<img src="./images/hung_spheres.png" align="left" style="width: 300px";/> Two small spheres (mass, $m$) are attached to insulating strings (length, $L$) and hung from the ceiling as shown. How does the angle (with respect ot the vertical) that the string attached to the $-q$ charge ($\theta_1$) compare to that of the $-2q$ charge ($\theta_2$)? 1. $\theta_1 > \theta_2$ 2. $\theta_1 = \theta_2$ 3. $\theta_1 < \theta_2$ 4. ???? Note: CORRECT ANSWER: B Draw the FBD, Newton 3 is important here.
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## Classical Electromagnetism <img src="./images/jupiter.jpg" align="left" width="400px";/> <img src="./images/helium.png" align="right" width="400px";/> $\sim 10^8\mathrm{m} \longrightarrow \longrightarrow \longrightarrow\longrightarrow\longrightarrow\longrightarrow \sim 10^{-16}\mathrm{m}$ **24 orders of magnitude** Note: As far we know, the theory of Classical E&M works for 24 orders of magnitude. It breaks down on the subatomic scale (QM), but we really don't know how far it extends one the upper-end as it is consistent with relativity. Tests of Jupiter's magnetic field are consistent with predictions from Classical E&M.
## Electrostatics <img src="./images/efieldwiki.png" align="center";/>
5 charges, q, are arranged in a regular pentagon, as shown. What is the E field at the center? <img src ="./images/5charges.png" align="center" style="width: 250px";/> 1. Zero 2. Non-zero 3. Really need trig and a calculator to decide Note: CORRECT ANSWER: A
1 of the 5 charges has been removed, as shown. What’s the E field at the center? <img src ="./images/4charges.png" align="center" style="width: 400px";/> 1. $+(kq/a^2)\hat{y}$ 2. $-(kq/a^2)\hat{y}$ 3. 0 4. Something entirely different! 5. This is a nasty problem which I need more time to solve Note: CORRECT ANSWER: B Superposition!
If all the charges live on a line (1-D), use: $$\lambda \equiv \dfrac{\mathrm{charge}}{\mathrm{length}}$$ Draw your own picture. What's $\mathbf{E}(\mathbf{r})$?
To find the E-field at P from a thin line (uniform charge density $\lambda$): <img src ="./images/linecharge.png" align="right" style="width: 400px";/> $$\mathbf{E}(\mathbf{r}) = \dfrac{1}{4\pi\varepsilon_0}\int \dfrac{\lambda dl'}{\mathfrak{R}^2}\hat{\mathfrak{R}}$$ What is $\mathfrak{R}$? 1. $x$ 2. $y'$ 3. $\sqrt{dl'^2 + x^2}$ 4. $\sqrt{x^2+y'^2}$ 5. Something else Note: CORRECT ANSWER: D
$\mathbf{E}(\mathbf{r}) = \int \dfrac{\lambda dl'}{4\pi\varepsilon_0\mathfrak{R}^3}\vec{\mathfrak{R}}$, so: $E_x(x,0,0) = \dfrac{\lambda}{4\pi\varepsilon_0}\int \dots$ <img src ="./images/linecharge_coords.png" align="right" style="width: 400px";/> 1. $\int \dfrac{dy'x}{x^3}$ 2. $\int \dfrac{dy' x}{(x^2 + y'^2)^{3/2}}$ 3. $\int \dfrac{dy' y'}{x^3}$ 4. $\int \dfrac{dy' y'}{(x^2+y'^2)^{3/2}}$ 5. Something else Note: CORRECT ANSWER: B
What do you expect to happen to the field as you get really far from the rod? $$E_x = \dfrac{\lambda}{4\pi\varepsilon_0\}\dfrac{L}{x\sqrt{x^2+L^2}}$$ 1. $E_x$ goes to 0. 2. $E_x$ begins to look like a point charge. 3. $E_x$ goes to $\infty$. 4. More than one of these is true. 5. I can't tell what should happen to $E_x$. Note: CORRECT ANSWER: D (A and B)