# Electric potetntial for concentric spheres [closed]

I am studying electrostatics. The book I am using has this problem as an exercise for electric potential. The problem states that:

‘There are two metal hollow spheres of radius $a$ and $b$, $b$ being the larger sphere. The two spheres are concentric. The smaller one has a positive charge $+q$ and the larger one a negative charge $-q$. Find the electrostatic potential as a function of $r$ for $r>b$, $b>r>a$, $r<a$.'

## closed as off-topic by Alfred Centauri, Yashas, ZeroTheHero, John Rennie, Qmechanic♦Apr 12 '17 at 8:50

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• I think the positive charge on the inner sphere should pull the negative charge on the inner side and make the electric field inside zero. But i know i am wrong(well the ans given at the end of the book disagrees with me). I just want someone to shed some light on the physical picture whats happening here – Arya Apr 12 '17 at 2:13
• Please note that we don't answer homework or worked example type questions. Please read How do I ask homework questions on Physics Stack Exchange? and Are check-my-work questions on-topic? for "check my work" problems. – Yashas Apr 12 '17 at 3:08
• It wasn't a homework question. I was self studying. I was actually confused why shouldnt the negative chrage affect the region between the two spheres too. I know the result gauss' law gives me. But I was curious about the physical reason – Arya Apr 13 '17 at 2:49

The electric field inside is not zero. Well the word inside is ambiguous. I'm assuming by saying inside you mean $a<r<b$.

Anyway getting back to your question...
I'm assuming you're familiar with the Gauss law of electrostatics. If you aren't:

Gauss law simply assumes a surface (of any shape! though for our convenience we resort to symmetrical surfaces) and says that the electric field at any point in the surface would be zero if the net charge enclosed by that surface is zero.

In the first case when $r>b$ where $b$ is the radius of the larger sphere, the electric potential is simply zero.

(Note: It's imperative that you make yourself familiar with the definition of electric potential. Electric potential is defined as the work done to bring a unit positive charge to a particular point. Now if the field is zero, it's equivalent to saying that we don't see any charge and there is no charge against which work is to be done.) If you assume a surface that encloses any point $r>b$ and our two spheres, the electric field at that point would be zero (because one sphere has $+q$ while the other has $-q$ hence the net charge enclosed by the surface is zero. $+q-q=0$).

Now it is also known to you that the electric potential difference is given by ${\Delta}V=E{\delta}l$ (formula for potential difference not potential) where the symbols have their usual meaning (here $l$ would be decided by the distance you make the charge move not $r$, assuming you have taken the coordinate system at the centre of the spheres). Do take care of the signs. It's a bit difficult to explain but it is almost always clear by the context.

At any point between the spheres, the electric field would be due to the smaller sphere only (which can be doing out using Gauss law and the aforementioned formula) and the electric potential at the surface of the surface of the second sphere would be the electric field (which can be found using Gauss law) times $b-a$.

The electric potential inside the smaller sphere can be similarly found. Hope that helps.