What is the electric field around a conducting sphere?

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Hi all,
I am having some troubles with some Electromagnetism questions. First of all can someone please explain the basics behind Electric fields. The question asks "What is the electric field around a conducting sphere of radius R carrying a charge Q". Does this imply that because the sphere is conducting that the charge is spread evenly and therefore can be considered as a point charge at the centre of the sphere? Does the use of Gauss's law help to get the correct answer, i.e

E= q/(4 x pi x permittivity of free space x r^2)x the unit vector in the field direction

I am slightly confused about this one.

The next part of the question asks " What is the electric potential at the surface of the sphere if the potential infinately far away is zero". As far as I know the potential is essentially a 'band' where the electric field is constant, so therefore is the electris field at the distance of the sphere, and because infinately far away = zero then at the surface is simply minus the field strength?

Please any explanation would be great. thanks
 

Answers and Replies

  • #2
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First part sounds good:

Potential from a point away from infinity is simply:
[tex]V=\frac{kq}{r}[/tex],

this is indirectly from coulombs law:

[tex]E=\frac{kq}{r^2}[/tex]
[tex]V=E\cdot r=\frac{kq}{r}[/tex]

where [tex]k=\frac{1}{4\pi\epsilon_0}[/tex]

Does this help you?

Regards,
Sam

(Sorry about the mistake if anyone read it quick enough)!
 
Last edited:
  • #3
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An important point with this, is to notice that V = E.r (scalar product), is a scalar quantity. I have made many mistakes in the past on this topic thinking that Voltage had a direction (whoops)! :smile:

Note: I've left out the [tex]\cos\theta[/tex], oweing to the fact that the Electric field about a point charge (which this example can be considered), is radial.

Regards,
Sam
 
Last edited:

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