Given the two diff. eq's : $$\dfrac{1}{X}\dfrac{d^2X}{dx^2} = C_1 \qquad \dfrac{1}{Y}\dfrac{d^2Y}{dy^2} = C_2$$ where $C_1+C_2 = 0$. Given the boundary conditions in the figure, which coordinate should be assigned to the negative constant (and thus the sinusoidal solutions)? <img src="./images/cq_cartesian_bc_1.png" align="right" style="width: 400px";/> 1. x 2. y 3. $C_1 = C_2 = 0$ here 4. It doesn't matter. Note: * CORRECT ANSWER: B
Given the two diff. eq's : $$\dfrac{1}{X}\dfrac{d^2X}{dx^2} = C_1 \qquad \dfrac{1}{Y}\dfrac{d^2Y}{dy^2} = C_2$$ where $C_1+C_2 = 0$. Given the boundary conditions in the figure, which coordinate should be assigned to the negative constant (and thus the sinusoidal solutions)? <img src="./images/cq_cartesian_bc_2.png" align="right" style="width: 400px";/> 1. x 2. y 3. $C_1 = C_2 = 0$ here 4. It doesn't matter. Note: * CORRECT ANSWER: C * It's constant throughout!
When does $\sin(ka)e^{-ky}$ vanish? 1. $k = 0$ 2. $k = \pi/(2a)$ 3. $k = \pi/a$ 4. A and C 5. A, B, C Note: * CORRECT ANSWER: D
Suppose $V_1(r)$ and $V_2(r)$ are linearly independent functions which both solve Laplace's equation, $\nabla^2 V = 0$. Does $aV_1(r)+bV_2(r)$ also solve it (with $a$ and $b$ constants)? 1. Yes. The Laplacian is a linear operator 2. No. The uniqueness theorem says this scenario is impossible, there are never two independent solutions! 3. It is a definite yes or no, but the reasons given above just aren't right! 4. It depends... Note: * CORRECT ANSWER: A
What is the value of $\int_0^{2\pi} \sin(2x)\sin(3x)\;dx$ ? 1. Zero 2. $\pi$ 3. $2 \pi$ 4. other 5. I need resources to do an integral like this! Note: * CORRECT ANSWER: A * Orthogonal functions!
### Exact Solutions: $$V(x,y) = \sum_{n=1}^{\infty} \dfrac{4V_0}{n\pi}\dfrac{1}{\cosh\left(\frac{n\pi}{2}\right)}\cosh\left(\frac{n\pi x}{a}\right)\sin\left(\frac{n \pi y}{a}\right)$$ ### Approximate Solutions: ### (1 term; 20 terms) <img src="./images/saddle_potential.png" align="center" style="width: 400px";/> <img src="./images/saddle_potential_20.png" align="center" style="width: 400px";/>
### Separation of Variables (Spherical) <img src="./images/metal_in_ext_field.jpg" align="center" style="width: 500px";/>
Given $\nabla^2 V = 0$ in Cartesian coords, we separated $V(x,y,z) = X(x)Y(y)Z(z)$. Will this approach work in spherical coordinates, i.e. can we separate $V(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)$? 1. Sure. 2. Not quite - the angular components cannot be isolated, e.g., $f(r,\theta,\phi) = R(r)Y(\theta,\phi)$ 3. It won't work at all because the spherical form of Laplace's Equation has cross terms in it (see the front cover of Griffiths) Note: * CORRECT ANSWER: A