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## Homework Statement

A dielectric slab with a susceptibility ##\chi_e## rests on a conducting plate whose upper surface carries a free surface charge density ##\sigma##. Show that the polarization surface charge density ##\sigma_{pol}## on the lower face of the dialectic slab is:

##\sigma_{pol} = - \sigma \frac{\chi_e}{1+\chi_e}##

## Homework Equations

The polarization vector P is: ##P=\chi \epsilon_0 E##

The electric field between the surfaces is

##E=\frac{\sigma - \sigma_{pol}}{\epsilon_0} = \frac{\sigma}{\epsilon_0} \frac{1}{1+\chi}##

## The Attempt at a Solution

I believe in this case ##\sigma_{pol} = P##, so combining the two equations I get

##\sigma_{pol}=P=\chi_e \epsilon_0 \frac{\sigma}{\epsilon_0} \frac{1}{1+\chi}##

##= \sigma \frac{\chi_e}{1+\chi_e}##

However I did not get the minus sign in front of my expression. What is wrong?

I know that ##\sigma_{pol}## and ##\sigma## must be of opposite sign, but mathematically how do I bring in the negative sign? Is my approach to the problem correct?

Any help would be appreciated.