# Electrostatics Problem using Laplace's Equation

## Homework Statement

Prove that in a region free of electric charge, the value of the potential
at a point is equal to its average over any sphere centered at that point.

## Homework Equations

V(r) = 1/(4piR^2) Integral(V * da)

## The Attempt at a Solution

I defined a point outside the region where a point charge is located. At a random point inside the potential is kq/r. The point charge has the coordinate H,alpha(angle),gamma(angle).Now I made my auxiliary sphere and using that same point where the charge is located I got the Integral of 1/(4pi4r^2) q/(4piepsiolonsubnot) R^2 sin(phi) (H^2 +R^2 -2HR(cosalphacosthetasinalphasintheta+sinthetasinalphasingammasinphi+cosgammacosphi_^1/2 but I cannot integrate it.

I'm not 100% sure, but I think that there's a simple solution, if you consider the electric field, and evaluate potential differences through line intergrals.

A solid hint, is that the $$\vec{E} = \vec{0}$$ inside the region without charge. So you need to prove:

1. Any point inside the sphere has the same potential.
2. Any point P outside the sphere has the same potential as a point on the sphere S, which is the intersection of the line Origin - P with the sphere.

The rest will simply follow.

BvU
Homework Helper
Not a conducting sphere !

@BvU, doesn't have to be. Simply use Gauss' theorem and the fact that there are no charges within the sphere, and you'll get that the entire spheroid (or ball) is equipotential.

BvU
Homework Helper
Exercise is broader than that. Not equipotential (by equipotential I mean that the potential at a distance r from the point in question is the same for all directions).

berkeman
BvU
Homework Helper
Hello Faizan,
potential at a point is equal to its average over any sphere centered at that point
You apparently have decided you need to integrate over the surface of the sphere. Is that what the exercise says ?

Also, you take the potential field of a point charge as a given. Is that genral enough ? You might have to explain why.

And: don't make things more complicated than needed: for your attempt a charge at the origin and a sphere on an axis is good enough. Note the rotation symmetry makes the azimuthal integration superfluous.

And, like Alex, I have the nagging feeling the consequences of a simple mathematical observation (##\Delta V = 0##) should do the trick, but my coffee hasn't worked yet

Delta2
Homework Helper
Gold Member
Not sure at all I might be wrong, but the way I interpret the statement of this problem is that the volume of the sphere of integration is completely free of charge density, and that the charge density that creates the potential and the electric field , is in some distant region far away from the sphere of integration.

@Delta² I think you're right. Maybe, we could reduce that case to when the charge is. inside the sphere by the method of images in reverse. (we Typically replace a conduting sphere with a charge (or two) inside it, but here we could do the opposite.

Delta2
BvU
Homework Helper
I am certain ##\Delta^2## is interpreting the exercise correctly (his reading and mine happen to coincide).
No disturbance of the field is 'allowed' -- as a conducting sphere would cause. We have to make do with a divergence-free ##\vec E## (which I seem to remember implies flux in = flux out ....)

As a welcoming present for @Faizan Samad (still herer ?) an example (right-click the formula and | Show Math as | ##\TeX## commands to see source) of his expression (not 'corrected'):

1/(4pi4r^2) q/(4piepsiolonsubnot) R^2 sin(phi) (H^2 +R^2 -2HR(cosalphacosthetasinalphasintheta+sinthetasinalphasingammasinphi+cosgammacosphi_^1/2

with the incantation

$${1\over 4\pi 4r^2} {q \over 4 \pi\varepsilon_0 R^2} \sin\phi \\ \left (H^2 +R^2 -2HR\left ( \cos\alpha\cos\theta\sin\alpha\sin\theta +\sin \theta \sin\alpha\sin\gamma\sin\phi +\cos\gamma\cos\phi\right )^{1/2} \right )$$

this becomes

$${1\over 4\pi 4r^2} {q \over 4 \pi\varepsilon_0 R^2} \sin\phi \\ \left (H^2 +R^2 -2HR\left ( \cos\alpha\cos\theta\sin\alpha\sin\theta +\sin \theta \sin\alpha\sin\gamma\sin\phi +\cos\gamma\cos\phi\right )^{1/2} \right )$$

##\ \int dA\ ## or ## \ \int dV\ ## that is the question ....

Delta2
Ray Vickson
Homework Helper
Dearly Missed
I'm not 100% sure, but I think that there's a simple solution, if you consider the electric field, and evaluate potential differences through line intergrals.

A solid hint, is that the $$\vec{E} = \vec{0}$$ inside the region without charge. So you need to prove:

1. Any point inside the sphere has the same potential.
2. Any point P outside the sphere has the same potential as a point on the sphere S, which is the intersection of the line Origin - P with the sphere.

The rest will simply follow.

This seems false. Consider the field due to a charge ##q## located at ##(0,0,0)## and look at the region ##R## inside and on the surface of a sphere of radius 1, centered at ##(2,0,0)##. There is no charge in ##R## but the electric field is not zero in ##R##, and different points in ##R## can have different potentials.

Delta2
@Ray Vickson, @BvU caught that mistake earlier. I actually misread the question.

BvU
vela
Staff Emeritus
Homework Helper
And, like Alex, I have the nagging feeling the consequences of a simple mathematical observation (##\Delta V = 0##) should do the trick, but my coffee hasn't worked yet
How'd you get ##\Delta V = 0##?

I think you're right. Maybe, we could reduce that case to when the charge is. inside the sphere by the method of images in reverse. (we Typically replace a conduting sphere with a charge (or two) inside it, but here we could do the opposite.
I don't think there's any physical sphere, conducting or not, in the problem.

Ray Vickson
Homework Helper
Dearly Missed
How'd you get ##\Delta V = 0##?

I don't think there's any physical sphere, conducting or not, in the problem.

Is #7using ##\Delta## in place of the notation ##\nabla^2?## Some writers do that.

vela
Staff Emeritus
Homework Helper
Ah, that would make sense.

Delta2
Homework Helper
Gold Member
I think this probably should be solved using some identities and theorems (divergence theorems like) from vector calculus and using the property that ##\nabla^2V=0##, without considering at all the charge density of the source and without performing real integration.

BvU
BvU