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Electric Field/Charge

  • Thread starter Air
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  • #1
Air
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Hello, I will post questions on this thread on Electric Field/Charge and other topics related to it. I'm studying it and am very weak at it thus any help will be appreciated.

For now, here's one question.
  • I've come across two equations for E which are E = (Kq)/(r) and E = (Kq)/(r^2). I don't understand why it is r and sometimes (r^2). Are they specific for some cases? Also, which is used more often?

Thanks in advance.
 
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Answers and Replies

  • #2
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Well the first equation you have stated is for Potential, and the second equation is electric field strength for a radial field.
 
  • #3
Doc Al
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The second equation, E = (Kq)/(r^2), gives the field from a point charge. The first equation, E = (Kq)/(r), is not dimensionally correct, so I suspect you are mistaking it for another equation. (Perhaps the potential from a point charge: U = (Kq)/(r).)

[RoryP beat me to it. :smile:]
 
  • #4
Air
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What is the potential used for? When is it used in context?

I know that it's related to Work by W=Ua - Ub.Work is the energy required to move a charge.
__________________
Also, for coulombs law, there is a unit vector (r) in the equation. Isn't unit vector value always one?

(Thanks for the help so far :biggrin:)
 
  • #5
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Hmm not entirely sure, i know the units for Potential are Volts. And i also know that if you integrate field strength dr between the limits infinity and current position you end up with Potential. =]
Convieniently the laws for universal gravitation run in parallel to the ones for electric fields, so if you have knowledge of the gravitational ones then it could be helpful =]
 
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  • #6
Air
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I only know the gravitational theory briefly but I understood what you have said. :smile:

__________________
Another question:

I understand that for a charged conducting sphere, the charge is built up on the surface thus there is no charge enclosed inside (if we are considering Gauss surface inside). When they talk about charged surface sphere, what is the structure like? Is is hollow or solid and is there no charge enclosed for both situations?
 
  • #7
Doc Al
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What is the potential used for? When is it used in context?

I know that it's related to Work by W=Ua - Ub.Work is the energy required to move a charge.
Read about electric potential and potential energy here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/volcon.html#c1"

Also, for coulombs law, there is a unit vector (r) in the equation. Isn't unit vector value always one?
The magnitude of a unit vector is always one, but its direction varies. A unit vector is used to specify the direction of the force (or other vector).
 
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  • #8
Doc Al
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I understand that for a charged conducting sphere, the charge is built up on the surface thus there is no charge enclosed inside (if we are considering Gauss surface inside). When they talk about charged surface sphere, what is the structure like? Is is hollow or solid and is there no charge enclosed for both situations?
A conducting sphere can be solid or hollow. For the electrostatic case, there is never any net charge within the material of the conductor, only on its surface.
 
  • #9
Air
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Thanks Doc_Al. Amazing help as always. :smile:

__________________
Another question:

Isn't the formula for Gauss law (When considering angle), EAcosx, so If the Gaussian surface is perpendicular, there would be no field as cos(90)=0. Why is it that when considering infinite sheet of charge, we draw the cyclindrical Gaussian surface, we say that the E is perpendicular to the sheet?
 
  • #10
Doc Al
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Isn't the formula for Gauss law (When considering angle), EAcosx, so If the Gaussian surface is perpendicular, there would be no field as cos(90)=0.
That's a formula for finding the electric flux through a surface. Realize that the angle is measured from the normal to the surface. So if the field is perpendicular to the surface, the angle it makes to the normal is θ = 0, not 90.
Why is it that when considering infinite sheet of charge, we draw the cyclindrical Gaussian surface, we say that the E is perpendicular to the sheet?
E is perpendicular to the sheet, and thus makes an angle of θ = 0 with the normal to the surface of the flat ends of the Gaussian surfaces. This should make sense, since when the field is perpendicular to the surface the flux is maximum (cos0 = 1).

Read: http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/elesht.html#c1"
 
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  • #11
Air
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  • #12
Doc Al
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I realised that there is a for a sphere of uniform charge, the electric field inside is not zero and a conducting sphere has zero electric field inside.
Good. The two charge distributions are very different. (Yet can often have the same electric field when r > a.)
Can the sphere of uniform charge also be hollow and solid, just like the other?
If you are asking if a sphere of uniform charge density can have a hollowed out portion--sure, why not?
 
  • #13
Air
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If you are asking if a sphere of uniform charge density can have a hollowed out portion--sure, why not?
So you cannot have a shell of uniformly charged density. The electric field inside would be zero for such a case. Is that correct?
 
  • #14
Doc Al
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So you cannot have a shell of uniformly charged density.
Why not? (Did you mean can, not cannot?)
The electric field inside would be zero for such a case. Is that correct?
Yes. If you have a spherically symmetric shell of charge, the field within the hollow part of the shell will be zero.
 

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