This confusion is about the following problem:
Suppose the entire region below the plane z=0 is filled with uniform linear dielectric material of susceptibility ##χ_e##. Calculate the force on a point charge q situated a distance d above the origin.
Now, this problem is done in Griffiths' Electrodynamics book( Example 4.8). Its clear how to do this problem. But problem comes from a point of surface charge density ##\sigma_b## on the surface of the dielectric.
Let me clear myself.
We need to know the surface bound charge. The sign of the charge will be negative. There will be no volume charge as the dielectric is considered to be linear and homogeneous. Now, $$\sigma_b=P.n=P_z=\epsilon_0 \chi_e E_z$$
Where ##E_z## is the z component of the total field just inside the dielectric at z=0. This field is due to in part to q and in part to the bound charge itself. The z component of the field due to the bound charge is ##-\sigma_b/2 \epsilon_0##. From here we can get the total electric field ##E_z## and then the bound charge. Now using method of image we can get the potential at any point anywhere may be for z>0 or z<0.
But my confusion arises from the 2 in the denominator of ##\sigma_b/2 \epsilon_0##.This 2 comes when the perpendicular component of the electric field above and below the surface is eual in magnitude. So that using Gauss's law we can have a 2 in the denominator. But in this case, the field due to the bound surface charge is not the same in the dielectric and in the vacuum because of different permittivity (##\epsilon## and ##\epsilon_0## respectively). So we cant write a 2 in the denominator. Rather I think this surface charge will not be equally divided.
I hope I have made myself clear. I better request the reader to go through the Example 4.8 in Griffiths's Electrodynamics book.
Suppose the entire region below the plane z=0 is filled with uniform linear dielectric material of susceptibility ##χ_e##. Calculate the force on a point charge q situated a distance d above the origin.
Now, this problem is done in Griffiths' Electrodynamics book( Example 4.8). Its clear how to do this problem. But problem comes from a point of surface charge density ##\sigma_b## on the surface of the dielectric.
Let me clear myself.
We need to know the surface bound charge. The sign of the charge will be negative. There will be no volume charge as the dielectric is considered to be linear and homogeneous. Now, $$\sigma_b=P.n=P_z=\epsilon_0 \chi_e E_z$$
Where ##E_z## is the z component of the total field just inside the dielectric at z=0. This field is due to in part to q and in part to the bound charge itself. The z component of the field due to the bound charge is ##-\sigma_b/2 \epsilon_0##. From here we can get the total electric field ##E_z## and then the bound charge. Now using method of image we can get the potential at any point anywhere may be for z>0 or z<0.
But my confusion arises from the 2 in the denominator of ##\sigma_b/2 \epsilon_0##.This 2 comes when the perpendicular component of the electric field above and below the surface is eual in magnitude. So that using Gauss's law we can have a 2 in the denominator. But in this case, the field due to the bound surface charge is not the same in the dielectric and in the vacuum because of different permittivity (##\epsilon## and ##\epsilon_0## respectively). So we cant write a 2 in the denominator. Rather I think this surface charge will not be equally divided.
I hope I have made myself clear. I better request the reader to go through the Example 4.8 in Griffiths's Electrodynamics book.
