Electrostatic Energy

$δ W = \int Φ δ ρ d V, d q = δ ρ d V$ Recall: $δ ρ = \nabla \cdot δ \vec{D}$ . Then, $δ W = \int Φ (\nabla \cdot δ \vec{D}) d V = \int \nabla \cdot (Φ δ \vec{D}) d V - \int δ \vec{D} \cdot \nabla Φ d V = \int Φ δ \vec{D} \cdot \hat{n} d a - \int δ \vec{D} \cdot \nabla Φ d V$ . If we choose a very large area (consider a sphere of radius $R$ ), $Φ \propto \frac{1}{R}$ and $\vec{D} \propto \frac{1}{R^{2}}$ . The surface integration is then proportional to $\frac{1}{R}$ . Thus, it it negigible.

The second term, $δ W = \int δ \vec{D} \cdot \vec{E} d V$ . For a linear, isotropic, dielectric. $δ \vec{D} = ε δ \vec{E}$ . So, $δ \vec{D} \cdot \vec{E} = δ (\frac{1}{2} ε δ \vec{E} \cdot \vec{E})$ . Then, $δ W = δ \int \frac{ε}{2} \int \vec{E} \cdot \vec{E} d V$ . So, $W = \frac{1}{2} \vec{D} \cdot \vec{E} = \frac{ε}{2} E^{2}$ .

Consider a parallel plate capacitor width $d$ and area $A$ , with a potential $V$ . Each plate has charge $Q$ . A dielectric is inserted between $ε$ . What is the total energy stored?

$Q = C V$ . Note: $V = E d$ and $Q = ε \nabla \cdot \vec{E} = A ε E$ . So, $Q = \frac{ε A}{d} V$ . Then, $C = \frac{Q}{V} = \frac{ε A}{d}$ . $W = \frac{1}{2} C V^{2}$ .

$\frac{1}{2} ε E^{2} \Rightarrow W = \frac{1}{2} ε \frac{V^{2}}{d^{2}} A d = \frac{1}{2} C V^{2}$ .

Consider a system of $N$ plates.

Assume that the charge $Q$ is constant in the system. If we move a plate. $E_{t o t a l} = T + W$ . $δ W = - \vec{F} \cdot δ \vec{x}, 0 = δ F_{t o t a l} = \vec{F} \cdot δ \vec{x} + δ W$ . So, $\vec{F} = - (\nabla W)_{Q}$ .

Consider a system of fixed potentials, $V_{i}$ . $δ E_{t o t a l} = δ T + δ W$ . Call the $δ E_{t o t a l} = δ W_{b}$ for the energy in the batteries. $W = \sum_{i} \frac{1}{2} Q_{i} V_{i}$ . $δ W = \frac{1}{2} \sum_{i} V_{i} δ Q_{i}$ . $δ W_{b} = \sum_{i} δ Q_{i} V_{i}$ . Note, the energy is twice the energy that is supplied by the potential. $δ W = 2 δ W_{b}$ . Thus, $\vec{F} \cdot δ \vec{W} = δ W \Rightarrow \vec{F} = (\nabla W)_{V}$ .

Consider a system of dielectrics and conductors, $(Q_{i}, V_{i}, ε_{i})$ . $W = \frac{1}{2} \int \vec{D} \cdot \vec{E} d^{3} x$ . $w = \frac{1}{2} \vec{D} \cdot \vec{E}$ . For a linear dielectric, $w = \frac{1}{2} ε | E |^{2} \propto I$ . $ε | E |^{2} = N ℏ ω$ .

Consider if $Q_{t o t a l}$ is constant. Then, $δ E_{t o t a l} = 0 = T + W$ . If we move a plate, $0 = δ E_{t o t a l} = δ T + δ W = δ T - \vec{F} \cdot δ \vec{x}$ , $F_{i} = - {(\frac{\partial W}{\partial x_{i}})}_{Q}$ .

Consider if $V_{t o t a l}$ is constant. Then, we have energy exchange between system and environment (i.e. through batteries). $δ E_{t o t a l} = \vec{F} \cdot δ \vec{x} + δ W = δ W_{b}$ . So, $δ W_{b} = \sum_{i} δ Q_{i} V_{i}$ . $W = \frac{1}{2} \sum_{i} Q_{i} V_{i}$ . $δ W = \frac{1}{2} \sum_{i} δ Q_{i} V_{i} = \frac{1}{2} δ W_{b}$ . So, $F_{i} = {(\frac{\partial W}{\partial x_{i}})}_{V}$ .

See PQ31, $E = \frac{V}{d}$ . $w = \frac{1}{2} ε E^{2}, \frac{1}{2} ε_{0} E^{2}$ . $W = \frac{1}{2} (ε \frac{V^{2}}{d^{2}} x w d + ε_{0} \frac{V^{2}}{d^{2}} (l - x) w d) = \frac{V^{2}}{2 d} (ε x + ε_{0} (l - x)) w$ . Note, we have $σ_{d} = \frac{ε V}{d}, σ_{0} = \frac{ε_{0} V}{d}$ in each region.

$C V = Q$ . Capacitors add in parallel.

If we had a constant $Q$ , the capacitance doesn’t change. So, $W = \frac{1}{2} \frac{Q^{2}}{C} = \frac{Q^{2}}{2} {(\frac{d}{(ε x + ε_{0} (l - x)) w})}^{2}$ . We can get $F_{x} = - {(\frac{\partial W}{\partial x})}_{Q} = \frac{Q^{2}}{2} \frac{1}{C^{2}} \frac{d C}{d x} == \frac{1}{2} \frac{Q^{2}}{C^{2}} (ε - ε_{0}) \frac{w}{d} = \frac{1}{2} (ε - ε_{0}) \frac{w}{d} V (x)^{2}$ . Here, we have $σ_{d} (x) = \frac{ε V (x)}{d}, σ_{0} = \frac{ε_{0} V (x)}{d}$ .