Exercise 3.4.9: Point charge in a grounded right-angle corner

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 3.4.9

Point charge in a grounded right-angle corner

cf. the image principle

Two grounded conducting half-planes, $x = 0$ ( $y > 0$ ) and $y = 0$ ( $x > 0$ ), form a right-angle corner. A point charge $q = 1$ nC is placed at $(d, d, 0)$ , $d = 1$ cm. (a) Find the image charges and the potential in the region $x > 0$ , $y > 0$ . (b) Find the force on $q$ . (c) What is the total charge induced on the two planes?

Adapted from: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 4-3.

Grounded right-angle corner with the charge q at (d,d) and its three images -q, -q and +q

Try it yourself first.

Strategy

Use the image recipe plane by plane: an image in each plane, and then an image of the images, until both planes are at $V = 0$ . All images must lie outside the corner region. The force on $q$ is the vector sum of the forces from the three images.

(a)

The image $- q$ at $(- d, d)$ makes the plane $x = 0$ an equipotential, the image $- q$ at $(d, - d)$ does the same for $y = 0$ . But each of these destroys the condition on the other plane, which is restored by a third image $+ q$ at $(- d, - d)$ . The four charges are antisymmetric with respect to both planes, so $V = 0$ on $x = 0$ and on $y = 0$ , and in the corner region only the real charge is present. By the uniqueness theorem:

Answer

V (x, y, z) R_{1, 4}^{2} R_{2, 3}^{2} = \frac{q}{4 π ε _{0}} [\frac{1}{R _{1}} - \frac{1}{R _{2}} - \frac{1}{R _{3}} + \frac{1}{R _{4}}], = (x \mp d)^{2} + (y \mp d)^{2} + z^{2}, = (x \pm d)^{2} + (y \mp d)^{2} + z^{2}

Exercise 3.4.9: field lines (orange) and equipotentials (blue) of a point charge in a grounded right-angle corner, computed from q and its three images (x, y in units of d). Every field line ends perpendicularly on one of the two planes, and the corner itself is a region of very weak field. — **Figure 3.4.34.** Exercise 3.4.9: field lines (orange) and equipotentials (blue) of a point charge in a grounded right-angle corner, computed from $q$ and its three images ( $x$ , $y$ in units of $d$ ). Every field line ends perpendicularly on one of the two planes, and the corner itself is a region of very weak field.

(b)

With $k = 1/ (4 π ε_{0})$ , $k q^{2} / d^{2} = 9 \times 1 0^{9} \times 1 0^{- 18} /1 0^{- 4} = 90.0$ μN. The two images $- q$ at distance $2 d$ attract with $k q^{2} / (4 d^{2})$ each, along $- a_{x}$ and $- a_{y}$ , and the image $+ q$ at distance $22 d$ repels with $k q^{2} / (8 d^{2})$ along $(a_{x} + a_{y}) / 2$ :

F = \frac{k q ^{2}}{d ^{2}} (- \frac{1}{4} + \frac{1}{8 2}) (a_{x} + a_{y}) = - 0.162 \frac{k q ^{2}}{d ^{2}} (a_{x} + a_{y}),

Answer

F = - 14.5 (a_{x} + a_{y}) μ N, ∣ F ∣ = 20.6 μ N (towards the corner)

(c)

The four charges add up to zero and their dipole moment is also zero (they form a quadrupole), so far from the corner the field falls off as $1/ r^{4}$ and no flux escapes to infinity: every field line leaving $q$ ends on the planes (Figure 3.4.34). Hence $Q_{ind} = - q = - 1$ nC, shared equally by the two planes by symmetry.

Check

If the charge were moved far from the corner along the plane $y = 0$ (keeping its height $d$ ), the other plane would hardly matter and the force should approach that of a single plane, $k q^{2} / (4 d^{2})$ : indeed the images $- q$ and $+ q$ behind the distant plane $x = 0$ then nearly cancel each other.

Remark

The method works for a corner of angle $π / n$ with $2 n - 1$ images ( $n = 2$ here: three). For $6 0^{\circ}$ five images are needed, for $4 5^{\circ}$ seven. For an angle that is not of the form $π / n$ the images would fall inside the physical region and the method fails.