of the manifestations of electromagnetic force which is one of the four fundamental forces present in nature are the electric fields and magnetic fields.

Question

Example 3.2. A point charge q is situated a distance a from the center of a
grounded conducting sphere of radius R (Fig. 3.12). Find the potential outside
the sphere.
V=0
FIGURE 312
FIGURE 3.13
Solution
Examine the completely different configuration, consisting of the point charge q
together with another point charge
--.
(3.15)
placed a distance
(3.16)
to the right of the center of the sphere (Fig. 3.13). No conductor, now-just the
two point charges. The potential of this configuration is
Vr) =
(3.17)
where s and e' are the distances from q and q', respectively. Now, it happens (see
Prob. 3.8) that this potential vanishes at all points on the sphere, and therefore fits
the boundary conditions for our original problem, in the exterior region."
Conclusion: Eq. 3.17 is the potential of a point charge near a grounded con-
ducting sphere. (Notice that b is less than R, so the "image" charge q' is safely
inside the sphere-you cannot put image charges in the region where you are cal-
culating V; that would change p, and you'd be solving Poisson's equation with
"Tis solution is due to William Thomson (later Loed Kelvin), who published ik in 1848, when he
was just 24. It was apparently inspired by a theorem of Apollonius (200 BC) that says the locus of
points with a fed ratio ef distances from two given points is a spbere. See .C. Maswell "Treatise on
Electricity and Magnetism, Vel. 1," Dover, New York, p. 245. I thank Gabriel Karl for this interesting
history.
3.2 The Method of Images
129
the wrong source.) In particular, the force of attraction between the charge and
the sphere is
q*Ra
4xeo (a - Ra
1
(3.18)
4T€0 (a - b)

Problem 3.9 In Ex. 3.2 we assumed that the conducting sphere was grounded
(V = 0). But with the addition of a second image charge, the same basic model
will handle the case of a sphere at any potential Vo (relative, of course, to infin-
ity). What charge should you use, and where should you put it? Find the force of
attraction between a point charge q and a neutral conducting sphere.
Chapter 3 Potentials
!
Problem 3.10 A uniform line charge à is placed on an infinite straight wire, a dis-
tance d above a grounded conducting plane. (Let's say the wire runs parallel to the
x-axis and directly above it, and the conducting plane is the xy plane.)
(a) Find the potential in the region above the plane. [Hint: Refer to Prob. 2.52.]
(b) Find the charge density o induced on the conducting plane.

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