Which charge distributions below produce a potential that looks like $\frac{C}{r^2}$ when you are far away? <img src="./images/multipole_charge_configs_1.png" align="center" style="width: 600px";/> E) None of these, or more than one of these! (For any which you did not select, how DO they behave at large r?) Note: * CORRECT ANSWER: E (Both C and D)
Which charge distributions below produce a potential that looks like $\frac{C}{r^2}$ when you are far away? <img src="./images/multipole_charge_configs_2.png" align="center" style="width: 600px";/> E) None of these, or more than one of these! (For any which you did not select, how DO they behave at large r?) Note: * CORRECT ANSWER: E (Both B and D)
In terms of the multipole expansion $V(r) = V(mono) + V(dip) + V(quad) + \dots$, the following charge distribution has the form: <img src="./images/multipole_charge_configs_3.png" align="center" style="width: 600px";/> 1. $V(r) = V(mono) + V(dip) +\;$ higher order terms 2. $V(r) = V(dip) +\;$ higher order terms 3. $V(r) = V(dip)$ 4. $V(r) =\;$ only higher order terms than dipole 5. No higher terms, $V(r) = 0$ for this one. Note: * CORRECT ANSWER: D
Which of the following distributions could have a dipole contribution to the potential far from the charges? <img src="./images/distributionA.png" align="left" style="width: 200px";/> <img src="./images/distributionB.png" align="center" style="width: 200px";/> <img src="./images/distributionC.png" align="left" style="width: 200px";/> D. None E. More than one! Note: * CORRECT ANSWER: E * We didn't specify where the origin!
In which situation is the dipole term the leading non-zero contribution to the potential? <img src="./images/multipole_charge_configs_4.png" align="center" style="width: 600px";/> 1. 1 and 3 2. 2 and 4 3. only 5 4. 1 and 5 5. Some other combo Note: * CORRECT ANSWER: D
Consider a single point charge at the origin. It will have ONLY a monopole contribution to the potential at a location $\mathbf{r} = \langle x,y,z\rangle$. As we have seen, if we move the charge to another location (e.g., $\mathbf{r}' = \langle 0,0,d \rangle$), the distribution now has a dipole contribution to the potential at $\mathbf{r}$! What the hell is going on here? 1. It's just how the math works out. Nothing has changed physically at $\mathbf{r}$. 2. There is something different about the field at $\mathbf{r}$ and the potential is showing us that. 3. I'm not sure how to resolve this problem.
## Polarization <img src="./images/diel.gif" align="center" style="width: 400px";/>
<img src="./images/charge_near_block.png" align="right" style="width: 400px";/> A stationary point charge $+Q$ is near a block of polarization material (a linear dielectric). The net electrostatic force on the block due to the point charge is: 1. attractive (to the left) 2. repulsive (to the right) 3. zero Note: * CORRECT ANSWER: A
<img src="./images/sphere_p0_z.png" align="right" style="width: 300px";/> The sphere below (radius $a$) has uniform polarization $\mathbf{P}_0$, which points in the $+z$ direction. What is the total dipole moment of this sphere? 1. zero 2. $\mathbf{P}_0 a^3$ 3. $4\pi a^3 \mathbf{P}_0/3$ 4. $\mathbf{P}_0$ 5. None of these/must be more complicated Note: * CORRECT ANSWER: C
<img src="./images/cube_p0_z.png" align="right" style="width: 300px";/> The cube below (side $a$) has uniform polarization $\mathbf{P}_0$, which points in the $+z$ direction. What is the total dipole moment of this cube? 1. zero 2. $a^3 \mathbf{P}_0$ 3. $\mathbf{P}_0$ 4. $\mathbf{P}_0/a^3$ 5. $2 \mathbf{P}_0 a^2$ Note: * CORRECT ANSWER: B