- #1

gruba

- 206

- 1

## Homework Statement

Plate capacitor with distance between plates 'd=4mm' is fully filled with homogeneous linear dielectric. Capacitor is connected to the source of DC voltage, first stationary state is established, then the capacitor is separated from DC voltage source. After that, dielectric is fully eliminated from capacitor and new stationary state is established. Difference between voltage in second and first stationary state is '(delta)U=144V'. Calculate intensity of polarization vector 'P' of a dielectric in the first stationary state (capacitor is separated from DC voltage source, and dielectric is not eliminated).

## Homework Equations

((integral)DdS)=Q

D=e

_{0}E+P

(delta)U=U

_{2}-U

_{1}

## The Attempt at a Solution

Law of conservation of charge is valid, so Q=const.

In the first stationary state, we have a plate capacitor with dielectric, so we use Gauss law for dielectrics:

((integral)DdS)=Q

D=Q/S=e

_{0}E+P (1)

In the second stationary state, we have a plate capacitor without dielectric (vacuum), so we use Gauss law for vacuum:

((integral)EdS)=Q/e

_{0}

e

_{0}E=Q/S (2)

Combining (1) and (2) gives

P=e

_{0}(E

_{0}-E)=e

_{0}*(delta)U/d

I get the intensity of polarization vector P=318.6 nC/m

^{2}

What puzzles me is that we need to find intensity of polarization vector in the FIRST stationary state, so we need to find voltage U

_{1}in that state? Also, is there intensity of polarization vector if there are no dielectrics?

Thanks.