What is the solution for the integral of 2x?

[Decadohedron]

Homework Statement

Given an E field, determine if it's a possible electrostatic field. If so, determine a potential

Homework Equations

∇⋅E

∇×E

The Attempt at a Solution

[/B]
Just more of a clarification, since my friend and I both attempted this question differently.

I took the divergence and the curl of the given E field.
The divergence wasn't 0, so I said that this isn't an electrostatic field and I didn't determine a potential.

My friend took the curl, it was 0. So they found a potential for it.

So this confused me, and I am now asking the question.

Thanks for your help!

 

[TSny]

I took the divergence and the curl of the given E field.
The divergence wasn't 0, so I said that this isn't an electrostatic field and I didn't determine a potential.

The divergence of E is zero only at points where there is no electric charge density.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/diverg.html

 

[SammyS]

Homework Statement

Given an E field, determine if it's a possible electrostatic field. If so, determine a potential

Homework Equations

∇⋅E

∇×E

The Attempt at a Solution

[/B]
Just more of a clarification, since my friend and I both attempted this question differently.

I took the divergence and the curl of the given E field.
The divergence wasn't 0, so I said that this isn't an electrostatic field and I didn't determine a potential.

My friend took the curl, it was 0. So they found a potential for it.

So this confused me, and I am now asking the question.

Thanks for your help!

What have you learned in your course?

Does the gradient of the potential that your friend found give the field?

 

[Decadohedron]

The divergence of E is zero only at points where there is no electric charge density.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/diverg.html

Which should mean that I can be at some random r far away from the charge density and it's not an electrostatic field.

 

[Decadohedron]

What have you learned in your course?

Does the gradient of the potential that your friend found give the field?

It does, because it's a simple integral, but that's not where the confusion lies...

From the prof's mouth:
"Both the Divergence AND the Curl of an E field must be 0 for a field to be Electrostatic"
Then, to me, if the divergence isn't 0 -> not electrostatic -> no need to do the following, since this is what the question was asking:

V = -∫E⋅dl

I guess I'll do it because it's a ridiculously easy integral anyway - but I still have a hard time accepting that I have to do it in the first place since the field given doesn't meet both conditions.

 

[SammyS]

It does, because it's a simple integral, but that's not where the confusion lies...

From the prof's mouth:
"Both the Divergence AND the Curl of an E field must be 0 for a field to be Electrostatic"
Then, to me, if the divergence isn't 0 -> not electrostatic -> no need to do the following, since this is what the question was asking:

V = -∫E⋅dl

I guess I'll do it because it's a ridiculously easy integral anyway - but I still have a hard time accepting that I have to do it in the first place since the field given doesn't meet both conditions.

Perhaps your professor has a special definition of an "Electrostatic E field".

I general, all that is required for there to be an 'electrostatic potential' associated with an electric field, $\ \vec E \,,\$ is that its curl be 0. I.e. $\ \nabla \times \vec E =0 \,.$. If the curl of E is zero, then there is a potential function such that $\ \nabla \cdot V = \vec E \,,\$ and the difference in electric potential in going from location a to location b is given by $\displaystyle \ V_b - V_a = - \int_a^b \vec E \cdot d\vec \ell \,,\$ independent of the Path taken.

See Wikipedia: Electric potential
.

 

[TSny]

From the prof's mouth:
"Both the Divergence AND the Curl of an E field must be 0 for a field to be Electrostatic"
Then, to me, if the divergence isn't 0 -> not electrostatic

The prof's statement is not true.

An electrostatic field is an E field produced by charges at rest. The curl of an electrostatic field will be zero at all points of space (including points where the charge density ρ is not zero). So, if you find that the curl of an E field is nonzero, then the field cannot be an electrostatic field.

The divergence of an E field will be zero for points where ρ = 0, but the divergence of the field will be nonzero at any point where ρ $\neq$ 0. This applies not only to electrostatic fields but to any E field. So, if you are given an E field such that the divergence is nonzero at some point, then you cannot tell from this whether or not the field is an electrostatic field. All you could say from this is that there is a nonzero charge density at points where the divergence is nonzero.

Consider the E field produced by a uniformly charged sphere at rest. The E field inside and outside the sphere would be an electrostatic field. The divergence of the field will be zero at any point outside the sphere and the divergence will be nonzero at any point inside the sphere.

 

[Decadohedron]

Thanks for the clarification dudes!

Now what's the integral of 2x *ponders*
The internet needs sarcasm font.