Gauss' Law and Conductors

  • Thread starter Ignitia
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  • #1
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Homework Statement


A point charge q=−5.0×10−12 C is placed at the center of a spherical conducting shell of inner radius 3.5cm and outer radius 4.0 cm. The electric field just above the surface of the conductor is directed radially outward and has magnitude 8.0 N/C. (a) What is the charge density on the inner surface of the shell? (b)What is the charge density on the outer surface of the shell? (c) What is the net charge on the conductor?


Homework Equations


E=K (Q/r2)

K: Coloumb's Constant, so

(E*r2)/K = Q

with Q being the derivative of the enclosed q=−5.0×10−12 C

and

σ= Q/S

With S = ∫ 4πr2dr

3. The Attempt at a Solution


So if I plug in the values accordingly, I should get:

a) r = 3.5
σ= Q/S
σ= ((E*r2)/K)/∫ 4πr2dr (integrate from 0 to 3.5)


b) r = 4.5
σ= Q/S
σ= ((E*r2)/K)/∫ 4πr2dr (integrate from 3.5 to 4.5)

c) Should just be the same as q=−5.0×10−12 C

Before I plug this in to a calculator and get the value, is this correct?

Edit: The integrate sign isn't showing up, so the little circle is actually that.
 

Answers and Replies

  • #2
Doc Al
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No need for any (non-trivial) integration. Start here: Imagine a Gaussian surface just outside the conducting sphere. What total charge must it contain?
 
  • #3
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Well, if E=8.0 N/C then it's

|E*dA
E * |dA
E*A = 8.0*4πr^2 (with a sphere Gaussian surface)

I am still confused, however.
 
  • #4
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Hold on, I think I got it. Treat the surfaces of the conductors as Gaussian Spheres, with respect to the r and given charges? Is that it?

A) q=−5.0×10−12 C is the charge to the surface so its just q/(4*pi*r^2) with r=3.5
B) E=8.0N/C so plug that into E=K q/r^2 to get the second q, and then q/(4*pi*r^2) with r=4.5

So how do you get C?

Edit: Actually, net charge would be 0 on the conductor, wouldn't it?
 
Last edited:
  • #5
Doc Al
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Edit: Actually, net charge would be 0 on the conductor, wouldn't it?
No.

Once again, start here: You have the field just outside the sphere, thus at r = 4.0 cm. Use Gauss' law for a spherical surface of that radius to find the total charge contained within. Once you have the total charge, you can figure out what must be the charge on the conducting sphere, and then how that charge is distributed on the inner and outer surfaces.
 
  • #6
SammyS
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Hold on, I think I got it. Treat the surfaces of the conductors as Gaussian Spheres, with respect to the r and given charges? Is that it?

A) q=−5.0×10−12 C is the charge to the surface so its just q/(4*pi*r^2) with r=3.5
B) E=8.0N/C so plug that into E=K q/r^2 to get the second q, and then q/(4*pi*r^2) with r=4.5

So how do you get C?

Edit: Actually, net charge would be 0 on the conductor, wouldn't it?
I don't see Gauss's Law stated anywhere in this thread. (Did I overlook it? I'm fairly old.)

For A) Be careful of units. r = 3.5cm, not 3.5m .

For B) Use Gauss's Law directly to simplify the calculation. (B.T.W.) OP states that r = 4.0cm, not 4.5cm.
... but what you have is otherwise correct. - - - Maybe call this charge Q, or QTotal or ..., but something other than q.​

What is this "C" you refer to?

And:
Don't assume the net charge is zero.
 

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