# Relationship between applied electric field and resulting surface charge desnity

1. Dec 6, 2014

### sagarbhathwar

I had this problem in my exam. It required me to calculate relative permittivity of a dielectric slab which acquires a surface charge density of 800micro Coulomb/m^2 when subjected to an electric field of 10^8 V/m. Applying E=sigma/(2epsilon), I got relative permittivity as 0.45 which is less than 1. Considering its only a hypothetical question, is my approach right or is it wrong? How can I solve this?

2. Dec 6, 2014

### ehild

The surface charge density induced in an electric field does not mean that the dielectric slab acquires net charge. Read http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html
The permittivity of the dielectric is equal to ε=D/E where D is the electric displacement and E is the electric field. D=ε0E+P , P is the polarization density. The normal component of D is the same inside and outside the dielectric at the surface of the dielectric slab.
What is D outside? How is P related to the surface charge density?

Last edited: Dec 6, 2014
3. Dec 7, 2014

### rude man

Bit confused as to what the problem says. I suppose the idea is the E field outside the slab is 10^8 V/m & you can solve the problem that way. Use a gaussian surface (e.g. "pill box") going from the slab into the external 10^8 V/m field.

But just FYI, if you started with a 10^8 V/m E field and then stuck a slab into it the external E field would itself be altered (raised), depending on the relative thicknesses of the slab and the (presumably) parallel plate separation with E = 10^8 V/m before the slab was introduced, resulting in a higher surface charge density.

4. Dec 7, 2014

### sagarbhathwar

The problems says certain charge density is acquired when subjected to electric field. Nothing else at all

5. Dec 7, 2014

### rude man

Sure. Just do the problem per my first paragraph in my previous post.

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