Stray HW3 hints

(1) We set this one up in class. Put one charge at the origin, and write the distance from the second charge to the field point in terms of the distance to the first charge and the separation of the charges. E.g., if b is a vector pointing from one charge to the other,

\vec{r}^{\prime}= \vec{r}-\vec{b}\\
\hat{r} \equiv \frac{\vec{r}}{|\vec{r}|}\\
\vec{E}_1 = \frac{kq_1q_2}{r^2}\,\hat{r} = \frac{kq_1q_2\vec{r}}{r^3}\\
\vec{E}_2 = \frac{kq_1q_2}{r^{\prime\,}^2}\,\hat{r}^{\prime} = \frac{kq_1q_2 \left(\vec{r}-\vec{b}\right)}{|\vec{r}-\vec{b}|^3}

Do not forget that your volume element in spherical coordinates has an r-squared, or you will get a nasty integral.

(2) The field is nonzero only between the two spheres ... the result is the energy of a spherical capacitor, if you want to check your answer.

(3) Set this up in class. Solve for the potential at an arbitrary point, and either (a) show that at least 4 points from the same ellipse give the same potential, or (b) plug in the equation for an ellipse and show that it results in V=constant.

(4) See previous hint ...

(5) You can use the boundary conditions we derived for this one. From one side of a given plate to the other, the difference in the normal components of E has to give you the surface charge density.

E_{\mathrm{above}} - E_{\mathrm{below}} = \sigma/\epsilon_0

Above the top plate, or below the bottom plate, the field is zero. Thus, the field in between any two plates can be related to the surface charge on the upper or lower plate. How does that relate to the charge on a given side of the middle plate? Two parallel plates will have the same charge on the sides that are facing. Just like we talked about on Wednesday, if one plate has charge Q, it will induce -Q on the adjacent plate.

The only thing left is to realize that since the electric field must be constant inside (infinite plate, no spatial dependence!), the electric potential must just be

V = Ed = \sigma/\epsilon_0

This gets you the surface charge. Once you have that, you know the field. Once you know the field, you can use the method of problem 1 to find the total energy. Which can then be optimized. I bet you can guess the optimum spacing already though.

(6) Capacitance is just total charge divided by potential for a given conductor. If you know the potential for a prolate spheroid with a charge Q (problem 3!), C=Q/V. You can fancy that up by noting that the eccentricity (epsilon) is just d/a. After that, note

\ln{\left(\frac{1+x}{1-x}\right) = \ln{\left(1+x)} - \ln{\left(1-x)} \approx 2x

7. Here's the basic scheme for finding the coefficients. The self capacitance terms C_ii is just the capacitance of the system when the two elements are joined together, so they have the same charge and the potential of both is zero. The C_ij off-diagonal terms are found by setting the potential of one of the conductors to zero, and having a charge Q on the other. You'll need the formula for the capacitance of a spherical capacitor, which you can find in your text or online.

We'll go over that in class on Friday, but here's what you should come up with:

C_{11}=\frac{ab}{k\left(b-a\right)} = -C_{21}=-C_{12} \\ 
C_{22}=\frac{b^2}{k\left(b-a\right)}

8. Textbook problem. Try checking in some textbooks :-) Feynman (Vol. II) does a great job.