A metal sphere of radius R = 10.0 cm stands on an insulating stand in a large empty room…

A metal sphere of radius $R = 10.0\ \text{cm}$ stands on an insulating stand in a large empty room. A thin wire joins it to the earth, so the sphere stays at zero potential. A small bead with a charge $q = +2.0\ \text{nC}$ hangs on a silk thread with its centre $d = 30.0\ \text{cm}$ from the centre of the sphere, as in the figure. The bead is far from the walls and from the wire, and its size can be ignored. Take $\dfrac{1}{4\pi\varepsilon_0} = 9.0 \times 10^{9}\ \text{N m}^2\text{/C}^2$ .

A metal sphere of radius R = 10.0 cm, centre O, stands on an insulating stand. A wire from the sphere runs down to an earth symbol. To the right of the sphere a small bead with charge q = +2.0 nC hangs on a silk thread, on the horizontal line through O. A dimension line marks d = 30.0 cm from O to the bead.

When the bead is brought in, charge flows along the wire until every point of the sphere is at zero potential. Outside the sphere, the field of this induced charge turns out to be exactly the field of a single point charge placed inside the sphere, on the line from the centre to the bead. You may use that fact, but you must find the size and the place of that point charge yourself.

What is the total charge on the sphere?
What is the size of the force on the bead?
The induced charge is not spread evenly. What is the ratio of the surface charge density at the point of the sphere nearest the bead to that at the point farthest from it?
The bead is now carried slowly, along the line through the centre, to a very great distance. How much work must be done against the electric force?
Start again with the wire taken away before the bead is brought in, so that the sphere is insulated and carries no net charge. With the bead again $30.0\ \text{cm}$ from the centre, what is the size of the force on it?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Replace the sphere by an image charge $q'$ at a distance $b$ from its centre. Then the force, the charge density and the work all come from two point charges.

Step 1. The image. Zero potential at the nearest point ( $d - R$ from the bead) and the farthest point ( $d + R$ ) gives $\dfrac{q}{d - R} + \dfrac{q'}{R - b} = 0$ and $\dfrac{q}{d + R} + \dfrac{q'}{R + b} = 0$ . Solving, $b = \dfrac{R^2}{d} = 3.33\ \text{cm}$ and $q' = -\dfrac{qR}{d}$ . Every other point of the sphere is then at zero potential too, because its distances to $q$ and to $q'$ are always in the ratio $d : R$ .

Step 2. Charge on the sphere (a). Gauss's law over a surface just outside the sphere: the sphere's charge has the field of $q'$ , so it equals $q' = -2.0\ \text{nC} \times \dfrac{10.0}{30.0} = -0.67\ \text{nC}$ .

Step 3. Force (b). The bead is $d - b = 26.67\ \text{cm}$ from the image:

\displaystyle F = \frac{kq|q'|}{(d - b)^2} = \frac{kq^2Rd}{(d^2 - R^2)^2} = 1.69 \times 10^{-7}\ \text{N}

towards the sphere.

Step 4. Charge density (c). The field just outside the surface gives $\sigma = \varepsilon_0E$ , which works out to $\sigma = -\dfrac{q(d^2 - R^2)}{4\pi R\,s^3}$ , where $s$ is the distance from the bead. So $\sigma \propto 1/s^3$ , and the ratio is $\left(\dfrac{d + R}{d - R}\right)^3 = 2^3 = 8.0$ .

Step 5. Work (d). At a distance $r$ the force is $\dfrac{kq^2Rr}{(r^2 - R^2)^2}$ . So

\displaystyle W = \int_d^\infty \frac{kq^2Rr\,dr}{(r^2 - R^2)^2} = \frac{kq^2R}{2(d^2 - R^2)} = 2.25 \times 10^{-8}\ \text{J}

Step 6. The neutral sphere (e). Add $+qR/d$ at the centre. The sphere stays at one potential, and its net charge is zero. The force is

\displaystyle F = kq\,\frac{qR}{d}\left[\frac{1}{(d - b)^2} - \frac{1}{d^2}\right] = 3.54 \times 10^{-8}\ \text{N}

still towards the sphere.

Answer: (a) $-0.67\ \text{nC}$ . (b) $1.69 \times 10^{-7}\ \text{N}$ . (c) $8.0$ . (d) $2.25 \times 10^{-8}\ \text{J}$ . (e) $3.54 \times 10^{-8}\ \text{N}$ .

Check. Far away the neutral sphere is a small induced dipole, and the force should tend to $\dfrac{2kq^2R^3}{d^5}$ . Here that gives $2.96 \times 10^{-8}\ \text{N}$ , below the exact value: at only $d = 3R$ the bead is not yet far away, and the higher terms still add to the pull.

Watch out. The work is not $\dfrac{kq|q'|}{d - b} = 4.50 \times 10^{-8}\ \text{J}$ , the energy of two fixed charges. The image is not a real charge that stays put: it moves and grows as the bead comes closer, and the integral gives exactly half that value.

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