<img src="./images/capacitor_gap_connected.png" align="center" style="width: 500px";/> A parallel plate capacitor is attached to a battery which maintains a constant voltage difference V between the capacitor plates. While the battery is attached, the plates are pulled apart. The electrostatic energy stored in the capacitor 1. increases. 2. decreases. 3. stays constant. Note: * CORRECT ANSWER: B * Potential same; field is reduced; but shows up squared while d is increased, overall goes down

## Exam Rewrites * Circled grade is the grade you will earn if you do the rewrite * To earn the circled grade: * Complete the parts that you didn't earn full credit * Write one paragraph per problem (max 4 paragraphs) about what you didn't understand at the time of the exam, what you did to correct that understanding, and how that relates to the solution you've written up.

### Laplace's Equation <img src="./images/laplace.png" align="center" style="width: 900px";/>

<img src="./images/region_w_no_charge.png" align="right" style="width: 200px";/> A region of space contains no charges. What can I say about $V$ in the interior? 1. Not much, there are lots of possibilities for $V(r)$ in there 2. $V(r)=0$ everywhere in the interior. 3. $V(r)=$constant everywhere in the interior Note: * CORRECT ANSWER: A * Without the boundary information, we can't solve the problem. * But once we have BCs, we have a unique answer.

<img src="./images/region_with_no_charge_Vset.png" align="right" style="width: 200px";/> A region of space contains no charges. The boundary has V=0 everywhere. What can I say about $V$ in the interior? 1. Not much, there are lots of possibilities for $V(r)$ in there 2. $V(r)=0$ everywhere in the interior. 3. $V(r)=$constant everywhere in the interior Note: * CORRECT ANSWER: B * Uniqueness argues it must be B; satisfies all the results and the BCs

For the 1D Laplace problem ($\nabla^2 V = \partial^2 V/\partial x^2 = 0$), we can choose the following ansatz: 1. $k_0\,x$ 2. $k_0\,x+k_1$ 3. $k_0\,x^2+k_1\,x+k_2$ 4. Can't tell Note: * CORRECT ANSWER: B * It has to be a linear equation

<img src="./images/cubical_lattice.png" align="center" style="width: 300px";/> If you put a positive test charge at the center of this cube of charges, could it be in stable equilibrium? 1. Yes 2. No 3. ??? Note: * CORRECT ANSWER: B * Earnshaw's theorem: no electrostatic system can be held in stationary stable equilibirum * Just nudge the charge and it runs away