A parallel-plate capacitor has plates of area separated by a gap

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Electric fields in matter · Electric fields in matter

The Boundary-Value Problem, Ch Energy and force in dielectrics

Problem

A parallel-plate capacitor has plates of area

S = 0.015\,\mathrm{m^2}

separated by a gap

d = 0.80\,\mathrm{mm}

, completely filled with a linear dielectric liquid of relative permittivity

\varepsilon_r = 4.0

. A battery holds the plates at a fixed potential difference

V = 300\,\mathrm{V}

. One plate can slide so as to change the gap; take

x

to be the instantaneous gap, currently equal to

d

. ((a)) Starting from the work done bringing free charge in from infinity with the dielectric settled at each step, derive

\delta W_{\rm free} = \int\mathbf{E}\cdot\delta\mathbf{D}\,d\tau

, and show that for this linear dielectric it integrates to

W_{\rm free} = \tfrac12\int\mathbf{D}\cdot\mathbf{E}\,d\tau

. Evaluate

W_{\rm free}

for the capacitor as described, and check your result against

\tfrac12C(x)V^2

. ((b)) By accounting for the work the battery supplies as

x

changes by a small amount at fixed

V

, derive

F_x = +(\partial W/\partial x)_V

for this capacitor, and use it to find the magnitude and direction of the force on the sliding plate at

x=d

. ((c)) Model one molecule of the liquid as a charge on a spring of stiffness such that

\mathbf{p}=\alpha\mathbf{E}

, with

\alpha = 2.0\times10^{-40}\,\mathrm{C\,m^2\,V^{-1}}

, sitting in the field acting on it,

E = V/d

, found in part (a). Show that the work of stretching this dipole as the field is switched on is

+\tfrac12\alpha E^2

, and combine it with the interaction energy

-\alpha E^2

to find the molecule's total energy in the field.