Understanding the Electric Field of a Charged Sphere

[Leo Liu]

Homework Statement

Just a random question about electricity.

Relevant Equations

$${E} = {{kQ} \over {r^2}}$$

This page claims that "[t]he electric field outside the sphere is given by: ${E} = {{kQ} \over {r^2}}$, just like a point charge". I would like to know the reason we should treat the sphere as a point charge, even if the charges are uniformly distributed throughout the surface of the conducting sphere. Also, is this statement valid for a non-conducting charged sphere? Thank you.

Update:
Reading this thread I realized that I can create a spherical Gaussian surface outside the conducting sphere and $\oint{E}\mathrm{d}A = {E4\pi r^2} = {Q \over \epsilon_0}$, which implies its electric field is the same as the electric field of a point charge, yet I still don't intuitively understand why it holds ture.

 

[jasonRF]

The main difference between the conducting sphere and the point charge is that the field inside the sphere is zero; the equation you derived is only true outside the sphere.

By the way, the electric field is a vector field, so you really should indicate the direction of the field.

jason

 

[Leo Liu]

The main difference between the conducting sphere and the point charge is that the field inside the sphere is zero; the equation you derived is only true outside the sphere.

By the way, the electric field is a vector field, so you really should indicate the direction of the field.

jason

Hi, thank you for replying. I understand that the strength of the field inside a conducting spherical shell is 0 since the excess charges tend to stay on the outer surface of a conductor and $dA$ from every direction cancels out. However, is this valid for an insultor given that the charges inside are uniformly distributed?

 

[DaveE]

Hi, thank you for replying. I understand that the strength of the field inside a conducting spherical shell is 0 since the excess charges tend to stay on the outer surface of a conductor and $dA$ from every direction cancels out. However, is this valid for an insultor given that the charges inside is uniformly distributed?

Yes. You can deconstruct a solid sphere of uniform charge distribution into a set of concentric shells. If you know it's true for the outside surface then it will also be true for the next-most outside shell, etc. Because of the superposition property of EM you can add solutions together (or subtract) to make a new solution.

 

[haruspex]

is this valid for an insultor given that the charges inside is uniformly distributed?

It is true for a charge uniformly distributed on a spherical shell, and hence also for a sphere in which the charge density depends only on radius. Similarly for any force field follwing an inverse square law, such as gravitation.
It follows that it is true for a conducting spherical shell in the absence of any other field. But, e.g., if there is a point charge in the vicinity then the charge will not be uniform.

 

[Leo Liu]

It is true for a charge uniformly distributed on a spherical shell, and hence also for a sphere in which the charge density depends only on radius. Similarly for any force field follwing an inverse square law, such as gravitation.
It follows that it is true for a conducting spherical shell in the absence of any other field. But, e.g., if there is a point charge in the vicinity then the charge will not be uniform.

In the case that a point charge is placed closed to the sphere, is it correct to say that the charge distribution on the surface cannot be uniform because the point charge will attract some charges with oppsite sign in the sphere (and repel the others)?

 

[hutchphd]

Yes and you can say that the field inside the conductor will be zero. The rest of the story gets more complicated (but do-able e.g. method of images)

 

[haruspex]

you can say that the field inside the conductor will be zero

Assuming the point charge is outside.

 

[PeroK]

Homework Statement:: Just a random question about electricity.
Relevant Equations:: $${E} = {{kQ} \over {r^2}}$$

This page claims that "[t]he electric field outside the sphere is given by: ${E} = {{kQ} \over {r^2}}$, just like a point charge". I would like to know the reason we should treat the sphere as a point charge, even if the charges are uniformly distributed throughout the surface of the conducting sphere. Also, is this statement valid for a non-conducting charged sphere? Thank you.

Update:
Reading this thread I realized that I can create a spherical Gaussian surface outside the conducting sphere and $\oint{E}\mathrm{d}A = {E4\pi r^2} = {Q \over \epsilon_0}$, which implies its electric field is the same as the electric field of a point charge, yet I still don't intuitively understand why it holds ture.

This is the "shell theorem", which applies to any inverse square field. Usually to Newtonian Gravity and Electrostatics. See, for example:

https://en.wikipedia.org/wiki/Shell_theorem

The mathematics is essentially the same for electrostatics.