Electric field at the center of a sphere

[Physicslearner500039]

Homework Statement

A sphere of radius r carries a surface charge of density σ = ar, where a is a constant vector, and r is the radius vector of a point of the sphere relative to its center. Find the electric field strength vector at the center of the sphere.

Relevant Equations

E = q/(4πεr^2)

My first impression was the electric field is 0 at the center of the sphere, but it turned out not the case.
My understanding when problems refer surface charge density, is that the charge exists only on the surface and it is hollow inside the sphere. Am i correct?
Using the electric field equation assuming width dr. The area of the strip is $4\pi r.dr$ is this correct?

$E = \int_0^r \sigma . 4\pi r dr/(4\pi \epsilon r^2)$
replacing $\sigma$
$E = \int_0^r a dr/\epsilon \\ E = ar/\epsilon$
I am not very confident of the method. Please advise.

 

[BvU]

Relevant Equations: NA

This can't be true

I am not very confident of the method.

Neither am I. In fact , I don't even see a method in your approach. Can you explain what you think your integral represents ?

 

[BvU]

Using the electric field equation assuming width dr. The area of the strip is $4\pi r\; dr$ is this correct?

Whwat coordinate system are you using ?
Re area: Compare a strip of your description at the equator with one at a pole !

 

[BvU]

Almost forgot: free advice, golden tip: make a clear sketch of your own !

 

[BvU]

Relevant Equations:: E = q/(4πεr^2)

Good to see you added one relevant equation. Note that $\vec E$ is a vector, though !
and is this r the same r as the radius of the sphere ?

My understanding when problems refer surface charge density, is that the charge exists only on the surface and it is hollow inside the sphere. Am i correct?

Doesn't say the sphere is hollow, but you may assume $\varepsilon=\varepsilon_0$, so it doesn't matter.

Also doesn't say the sphere is non-conducting, which I think is a painful omission.

 

[Physicslearner500039]

Thank you for the reply.

Note that $\vec E$ is a vector, though !

Yes my mistake

and is this r the same r as the radius of the sphere ?

Yes

Whwat coordinate system are you using ?

Not sure about the question. It is the normal x,y coordinates.
I read the following from the text

The sphere in question almost matches with above conditions. I am not sure what i am missing.

 

[BvU]

That's for a conducting sphere or shell: the charge distributes evenly over the surface.

It is the normal x,y coordinates

no it is not: if you write $$E = \int_0^r \sigma . 4\pi r dr/(4\pi \epsilon r^2) $$ you integrate over r from 0 to r. That is not in Cartesian coordinates. (*)

Now you see what confusion you can create by using the symbol $r$ for a whole bunch of completely different beasts.​

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Note that sigma is only nonzero at $r$, not at $r<r$.

Now you see (again ) what confusion you can create by using the symbol $r$ for a whole bunch of completely different beasts.​

Proposal:
Since you master a bit of $\TeX$, use $R$ as the radius of the sphere, so that one irritant source of misunderstanding/error is eliminated !

And write $\sigma = \vec a \cdot\vec r$.

Where did you position your vector $\vec a$ ?

#4 post advice is relevant and indispensable for setting up a decent integral. I don't want to do the work for you, but for my own purposes I draw a sphere with radius R in a spherical (physics) coordinate system and align the z-axis (ok, $\theta = 0$ ) with the vector $\vec A$. (upper case reminds me it's like R, a constant).

Next: I know that an area $dS$ of the shell has charge $\sigma\, dS = \vec A \cdot \vec r \,dS$ and subsequently I work out those factors in terms of $|\vec A|, r\ (=R), \theta, \phi\ d\theta, d\phi$.

It also gives me an idea of the charge distribution over the sphere and its symmetry.

Which way is $\vec E$ at the origin pointing ? So for which component of $\vec E$ do I need an expression for $dE$ as a function of the above ?(*) Generally speaking a problem with a shell or a sphere benefits from working in spherical coordinates. The word says it already. In this exercise: definitely !