Question

A charged sphere of radius 'a' is surrounded by a dielectric shell with outer radius 'b' and inner radius 'a'. Using Spherical Gaussian surface, the D-field is calculated using the known free charge Q.

To calculate the electric field, the D-field is divided by permittivity. How the field outside the dielectric(r>b), is unaffected by the presence of dielectric. The field outside the dielectric is just as if it was calculated for point charge Q. How are the surface charges on the dielectric shell not affecting the E-field?

Answer 1

The electric field outside the dielectric is unaffected by its presence because it is due to the charge distribution on the conducting sphere. The surface charges on the dielectric shell do not affect the electric field outside because they produce an equal and opposite induced electric field.

Step-by-step explanation:

The electric field outside the dielectric, at a distance r > b, is unaffected by the presence of the dielectric and is just as if it was calculated for a point charge Q. This is because the electric field outside the dielectric is due to the charge distribution on the conducting sphere, which remains unchanged by the dielectric. The surface charges on the dielectric shell do not affect the electric field outside because they produce an equal and opposite induced electric field, which cancels out the effect of the surface charges.