Today I gave a revision lecture/tutorial for my module Advanced Electromagnetism. With the Examination Period starting on Friday, that was the last class I will do for that. One of the topics I’ve been asked to cover in revision was the Method of Images for electrostatics. Preparing for the class I came across this cute problem which I thought I’d share here:
The question concerns a charge +q placed at a distance d as shown above an infinite earthed conducting plane distorted by the presence of a hemispherical bulge with radius R.
- Using the method of images, or otherwise, calculate the potential at an arbitrary point above the conducting surface. (HINT: you need three image charges)
- Find the magnitude and direction of the electrostatic force on the charge.
If you’re feeling keen you might also find what fraction of the total induced on the conductor is on the hemispherical part.
Answers through the comments box please!
Well, nobody posted an answer so here’s an outline solution.
To solve this problem you need three image charges: one is of charge – q at z=-d to make the plane an equipotential. For an isolated sphere you need a charge of -qR/d at z=-R^2/d (the inverse point of the sphere). But this charge also has an effect on the plane, which you need to correct by placing another image charge of +qR/d at z=-R^2/d. That is, the solution for the potential is due to the original charge plus three image charges. Then the potential is just the sum of four point charges.
You can differentiate the answer to the first bit to get the force, or you could work out the force on the original charge directly by adding the forces in the z-direction from the three image charges, it being obvious by symmetry that there is no other component of the force. For d>R this results in a force which is downward, so the charge is pulled towards the conductor. I’ll leave that as an exercise!
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