### Pset 4, Abubakar Abid (It may take a second for the LaTeX-javascript library to load...)

6.1) In summation notation, we can expand the LHS as: $$= \overrightarrow{A} \times (\epsilon_{ijk} B_j C_k \hat{x}_i)$$ $$= \epsilon_{abc} A_b (\epsilon_{cjk} B_j C_k ) \hat{x}_a$$ $$= (\delta_{aj} \delta_{bk} - \delta_{ak} \delta_{bj}) A_b ( B_j C_k ) \hat{x}_a$$ $$= A_j ( B_j C_k ) \hat{x}_k - A_k ( B_j C_k ) \hat{x}_j$$ $$= (A \cdot B) C - B ( A \cdot C )$$ The identity follows from letting $$A = \nabla, B = \nabla, C = E$$

6.2 a) From the integral form of Gauss's Law, we have $$\int_S{\overrightarrow{E} \cdot d\overrightarrow{A}} = \frac{q}{\epsilon}$$ If the charge distribution is uniform and there are no charges between the plates, we can write: $$\int{\overrightarrow{E} \cdot d\overrightarrow{l}} \cdot A = \int{ \frac{q}{\epsilon} d\overrightarrow{l}}$$ $$V \cdot A = \frac{q}{\epsilon} d$$ $$\frac{A\epsilon}{d} = \frac{q}{V} \equiv C$$ 6.2 b) From $$q = CV = CdE$$ $$\frac{dq}{dt} = Cd\frac{dE}{dt}$$ $$I = Cd\frac{dE}{dt}$$ Using (a), $$I = A \epsilon \frac{dE}{dt}$$ $$I = A \frac{dD}{dt}$$ $$I = \int_S{ \frac{d\overrightarrow{D}}{dt} }$$ 6.2 c) Neglecting magnetic fields, we define $$U \equiv \frac{1}{2} \overrightarrow{E} \cdot \overrightarrow{D}$$ Integrating over the volume to get the total energy is easy: $$E_t = A d \frac{1}{2} |\overrightarrow{E}|^2 \epsilon$$ $$E_t = \frac{A}{2d} V^2 \epsilon$$ $$E_t = \frac{CV^2}{2}$$ 6.2 d) The product of voltage and amp-hours is energy, so using (c), we find that the area would have to be 8.1e8 m^2. With the same spacing, a stack of such capacitors would be 81,000 meters high.

6.3 a) Applying Ampere's Law (neglecting displacement current) and converting it to integral form using Stokes' Theorem, we have: $$\nabla \times \overrightarrow{H} = \overrightarrow{J}$$ $$\oint \overrightarrow{H} \cdot d\overrightarrow{l} = \int_S \overrightarrow{J} \cdot dA$$ Integrating over one meter, we find that the $$|H| = nI$$ 6.3 c) Neglecting magnetic fields, we define $$U \equiv \frac{1}{2} \overrightarrow{H} \cdot \overrightarrow{B}$$ Integrating over the volume to get the total energy: $$E_t = A l \frac{1}{2} |\overrightarrow{H}|^2 \mu$$ $$E_t = A l \frac{1}{2} n^2 |I|^2 \mu$$ Let $$N = nl$$, the number of turns. Then, $$E_t = A \mu \frac{N^2}{l} \frac{1}{2} |I|^2$$ Substituting for the inductance of a loop, $$E_t = \frac{L}{2} |I|^2$$ 6.3 c) The energy turns out to be 62.5 MJ by substitution into (b). The force is just that divided the relevant distance, or approximately 62.5 mega-Newtons.

6.4) The force per meter between two wires turns out to be: $$\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi d}$$ $$\frac{F}{l} = \frac{\mu_0 I^2}{2 \pi d}$$ $$\frac{F}{l} = \frac{\mu_0}{2 \pi} = 2 \cdot 10^-7 N$$ 6.5 a) The Poynting vector in free space will |E|^2/377 because |E|/|H| = 377 ohms in free space, so (1 kW/m^2) = |E|^2/377 -> 614 V/m

6.5 b) By analogous calculations, the fields would be 19,000 V/m and 19,000,000 V/m.