A Vertical Electrostatic Dipole above a Conducting Plane: 22-Elec-A7 Electromagnetics, May 2014, Q8

Question 8: A Vertical Electrostatic Dipole above a Conducting Plane (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A point electric dipole of moment $p = 10^{-8}$ C·m, oriented vertically and pointing upward, held $h = 2$ m above an infinite horizontal conducting plane. The medium is free space, $\epsilon_0 = 8.85\times10^{-12}$ F/m.

Find. The magnitude and direction of the electric field at the point on the conducting plane directly beneath the dipole.

Figure 8.1 — The real dipole and its image. Charge-by-charge mirroring leaves the image pointing the same way as the original, so the two on-axis fields add.

Approach. Replace the conductor by the image dipole, establish the image's orientation from first principles, add the two on-axis dipole fields at the surface point, and check the result against the induced surface charge.

Build the image charge by charge. Write the dipole as $+q$ at height $h + d/2$ and $-q$ at $h - d/2$. The image of a charge in a grounded plane is the opposite charge at the mirror position, so the image consists of $-q$ at $-(h+d/2)$ and $+q$ at $-(h-d/2)$.
Read off the image's orientation. In the image pair the positive charge sits above the negative one, exactly as in the original. The image is therefore a dipole of the same magnitude $p$ also pointing upward, located a distance $h$ below the plane:$$\vec{p}_{image} = +p\,\hat{z}\ \text{ at } z = -h.$$This is the opposite of the familiar horizontal-dipole case, where the image reverses.
On-axis field of a point dipole. Along the dipole axis, on either side, the field is parallel to $\vec{p}$ and twice the broadside value:$$\vec{E}_{axis} = \frac{1}{4\pi\epsilon_0}\frac{2p}{r^{3}}\,\hat{z}.$$The observation point lies on the common axis of both dipoles, at distance $h$ from each.
Add the two contributions. Both are at distance $h = 2$ m and both point upward, so they reinforce:$$\vec{E} = 2\times\frac{1}{4\pi\epsilon_0}\frac{2p}{h^{3}}\,\hat{z} = \frac{p}{\pi\epsilon_0 h^{3}}\,\hat{z}.$$
Evaluate. Substituting the numbers,$$E = \frac{10^{-8}}{\pi(8.85\times10^{-12})(2)^{3}} = \frac{10^{-8}}{2.224\times10^{-10}},$$$$\boxed{E = 45.0\ \text{V/m, directed vertically upward (normal to the plane, away from the conductor).}}$$
Check against the induced surface charge. At a conductor surface $E = \sigma_s/\epsilon_0$ along the outward normal, so$$\sigma_s = \epsilon_0 E = (8.85\times10^{-12})(45.0) = 3.98\times10^{-10}\ \text{C/m}^{2} = 398\ \text{pC/m}^{2}.$$The sign is positive, which is exactly what the geometry demands: the dipole's negative end is the one facing the plane, so it draws positive charge up beneath it.

The field is normal to the plane, as it must be on any conductor: a tangential component would drive surface current indefinitely in an electrostatic problem. Directly under the dipole the tangential parts of the real and image fields cancel by symmetry in any case, and only the vertical parts survive — which is why this particular point is the one the examiner chose.

Final Results
Quantity	Symbol	Value
Dipole moment	$p$	$10^{-8}$ C$\cdot$m
Height above the plane	$h$	2 m
Image dipole	$\vec{p}_{image}$	$+p\,\hat{z}$ at depth $h$ (same direction)
Field from the real dipole	$E_{real}$	22.5 V/m upward
Field from the image dipole	$E_{image}$	22.5 V/m upward
Total field on the plane	$E$	45.0 V/m, vertically upward
Induced surface charge density	$\sigma_s$	$+398$ pC/m$^{2}$

Topic. Electrostatic image theory for a point dipole above a grounded conducting plane.

Key relations. Image of $+q$ at height $h$ is $-q$ at $-h$; a vertical dipole images as a vertical dipole of the SAME sense, a horizontal dipole images reversed; on-axis field $E = 2p/(4\pi\epsilon_0 r^{3})$; $E = \sigma_s/\epsilon_0$ at a conductor surface.

Why this works. The grounded plane forces the potential to zero everywhere on it, and the image construction is simply the charge distribution that reproduces that boundary condition while leaving the region above the plane unchanged. Applying it charge by charge — rather than trying to guess how a dipole reflects — settles the orientation unambiguously, and the result differs for vertical and horizontal dipoles. Once the image is in place the conductor can be deleted and the problem is two point dipoles in free space, both viewed end-on.

Common pitfall. The dominant error is assuming the image dipole is inverted, which cancels the field instead of doubling it and gives zero. Almost as common is using the broadside dipole field $p/(4\pi\epsilon_0 r^{3})$ instead of the on-axis value, which halves the answer to 22.5 V/m. Forgetting the image altogether gives the same 22.5 V/m by a different route.

Source. Hayt & Buck, Engineering Electromagnetics, 9th ed., §5.5 (method of images); Sadiku, Elements of Electromagnetics, 7th ed., §6.6.

Question 8 of 8: A Vertical Electrostatic Dipole above a Conducting Plane

Question 8: A Vertical Electrostatic Dipole above a Conducting Plane (20 marks)