Force on a Layer of Variable Charge
1. Concept: Maxwell Stress / Direct Integration
The force on a distribution of charge can be found by integrating $dF = dq E$. Consider a layer of thickness $dx$. The charge in this layer is $dq = \rho A dx$. From Gauss’s Law, $\frac{dE}{dx} = \frac{\rho}{\epsilon_0}$, so $\rho = \epsilon_0 \frac{dE}{dx}$.
2. Integration
Substitute $\rho$ into the force element:
$$ dF = (\epsilon_0 \frac{dE}{dx} A dx) E = \epsilon_0 A E dE $$To find the total force per unit area ($f = F/A$), integrate from the left face (field $E_1$) to the right face (field $E_2$):
$$ \frac{F}{A} = \int_{E_1}^{E_2} \epsilon_0 E dE $$ $$ \frac{F}{A} = \epsilon_0 \left[ \frac{E^2}{2} \right]_{E_1}^{E_2} $$3. Final Expression
$$ \frac{F}{A} = \frac{1}{2}\epsilon_0 (E_2^2 – E_1^2) $$
This result represents the difference in the “Maxwell stress” (electric pressure) acting on the two sides of the layer.