Question 3.9: In Ex. 3.2 we assumed that the conducting sphere was grounde...

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 V_{0} (relative, of course, to infinity). 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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Place a second image charge, q^{\prime \prime}, at the center of the sphere; this will not alter the fact that the sphere is an equipotential , but merely increase that potential from zero to  V_{0}=\frac{1}{4 \pi \epsilon_{0}} \frac{q^{\prime \prime}}{R} ;

q^{\prime \prime}=4 \pi \epsilon_{0} V_{0} R at center of sphere  .

For a neutral sphere,  q^{\prime}+q^{\prime \prime}=0 .

F=\frac{1}{4 \pi \epsilon_{0}} q\left(\frac{q^{\prime \prime}}{a^{2}}+\frac{q^{\prime}}{(a-b)^{2}}\right)=\frac{q q^{\prime}}{4 \pi \epsilon_{0}}\left(-\frac{1}{a^{2}}+\frac{1}{(a-b)^{2}}\right)

 

=\frac{q q^{\prime}}{4 \pi \epsilon_{0}} \frac{b(2 a-b)}{a^{2}(a-b)^{2}}=\frac{q(-R q / a)}{4 \pi \epsilon_{0}} \frac{\left(R^{2} / a\right)\left(2 a-R^{2} / a\right)}{a^{2}\left(a-R^{2} / a\right)^{2}}

 

=-\frac{q^{2}}{4 \pi \epsilon_{0}}\left(\frac{R}{a}\right)^{3} \frac{\left(2 a^{2}-R^{2}\right)}{\left(a^{2}-R^{2}\right)^{2}}

(Drop the minus sign, because the problem asks for the force of attraction.)

3.9

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