What is the magnitude of Electric Field at r=R/2

  • Thread starter jaydnul
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  • #1
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Homework Statement


The first question asks: What is the electric field when the radius of a sphere is R. For this first question I put [itex]E=k\frac{Q}{R^2}[/itex] and got the question correct.

The second question simply states: What is the electric field at r=R/2. I used Coulomb's law, Gauss' law, and common sense and every time got: [itex]E=k\frac{4Q}{R^2}[/itex] but there are no choices that match my answer.

Also there are no choices of [itex]E=\frac{Q}{ε_0πR^2}[/itex]

Am I doing something wrong?
 

Answers and Replies

  • #2
TSny
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Hello.

It will help if you state the problem with all the given information. For example, I am guessing that the sphere carries electric charge and that there are no other charges present. If the sphere does carry charge, how is that charge distributed? Is it uniformly spread over the surface of the sphere? Or is it uniformly spread throughout the volume of the sphere? Or spread nonuniformly?

It would also help if you would show how you got your answer. Otherwise, we cannot tell if you are doing something wrong.
 
  • #3
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It's a uniformly charged insulating sphere. Asks for the magnitude of electric field on the surface of the sphere. That's all it says.

So I did:
[tex]E*4π(\frac{R}{2})^2=\frac{Q_{encl}}{ε_0}[/tex]
Which simplifies to:
[tex]E=k\frac{4Q}{R^2}[/tex]
Right?
 
  • #4
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Is the charge present only on the surface?
What is your gaussian surface and what charge does it enclose?
 
  • #5
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That's literally everything the problem states...
 
  • #6
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Assuming its a uniform volume charge density-
What is your gaussian surface? What charge does it enclose?
Hint: Find the charge density for second part.
 
  • #7
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I thought the Gaussian surface was the surface of the sphere?

Charge density would be:
[tex]S=Eε_0[/tex]
And the charge it encloses would be [itex]Q[/itex]
 
  • #8
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You need to read up on Gaussian surface...try this http://maxwell.ucdavis.edu/~electro/flux/gaussian_surf.html orhttp://www.phy.duke.edu/~schol/phy152/faqs/faq4/node5.html
Note that for symmetric surfaces it has to pass through the point in question if you want the electric flux there and hence find the field.
Also charge density is given by Q/V where Q is charge and V the Volume.
 

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