# Calculating Electric field using Gauss law

I solved a question in electricity using the Gauss law, but comparing with others revealed the possibility that my answer is incorrect.

We have a Sphere with radius $$R$$ and Charge density $$\rho$$. The center of the sphere is located in the origin. Now inside of that sphere, there's a smaller sphere, with 0 density (no charge), whos center is located in distance $$\vec{d}$$ from the origin.

For simplicity you may imaging that $$\vec{d}=y\hat{y}$$ or something along these lines.

The question is now what's the electric field (outside of the spheres, inside the big sphere, and inside the small sphere)?

This question can be solved using gauss law, but I'm unsure if inside of the small sphere you have $$E=0$$ or something else, because if you choose any close surface $$S$$, you get that there is no charge inside of that envelope.

If it isn't 0, should we treat the big sphere as one with density of $$\rho$$ and the smaller sphere with $$- \rho$$? If so, then why?

• I'm guessing that the large sphere is a non-conductor with a charge/volume represented by the blue shading. Commented Nov 22, 2020 at 19:18
• @R.W.Bird not sure what do you mean by a non-conductor (know what it is, but not how it's relevant here), and yes the blue shading means we have charge here (note that although this is a 2d picture, we are talking about 3d spheres) Commented Nov 22, 2020 at 19:28
• @snatchysquid is there sphere a conductor or not?
– Babu
Commented Nov 22, 2020 at 20:09
• @Buraian not sure what do you mean by "conductor", from the comments on your answer I can indeed say this is a big sphere with fixed charge distribution, with a spherical region without any charge (blue is charged, white is the spherical non-charged region) Commented Nov 22, 2020 at 20:43
• Basically if the charge is contained in a metallic region.. What is the nature of the boundary?
– Babu
Commented Nov 22, 2020 at 20:51

This problem doesn't have spherical symmetry, so there's no way to use Gauss's Law directly. (If I understand what you wrote correctly, you appear to be implicitly assuming that 0 charge enclosed in a region implies that $$E = 0$$ on its boundary, but this is only the case if there is sufficient symmetry; in general, all Gauss's law says is that the flux through the boundary of some region is equal to the charge enclosed.)

Like you say though, you can treat the problem as the superposition of a sphere of uniform density $$\rho$$ and a smaller sphere of uniform density $$-\rho$$, then use Gauss's Law on each of these systems separately, since they do have spherical symmetry (albeit with different origins).

Hope this helps. Happy to elaborate further if necessary.

Clarifications:

Superposition is just the idea that if you have two charge distributions $$\rho_1$$ and $$\rho_2$$ that alone create electric fields $$E_1$$ and $$E_2$$, then the charge distribution $$\rho_1 + \rho_2$$ will have electric field $$E_1 + E_2$$. This may seem like a trivial statement, but as you can see in the above problem it turns out to be quite powerful (and yes, here $$\rho_1$$ and $$\rho_2$$ correspond to two spheres which overlap in the region with zero charge, since $$\rho_1 + \rho_2 = \rho -\rho = 0$$ here).

Spherical symmetry occurs when all directions from the origin are indistinguishable in a system. So a uniformly charged sphere has spherical symmetry, but in the above problem there isn't spherical symmetry since the $$y$$ direction is distinguishable from the other directions (since there is a hole in the charge distribution.

If you have spherical symmetry, then you can assume that the electric field at radius $$r$$ from the origin points radially and only depends on $$r$$ (i.e., not on $$\phi$$ or $$\theta$$). This allows you to convert the surface integral in Gauss's Law to the form $$\int_S \mathbf{E} \cdot \mathbf{da} = 4\pi r^2|E|,$$ which then allows you to calculate the electric field directly.

But it's really important to realize that this only applies when there is sufficient symmetry (which means spherical symmetry here, but could also mean cylindrical or planar symmetry in other problems). Otherwise, the surface integral cannot be simplified further, so you have to be more clever, which in the case of the above problem means using superposition to reduce the problem to two which do have sufficient symmetry to simplify the surface integral.

• So, you are subtracting the contribution to the field from the charge which is not in the hole. Commented Nov 22, 2020 at 19:21
• Why do we need sufficient symmetry? and why isn't it sufficient in this case? (i.e. how can I tell when something is symmetrical enough) Commented Nov 22, 2020 at 19:30
• by superposition you mean we have 2 spheres on top of each other (overlapping in the area of the smaller sphere)? I'm not sure I understand the idea of superposition. Commented Nov 22, 2020 at 19:31
• I will edit my post above to answer these questions. Commented Nov 22, 2020 at 19:33
• I still have some things to figure out, but I'm in the right direction I believe. Thank you! Commented Nov 22, 2020 at 21:45