Capacitor with a Dielectric — Question 7

Question 7

A partially inserted dielectric experiences an inward force under either constraint

A dielectric of relative permittivity $\kappa>1$ slides distance x into a rectangular plate gap d. Plate width is w and length L; $0<x<L$. Use $C(x)=\epsilon_0w[L+(\kappa-1)x]/d$ and neglect edge corrections.

1. Find the inward electric force at fixed voltage V.

2. Find it at fixed charge Q and compare how each depends on insertion depth.

Question 7 - Solution

A partially inserted dielectric experiences an inward force under either constraint

(a) Method and calculation

$$C\prime(x)=\epsilon_0w(\kappa-1)/d.$$$$F_{x,V}=\tfrac12V^2C\prime(x)=\frac{\epsilon_0w(\kappa-1)V^2}{2d}>0.$$

(b) Interpretation and further calculation

$$F_{x,Q}=-\frac d{dx}\frac{Q^2}{2C(x)}=\frac{Q^2C\prime(x)}{2C(x)^2}>0.$$$$F_{x,V}\text{ is constant};\quad F_{x,Q}\propto[L+(\kappa-1)x]^{-2}.$$

Physical check

The force remains inward at fixed charge, but weakens as the increasing capacitance lowers the voltage.