# Electrostatics Problem using Laplace's Equation

In summary, the value of the potential at a point is equal to its average over any sphere centered at that point.

## 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.

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.

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
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 general 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

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
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
Alex Petrosyan said:
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
BvU said:
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##?

Alex Petrosyan said:
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.

vela said:
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.

Ah, that would make sense.

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
vela said:
How'd you get ##\Delta V = 0##?
##\Delta\equiv\nabla^2##
I don't think there's any physical sphere, conducting or not, in the problem.
exactly

## What is Laplace's Equation in electrostatics?

Laplace's Equation is a mathematical expression that describes the distribution of electric potential in a region with no charge density. It is a second-order partial differential equation that is used to solve for the electric potential in electrostatic problems.

## How is Laplace's Equation used in electrostatics problems?

Laplace's Equation is used to solve for the electric potential in a region with no charge density. It is often used in conjunction with boundary conditions to determine the electric potential at different points in space.

## What are the limitations of using Laplace's Equation in electrostatics problems?

Laplace's Equation is limited to problems with no charge density, which means it cannot be used to solve for the electric potential in situations with charged particles or conductors. It also assumes that the electric potential is continuous and differentiable throughout the region, which may not always be the case in real-world situations.

## What are some common techniques for solving electrostatics problems using Laplace's Equation?

Some common techniques for solving electrostatics problems using Laplace's Equation include separation of variables, the method of images, and the method of conformal mapping. These techniques can be used to simplify the equation and solve for the electric potential in different geometries and boundary conditions.

## How does Laplace's Equation relate to other equations in electrostatics?

Laplace's Equation is closely related to other equations in electrostatics, such as Gauss's Law and Poisson's Equation. In fact, Poisson's Equation is a more general form of Laplace's Equation that includes a term for charge density. Gauss's Law can also be derived from Laplace's Equation in certain situations.

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