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Force on a point charge due to constant sphere surface charge density.

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Kizaru
#1
Sep27-09, 04:54 PM
P: 45
1. The problem statement, all variables and given/known data
The surface of a sphere of radius a is charged with a constant surface density [tex]\sigma[/tex]. What is the total charge Q' on the sphere? Find the force produced by this charge distribution on a point charge q located on the z axis for z > a and for z < a.


2. Relevant equations



3. The attempt at a solution
For z>a, I found [tex]\vec{F}[/tex] = [tex]\frac{\normalsizeq\sigma\hat{z}}{\epsilon_{0}z^{2}}[/tex]

Now, during my calculations, there was one point where I evaluated the integral for [tex]\normalsize\vartheta[/tex] to be [tex]\normalsize\frac{1}{z}\left(\frac{z-r^{'}}{\left(z^{2}-r^{'}^{2}-2zr^{'}\right)^{1/2}+\frac{z+r^{'}}{\left(z^{2}+r^{'}^{2}+2zr^{'}\right)}[/tex]

My thinking is that when I simplify the fractions, I obtain |z-r| and |z+r| in the denominator if I want the positive square root, which is the z > a case. So would the z<a case simply mean the square roots would have negative signs out, ie -|z-r| and |z+r| (z+r is stil positive).

Edit: My latex syntax is messed up. But basically I got (z-r') / (z^2+r'^2-2zr')^(1/2) + (z+r')/(z^2+r'^2+2zr')^(1/2)
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gabbagabbahey
#2
Sep28-09, 12:43 AM
HW Helper
gabbagabbahey's Avatar
P: 5,003
Quote Quote by Kizaru View Post
3. The attempt at a solution
For z>a, I found [tex]\vec{F}[/tex] = [tex]\frac{\normalsizeq\sigma\hat{z}}{\epsilon_{0}z^{2}}[/tex]
Hmmm... isn't there a [itex]q[/itex] missing from this expression?...and [itex]\sigma[/itex] is a charge density, not a charge....so the units on this aren't quite right!

What did you get for the total charge on the sphere?

Now, during my calculations, there was one point where I evaluated the integral for [itex]\theta[/itex] to be

[tex]\frac{1}{z}\left[\frac{z-r'}{\left(z^{2}-r'^{2}-2zr'\right)^{1/2}}+\frac{z+r'}{\left(z^{2}+r'^{2}+2zr' \right)^{1/2}}\right][/tex]

My thinking is that when I simplify the fractions, I obtain |z-r| and |z+r| in the denominator if I want the positive square root, which is the z > a case. So would the z<a case simply mean the square roots would have negative signs out, ie -|z-r| and |z+r| (z+r is stil positive).
(I fixed your [itex]\LaTeX[/itex]...just click on the images to see the code I used)

First, aren't you integrating over [itex]r'[/itex]?...It shouldn't be present in your final result!

Second,

[tex]|z\pm a|=\left\{\begin{array}{lr}z\pm a &, z\geq \mp a\\a\pm z &, z<\mp a\end{array}\right.[/tex]

Third, if you've learned Gauss's Law, a clever choice of Gaussian Surface would have allowed you to avoid integrating entirely!
Kizaru
#3
Sep28-09, 07:03 AM
P: 45
Quote Quote by gabbagabbahey View Post
Hmmm... isn't there a [itex]q[/itex] missing from this expression?...and [itex]\sigma[/itex] is a charge density, not a charge....so the units on this aren't quite right!

What did you get for the total charge on the sphere?



(I fixed your [itex]\LaTeX[/itex]...just click on the images to see the code I used)

First, aren't you integrating over [itex]r'[/itex]?...It shouldn't be present in your final result!

Second,

[tex]|z\pm a|=\left\{\begin{array}{lr}z\pm a &, z\geq \mp a\\a\pm z &, z<\mp a\end{array}\right.[/tex]

Third, if you've learned Gauss's Law, a clever choice of Gaussian Surface would have allowed you to avoid integrating entirely!
Yes, I did forget a q in my answer for z > a.

The sphere charge density is on the surface, so I would be integrating over the surface area of the sphere (r is constant), correct?

We're learning Gauss' Law in the next chapter, which is why the tedious integral was carried out here.


Thanks for confirming my thoughts on where the z < a would come into play in the integral :)


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