# Electrostatic Energy

$$\delta W = \int \Phi \delta\rho dV,dq = \delta\rho dV$$ Recall: $$\delta\rho = \nabla\cdot\delta\vec{D}$$. Then, $$\delta W = \int \Phi(\nabla\cdot\delta\vec{D})dV = \int\nabla\cdot(\Phi \delta\vec{D})dV - \int \delta\vec{D}\cdot\nabla\Phi dV = \int \Phi\delta\vec{D}\cdot\hat{n}da - \int \delta\vec{D}\cdot\nabla\Phi dV$$. If we choose a very large area (consider a sphere of radius $$R$$), $$\Phi\propto \frac{1}{R}$$ 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, $$\delta W = \int\delta\vec{D}\cdot\vec{E}dV$$. For a linear, isotropic, dielectric. $$\delta \vec{D} = \varepsilon\delta\vec{E}$$. So, $$\delta\vec{D}\cdot\vec{E} = \delta \left(\frac{1}{2}\varepsilon \delta\vec{E}\cdot\vec{E}\right)$$. Then, $$\delta W = \delta\int\frac{\varepsilon}{2}\int\vec{E}\cdot\vec{E}dV$$. So, $$W = \frac{1}{2}\vec{D}\cdot\vec{E} = \frac{\varepsilon}{2}E^2$$.

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

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

$$\frac{1}{2}\varepsilon E^2\Rightarrow W = \frac{1}{2}\varepsilon \frac{V^2}{d^2}Ad = \frac{1}{2}CV^2$$.

Consider a system of $$N$$ plates.

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

Consider a system of fixed potentials, $$V_i$$. $$\delta E_{total} = \delta T + \delta W$$. Call the $$\delta E_{total} = \delta W_{b}$$ for the energy in the batteries. $$W = \sum_i\frac{1}{2}Q_iV_i$$. $$\delta W = \frac{1}{2}\sum_iV_i\delta Q_i$$. $$\delta W_b = \sum_i \delta Q_i V_i$$. Note, the energy is twice the energy that is supplied by the potential. $$\delta W = 2\delta W_b$$. Thus, $$\vec{F}\cdot\delta\vec{W} = \delta W\Rightarrow \vec{F} = (\nabla W)_V$$.

Consider a system of dielectrics and conductors, $$(Q_i,V_i,\varepsilon_i)$$. $$W = \frac{1}{2}\int \vec{D}\cdot\vec{E} d^3x$$. $$w = \frac{1}{2}\vec{D}\cdot\vec{E}$$. For a linear dielectric, $$w = \frac{1}{2}\varepsilon |E|^2 \propto I$$. $$\varepsilon|E|^2 = N\hbar\omega$$.

Consider if $$Q_{total}$$ is constant. Then, $$\delta E_{total} = 0 = T + W$$. If we move a plate, $$0 = \delta E_{total} = \delta T + \delta W = \delta T - \vec{F}\cdot\delta\vec{x}$$, $$F_i = -\left(\frac{\partial W}{\partial x_i}\right)_{Q}$$.

Consider if $$V_{total}$$ is constant. Then, we have energy exchange between system and environment (i.e. through batteries). $$\delta E_{total} = \vec{F}\cdot\delta\vec{x} + \delta W = \delta W_b$$. So, $$\delta W_b = \sum_i\delta Q_iV_i$$. $$W = \frac{1}{2}\sum_i Q_iV_i$$. $$\delta W = \frac{1}{2}\sum_i \delta Q_iV_i = \frac{1}{2}\delta W_b$$. So, $$F_i = \left(\frac{\partial W}{\partial x_i}\right)_{V}$$.

See PQ31, $$E = \frac{V}{d}$$. $$w = \frac{1}{2}\varepsilon E^2, \frac{1}{2}\varepsilon_0 E^2$$. $$W = \frac{1}{2}\left(\varepsilon\frac{V^2}{d^2}xwd + \varepsilon_0\frac{V^2}{d^2}(l-x)wd\right) = \frac{V^2}{2d}(\varepsilon x + \varepsilon_0(l-x))w$$. Note, we have $$\sigma_d = \frac{\varepsilon V}{d},\sigma_0 = \frac{\varepsilon_0 V}{d}$$ in each region.

$$CV=Q$$. Capacitors add in parallel.

If we had a constant $$Q$$, the capacitance doesn’t change. So, $$W = \frac{1}{2}\frac{Q^2}{C} = \frac{Q^2}{2}\left(\frac{d}{(\varepsilon x+\varepsilon_0(l-x))w}\right)^2$$. We can get $$F_x = -\left(\frac{\partial W}{\partial x}\right)_{Q} = \frac{Q^2}{2}\frac{1}{C^2}\frac{dC}{dx} = = \frac{1}{2}\frac{Q^2}{C^2}(\varepsilon-\varepsilon_0)\frac{w}{d} = \frac{1}{2}(\varepsilon-\varepsilon_0)\frac{w}{d}V(x)^2$$. Here, we have $$\sigma_d(x) = \frac{\varepsilon V(x)}{d},\sigma_0=\frac{\varepsilon_0 V(x)}{d}$$.

Created: 2023-06-25 Sun 02:32

Validate