The divergence theorem? Gauss’ law? If you’ve had college physics, at least the second is familiar to you. Now those two things are really fascinating, and I’m going to put a math-lite layman’s explanation in a future post. But hey, it’s spring semester finals time and that means all you people in a second-semester physics class are going to have at least a problem or two about this. Shall we practice one, straight from the final I just finished grading? Of course!
A nonconducting sphere of radius R has a uniform charge density ?. At the center of the sphere is a point particle with charge Q. What is the electric field E at points r < R?
This one’s pretty easy, but a quite large number of students missed it at least partially. Try it for yourself, then check the answer below.
We have spherical symmetry, and that immediately suggests we can use a sphere as our Gaussian surface. So draw the imaginary Gaussian sphere with radius r inside the actual charged sphere, and write down Gauss’ law.
Now we know E is going to be emerging perpindicularly to our Gaussian surface, thus the dot product is just E da. And E is constant over the surface, so we just have
But the integral is now just the total area of the sphere: 4 pi r^2. Thus,
And the charge enclosed is the point charge Q plus however much of the solid sphere’s charge we enclose is. And that latter quantity is just the volume of the Gaussian sphere times the charge density ?. Substituting,
Divide to solve for E and we’re done!
If you wanted, you could go ahead and divide this out. But as it is, it’s a complete solution which will give you E at any point inside the sphere.
Note two more things. First, the radius of the sphere R is a red herring: we don’t need it to find the field inside. Second, that wouldn’t be true if r > R, because we’d no longer be including more and more charge in the Gaussian sphere as r increased. Then the total charge enclosed would just be Q + ?(4/3 ? R^3).
11 responses so far ↓
1 Kent // Sep 19, 2008 at 3:41 pm
Hey, what does rho stand for in the final eqn?
2 Kent // Sep 19, 2008 at 3:43 pm
oh… charge density. right.
3 gadagkar // Apr 20, 2009 at 10:42 pm
Sir,
What I would like to know is when exactly is a charge inside a sphere zero? Whether solid or hollow? Whether the charge is outside of the sphere only then? Or is it not zero when the charge is inside?
thanks!
paddy
4 edward li // Jan 28, 2010 at 1:16 pm
I think you should convert the rho back into Q/V so you can simplify the equation to Q/(2*pi*r^2*epsilon knot)
5 dog // Nov 7, 2010 at 7:17 am
thanks a lot!
6 josh s // Jan 24, 2011 at 12:35 pm
edward, that is not legit, the placed charged Q is separate from the density, you’d need to rescale and have a delta function to bring the density and the point charge together.
7 max // May 27, 2011 at 12:09 pm
Can’t you use:
E = (k*Q*r)/(R^3)
where k = 1/(4*pi*epsilon knot)?
8 Devasia Manuel // Jul 28, 2011 at 11:46 pm
I’ve got a dumb physics joke:
Some are cosmonauts, some are astronauts, others are ?-nauts
9 That's fine // Apr 14, 2012 at 4:54 am
What is the external electric field of a sphere and the potential due to the internal and the external point using Guass law.Thanks.
10 koolphysico // Jun 20, 2012 at 10:19 pm
sir,can u tell me why field inside a charged sphere is not infinity because it is very very close to some dQ charge———– and if u say the charge dQ is very small,can’t its value be a min of 1.602*10^-19? so the numerator will be of orders 10
6-10 and denominator will tend to ZERO
11 koolphysico // Jun 20, 2012 at 10:21 pm
sir,can u tell me why field inside a charged sphere is not infinity because it is very very close to some dQ charge———– and if u say the charge dQ is very small,can’t its value be a min of 1.602*10^-19? so the numerator will be of orders 10^-10
and denominator will tend to ZERO
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