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 V0 (relative, of course, to infinity). What charge should you use, and where should you put it? Find the

Question

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 [latex]V_{0}[/latex] (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.

Accepted Answer

Place a second image charge, [latex]q^{\prime \prime}[/latex], 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 [latex]V_{0}=\frac{1}{4 \pi \epsilon_{0}} \frac{q^{\prime \prime}}{R}[/latex] ;

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

For a neutral sphere, [latex]q^{\prime}+q^{\prime \prime}=0 [/latex] .

[latex]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)[/latex]

[latex]=\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}}[/latex]

[latex]=-\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}}[/latex]

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

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

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