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.