8.022 at ESG Fall 2004 Problem Set 07 due Friday October 29

Reading:

Purcell Chapter 4

Problems from the text:

4.3: We have to crunch numbers once in a while. 4.5: Once you convince yourself that the physical situation is that described, this is sort of simple. 4.11: Again with the number crunch. 4.19: Circuit design! Try the different possible circuit configurations, perhaps discovering some impossible ones. 4.22: The problem needs the result of Problem 4.18, which is optional but recommended. Maybe give 4.18 a try if 4.22 seems tricky. 4.26: This is a standard. The hint is the “obvious” way to do the problem, but expect to see another. 4.27: We’ll be using MAPLE to see how to get a machine to get that preposterous formula for R_{eq}, and perhaps even see why it’s not that preposterous. That is, the denominator is clearly (do I get to say “clearly”?) the determinant of something. The MAPLE worksheet we’ll be using may be downloaded from our web page. 4.32: Another standard, sort of neat math, and good preparation for 8.03.

**A**

From previous problem sets, you should by now have a reasonable expression for the scalar potential φ(r, θ) corresponding to a conducting sphere of radius R_{0} in the presence of an external (“ambient”) electric field E_{a} = E_{az} ẑ.

Now, find the same result another way, but using the method of Image Charges for a Conducting Sphere, as presented in class on Wednesday October 20, and using the result and notation in the online notes by that name.

That is, consider E_{a} to be the result of two very large point charges ±Q_{a} at distances R ≫ R_{0}, on opposite sides of the conducting sphere. The magnitude of the ambient field would then be |E_{az} ẑ| = 2 |Q_{a}| /R^{2}. Show that the resultant image charges constitute a dipole, and that in the limit as R → ∞, but keeping |E_{az} ẑ| constant (by adjusting Q_{a}) you reproduce the previous results.