Pre-Quantum Electrodynamics

• PM 10
• Gr 4.4.4

As for a conductor: a dielectric is attracted into an electric field. Calculations can be complicated: parallel plate capacitor with partially inserted dielectric: force comes from fringing field around edges.

Better: reason from energy. Pull dielectric out by \(dx\). Energy change equal to work done:

\[ dW = F_{me} dx \] where \(F_{me}\) is mechanical force exerted by external agent. \(F_{me} = -F\), where \(F\) is electrical force on dielectric. Electrical force on slab:

\[ F = -\frac{dW}{dx} \] Energy stored in capacitor:

\[ W = \frac{1}{2} C V^2 \tag{WcapV}\label{WcapV} \] Capacitance in configuration considered:

\[ C = \frac{\varepsilon_0 w}{d} (\varepsilon_r l - \chi_e x) \] where \(l\) is the length of the plates, and \(w\) is their width. Assume total charge \(Q\) on each plate is held constant as \(x\) changes. In terms of \(Q\),

\[ W = \frac{1}{2} \frac{Q^2}{C} \tag{WcapQ}\label{WcapQ} \] so

\[ F = -\frac{dW}{dx} = \frac{1}{2} \frac{Q^2}{C^2} \frac{dC}{dx} = \frac{1}{2} V^2 \frac{dC}{dx}. \] But \[ \frac{dC}{dx} = -\frac{\varepsilon_0 \chi_e w}{d} \] so

\[ F = -\frac{\varepsilon_0 \chi_e w}{2d} V^2. \]

Common mistake: to use WcapV (for \(V\) constant) instead of WcapQ (for \(Q\) constant) in computing the force. In this case, sign is reversed, \[ F = -\frac{1}{2} V^2 \frac{dC}{dx}. \] Here, the battery also does work, so

\[ dW = F_{me} dx + V dQ \] and

\[ F = -\frac{dW}{dx} + V \frac{dQ}{dx} = -\frac{1}{2} V^2 \frac{dC}{dx} + V^2 \frac{dC}{dx} = \frac{1}{2} V^2 \frac{dC}{dx} \] so like before but with the correct sign.