We derived the electric potential outside ($r>R$) the charged shell to be $$V(r) = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$$ What is it for $r<R$? 1. Zero 2. Constant 3. It changes but I don't know how yet 4. Something else Note: * Correct Answer: B

### Exam 1 Information * Exam 1 on Wednesday, October 4th (A149 PSS) - Across from FRIB (Wilson side) * 7pm-9pm - Arrive on time! * I will provide a formula sheet (posted on Piazza already) * You can bring one-side of a sheet of paper with your own notes. * 4 questions - True/False, Essay, Graphing, Calculations

### What's on Exam 1? * Identify whether conceptual statements about $\mathbf{E}$, $V$, $\rho$, and/or numerical integration are true or false. * Sketch and discuss delta functions in relation to charge density, $\rho$ * Calculate the electric field, $\mathbf{E}$, inside and outside a continuous distribution of charge and sketch the results * Calculate the electric potential, $V$, for a specific charge distribution and discuss what happens in limiting cases

<img src="./images/graph_shell.png" align="center" style="width: 400px";/> Could this be a plot of $\left|\mathbf{E}(r)\right|$? Or $V(r)$? (for SOME physical situation?) 1. Could be $E(r)$, or $V(r)$ 2. Could be $E(r)$, but can't be $V(r)$ 3. Can't be $E(r)$, could be $V(r)$ 4. Can't be either 5. ??? Note: * Correct Answer: B

We usually choose $V(r\rightarrow\infty) \equiv 0$ when calculating the potential of a point charge to be $V(r) = +kq/r$. How does the potential $V(r)$ change if we choose our reference point to be $V(R) = 0$ where $R$ is close to $+q$. 1. $V(r)$ higher than it was before 2. $V(r)$ is lower than it was before 4. $V(r)$ doesn’t change ($V$ is independent of choice of reference) Note: * CORRECT ANSWER: B * Show redefinition.

### Electrostatic Potential Energy <img src="./images/cathode_ray_tube.png" align="center" style="width: 600px";/>

<img src="./images/three_charges.png" align="right" style="width: 300px";/> Three identical charges $+q$ sit on an equilateral triangle. What would be the final $KE$ of the top charge if you released it (keeping the other two fixed)? 1. $\frac{1}{4\pi\varepsilon_0}\frac{q^2}{a}$ 2. $\frac{1}{4\pi\varepsilon_0}\frac{2q^2}{3a}$ 3. $\frac{1}{4\pi\varepsilon_0}\frac{2q^2}{a}$ 4. $\frac{1}{4\pi\varepsilon_0}\frac{3q^2}{a}$ 5. Other Note: CORRECT ANSWER: C

<img src="./images/three_charges.png" align="right" style="width: 300px";/> Three identical charges $+q$ sit on an equilateral triangle. What would be the final $KE$ of the top charge if you released *all three*? 1. $\frac{1}{4\pi\varepsilon_0}\frac{q^2}{a}$ 2. $\frac{1}{4\pi\varepsilon_0}\frac{2q^2}{3a}$ 3. $\frac{1}{4\pi\varepsilon_0}\frac{2q^2}{a}$ 4. $\frac{1}{4\pi\varepsilon_0}\frac{3q^2}{a}$ 5. Other Note: CORRECT ANSWER: A