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 \(\sigma_d(x) = \frac{\varepsilon V(x)}{d},\sigma_0=\frac{\varepsilon_0 V(x)}{d}\).