# Why is the electric field inside the hole non-zero?

Question: A sphere of radius $$a$$ is made of a nonconducting material that has a uniform volume charge density $$\rho$$. A spherical cavity of radius $$b$$ is removed from sphere which is a distance $$z$$ from the center of the sphere. Assume that $$a > z + b$$. What is the electric field in the cavity?

I understand the solution but what puzzles me is that if i apply the Gauss theorem to the cavity the electric field must be 0? many thanks in advance

• The important thing is that it is a non-conducting material. The charges cannot move when the cavity is created. Nov 16 '17 at 22:15
• You can't apply Gauss's law and get zero, because there is no relevant symmetries to argue that the field is uniform over any surface.
– Chris
Nov 16 '17 at 23:50
• Essentially the same question asked and answered here Nov 17 '17 at 0:14

Gauss' law here is not really useful in your setup because $$\oint \vec E\cdot d\vec S\ne \vert \vec E\vert \oint dS \tag{1}$$ i.e. the field is not radial on any Gaussian spherical surface with centre coinciding with the centre of your hole. It is only when $$\oint \vec E\cdot d\vec S = \vert \vec E\vert \oint dS =\vert \vec E\vert S = \frac{q}{\epsilon} \tag{2}$$ that you can then invert (2) to deduce $\vert \vec E\vert$ on the Gaussian, and (2) can be obtained only if $\vert \vec E\vert$ is constant on the surface so you can "pull it out" of the integral.

It is still true that $$\oint \vec E\cdot d\vec S=0$$ since a Gaussian sphere inside you hole encloses no charge, but because of (1) you cannot conclude that $\vert \vec E\vert=0$.

[I take it you know the field inside the hole is constant. You can show this using the superposition principle.]

• thanks for your prompt answer, so just to clarify my doubts, i thought that gauss theorem applied in my case means that when there is no charge inside the whole the electric field must be 0 inside regardless of any other electric filed from outside, but according to you we should take into consideration the other electric fields, please correct me of i m wrong, thanks!! Nov 16 '17 at 22:35
• No. You can only deduce the value of $\vert \vec E\vert$ if (2) holds, i.e. if the situation is so that the field is constant on the Gaussian surface. If you know it is constant (but of unknown magnitude) and you know its direction, you can use (2) to find $\vert \vec E\vert$. If it is not constant, (1) is valid and you cannot "isolate" for $\vert \vec E\vert$ since you know nothing of the rhs of (1) Nov 16 '17 at 22:37

I understand the solution but what puzzles me is that if i apply the Gauss theorem to the cavity the electric field must be 0?

Since there is no charge in the hollowed out volume, Gauss' law tells us that the flux of the electric field (through the surface bounding that volume) is zero, not that the electric field within the volume is zero.

Put another way, if every electric field line that enters the volume leaves the volume, the flux through the surface is zero (do you see why?) but the volume has non-zero electric field within.

Recall that electric field lines start (end) on positive (negative) charge. If there is no charge within the volume, no electric field line can originate / terminate there and thus, any field line that enters the region must exit the region.