# Why the electic field in the overlap region of solid sphere is not zero?

Two solid spheres with radius r and carrying uniform volume charge density +p and -p partially overlap. Why in the region where the spheres overlap, the net electric field is not zero but constant. I am asking this because:

• the net charge inside the overlap region is zero as there are equal number of positive and negative charges
• if a closed surface does not have any charge inside where an electric field line can terminate, then any electric field line entering the surface at one point must necessarily exit at some other point of the surface, then the electric flux through the surface is zero and hence electric field is zero.

Now, the overlap region can be imagined as a "closed surface" with no charge(as the net charge is zero) and we can say that the charge outside the overlap region is "external charge". As per Gauss's Law :

• The electric flux through a closed surface due to a charge outside that surface is zero.

• Charges are enclosed, but because the net charge included is zero, the net flux through the closed surface is also zero.

I am unable to understand where I am wrong; maybe in Gauss's law by closed surface they are talking about hollow closed surface.

## 1 Answer

You are incorrectly using Gauss' law: if there is no charge enclosed, it does not necessarily imply that $$\vec E=0$$, i.e. in other words it's now always true that $$\oint \vec E\cdot d\vec S=0\quad\Rightarrow \quad \vec E=0$$.

As a simple example, consider an empty box placed in a constant electric field: clearly then $$\oint \vec E\cdot d\vec S=0$$ as there is no charge enclosed in the box but the field is certainly not $$0$$. To use Gauss' law, you need to find a surface on which $$\vec E\cdot d\vec S=\vert \vec E\vert dS \cos\theta$$ is constant, so that $$\oint \vec E\cdot d\vec S = \vert \vec E\vert S$$, i.e. so that you can pull out the "constant $$\vert \vec E\vert \cos\theta$$" outside the integral sign. In the example of the box, $$\vec E\cdot d\vec S$$ is positive on one side, on another it's negative, and on 4 others the dot product is $$0$$ because $$\vec E$$ and $$d\vec S$$ are perpendicular. In your specific example, there is no reason to believe that the net $$\vert \vec E\vert$$ on a Gaussian surface around your overlap region will be constant, thus little chance that on your surface $$\oint \vec E\cdot d\vec S = \vert \vec E\vert S$$.

[Figure credit: E&M TIPERs, Hieggelke et al, Pearson Prentice Hall 2006]