Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Exercise 3.4.10

About the further exercises of Section 3.4

The ten exercises below are in the spirit of Problem 3.4.1, the home exercises and the worked examples. They use only $D = ε_{0} E + P = ε E$ , Gauss's law $\oint_{S} D \cdot d S = Q_{free}$ , the boundary conditions, the bound charges, $C = Q / V$ , $W = \frac{1}{2} \int D \cdot E d v$ and the image principle.

Exercise 3.4.10

Point charge near a grounded sphere

cf. the image principle

A grounded conducting sphere of radius $a$ is centred at the origin. A point charge $q$ is placed on the $z$ -axis at $z = d > a$ . (a) Show that an image $q^{'} = - q a / d$ at $z = b = a^{2} / d$ makes the sphere an equipotential $V = 0$ . (b) Find the force on $q$ and the total induced charge. (c) Evaluate for $a = 5$ cm, $d = 10$ cm, $q = 1$ nC, including $ρ_{S}$ at the nearest and farthest points of the sphere.

Adapted from: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 4-5.

Grounded sphere of radius a with the charge q at distance d on the z-axis and the image q' at distance b from the centre

Try it yourself first.

Strategy

Check the proposed image directly: compute the distances from a general point P of the sphere to $q$ and to $q^{'}$ and show that the two potentials cancel. Then everything follows as for the plane: force from the image, induced charge from Gauss's law, $ρ_{S} = ε_{0} E_{n}$ on the surface.

(a)

Take a point P on the sphere, $∣ OP ∣ = a$ , at the angle $θ$ from the $z$ -axis. By the law of cosines its distances to $q$ and to $q^{'}$ are given by

R^{2} R^{'2} = a^{2} + d^{2} - 2 a d cos θ, = a^{2} + b^{2} - 2 ab cos θ = \frac{a ^{2}}{d ^{2}} (d^{2} + a^{2} - 2 a d cos θ) = \frac{a ^{2}}{d ^{2}} R^{2},

using $b = a^{2} / d$ . Hence $R^{'} = a R / d$ and

V (P) = \frac{1}{4 π ε _{0}} [\frac{q}{R} + \frac{q ^{'}}{R ^{'}}] = \frac{1}{4 π ε _{0}} [\frac{q}{R} - \frac{q a / d}{a R / d}] = 0 for every θ .

The image lies inside the sphere, outside the region of interest, so by the uniqueness theorem the field outside the sphere is that of $q$ and $q^{'}$ .

(b)

The force is the attraction of the image, a distance $d - b = (d^{2} - a^{2}) / d$ away:

Answer

F = \frac{q q ^{'}}{4 π ε _{0} ( d - b ) ^{2}} = - \frac{q ^{2} a d}{4 π ε _{0} ( d ^{2} - a ^{2} ) ^{2}} [N]

(negative: towards the sphere). A closed surface just outside the sphere, not enclosing $q$ , receives no net flux from $q$ and the flux $q^{'} / ε_{0}$ from the image. By Gauss's law the charge inside it, i.e. the induced charge, is $Q_{ind} = q^{'} = - q a / d$ . It is smaller than $q$ in magnitude: the remaining field lines of $q$ go to infinity.

(c)

$q^{'} = - 0.5$ nC at $b = 2.5$ cm, and $d - b = 7.5$ cm:

F = - \frac{9 \times 1 0 ^{9} \times 1 0 ^{- 18} \times 0.05 \times 0.10}{( 0.0075 ) ^{2}} = - 0.80 μ N .

The surface charge follows from $ρ_{S} = ε_{0} E_{n}$ at $r = a$ :

ρ_{S} (θ) = - \frac{q ( d ^{2} - a ^{2} )}{4 π a ( a ^{2} + d ^{2} - 2 a d cos θ ) ^{3/2}},

ρ_{S} (0) ρ_{S} (π) = - 95.5 nC/ m^{2}, = - 3.54 nC/ m^{2},

and $\oint ρ_{S} d S = - 0.5$ nC $= q^{'}$ $✓$

Answer

q^{'} = - 0.5 nC at 2.5 cm, F = 0.80 μ N (attraction), ρ_{S} (0) / ρ_{S} (π) = 27

Exercise 3.4.10: magnitude of the (negative) induced surface charge density on the grounded sphere, from the point nearest the charge (\theta=0) to the farthest point (\theta=180^\circ). The ratio of the two extremes is 27. — **Figure 3.4.35.** Exercise 3.4.10: magnitude of the (negative) induced surface charge density on the grounded sphere, from the point nearest the charge ( $θ = 0$ ) to the farthest point ( $θ = 18 0^{\circ}$ ). The ratio of the two extremes is $27$ .

Check

For $d \to \infty$ the image charge $q^{'} = - q a / d \to 0$ : a very distant charge induces almost nothing on a small grounded sphere.

Remark

As $d \to a$ the image moves to the surface and $q^{'} \to - q$ : very close to a large sphere the charge sees a plane (Section 3.4.8). For an isolated, uncharged sphere one adds a second image $+ q a / d$ at the centre, so that the total charge vanishes while the surface remains an equipotential. The force is then smaller but still attractive: a neutral conductor always attracts a charge.

Exercise 3.4.10: Point charge near a grounded sphere