# Electric field in non-uniformly charged hollow sphere [closed]

My full question is

"A total charge $Q$ is uniformly distributed within a spherical shell with negligible thickness compared to its radius. A point charge $q$ ($q<< Q$) is taken away from the shell's north pole and replaced on its south pole without affecting the rest of the charge distribution. What is the electric field at the centre of the shell?"

I don't really know how to apply symmetry to this, in a normal uniformly charged shell the field inside would be zero, but for this..? Would using Gauss's law with non-uniform charge distribution be appropriate at all (even though no charge is enclosed at the centre)? Shall I treat the poles like point charges of 0 and +2q, is there some type of superposition thing I'm missing?

This has really stumped me.

## closed as off-topic by Kyle Kanos, ZeroTheHero, Jon Custer, Yashas, John RennieApr 17 '17 at 15:38

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• This is not an actual solution but just stating that using gauss's law won't be very helpful at all when you can't use symmetry arguments – Sumant Apr 15 '17 at 15:30
• Actually the statement $q<<Q$ suggests the removal of $q$ does not basically alter the field of the sphere. – ZeroTheHero Apr 15 '17 at 18:51
• Yeah I thought so, nice to actually get my head around it now – chennington Apr 15 '17 at 21:49

You can imagine the second case to be a superposition of the uniform spherical shell and two opposite point charges -q and q at the north and south pole respectively. The field inside the sphere would then be the field created by the -q/q dipole.

• I tried this but I don't think I really understand the physics of that. When the small charge is removed, doesn't that make the place where it was neutral? Like, how would it behave as a dipole? – chennington Apr 15 '17 at 15:18
• That's correct. The trick here is that we consider a neutral space as containing both a positive and a negative charge whose values that add up to zero. This allows us to treat a somewhat complicated problem as a superposition of two problems that are much easier to solve. The validity of this approach stems from the fact that the terms involving q, J in Maxwell's equations are linear, so that solutions of sub-distributions add up. This is what makes it possible to attribute the total field at a specific point as the sum of the fields created by each separate infinitesimal charge. – zap Apr 15 '17 at 15:38

The field inside the sphere is the superposition of two point charges with opposite signs, and a uniform spherical shell. The latter gives zero field so we just need the field due to two point sources.

For a given distance $r$ from the "pole", the field is

$$E=\frac{q}{4\pi \epsilon_0 r^2}$$

Pointing away from the positive charge. For the negative charge you get a field pointing towards the charge.

The total field is the vector sum of these two. Along the plane midway between the charges this field is perpendicular to the plane - at other points it will curve.

For a given point you should be able to write down the X and Y components of the field due to these two point charges and add them together. I believe this should be sufficient to get you to your answer.

• Yes, thank you so much, I thought this was the case but I guess I must revise the actual physics of what's happening here. Thanks again – chennington Apr 15 '17 at 21:52

The solution follows by considering the superposition of two cases.

1. when we just remove the charge (and the whole other body is kept as it is)
2. when we place that charge diametrically opposite to chargeless space.(this time only this charge is considered and not the sphere)
• Do you mind typing out your answer so that it can be read better, and indexed by search engines? – Glorfindel Apr 15 '17 at 15:33
• 1. If you are using images, please make at least sure they are rotated correctly with respect to the post - I have to tilt my head to read it. 2. Please type out your answer instead of posting it inside an image - a picture cannot be indexed by search engines. To format formulae, please use MathJax. – ACuriousMind Apr 15 '17 at 16:23
• Ahh, awesome, so much better to see this as just adding two fields, I wish I had just gone with my instinct! – chennington Apr 15 '17 at 21:53
• I've edited out the image because it was detracting from the quality of the post. – David Z Apr 16 '17 at 3:09