enter image description here I am reading Grifith Textbook and came across this example(Ex 4.5 in 4th edition). 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.

I am confused as to 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?


1 Answer 1


The fields of surface charges on the dielectric shell cancel. Think of the polarization in the dielectric shell as a charge $-q$ accumulated at radius $a$ and a charge $q$ accumulated at $b$. For radii greater than $b$ each causes a field of $\pm \frac{q}{4 \pi \varepsilon_0 r^2}$ (since both are spherically symmetric and cause a field that looks like that of a point charge) and adding them you get no net contribution to the electric field.


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