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

by Kizaru
Tags: charge, constant, density, force, point, sphere, surface
 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 $$\sigma$$. 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 $$\vec{F}$$ = $$\frac{\normalsizeq\sigma\hat{z}}{\epsilon_{0}z^{2}}$$ Now, during my calculations, there was one point where I evaluated the integral for $$\normalsize\vartheta$$ to be $$\normalsize\frac{1}{z}\left(\frac{z-r^{'}}{\left(z^{2}-r^{'}^{2}-2zr^{'}\right)^{1/2}+\frac{z+r^{'}}{\left(z^{2}+r^{'}^{2}+2zr^{'}\right)}$$ 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
HW Helper
P: 5,004
 Quote by Kizaru 3. The attempt at a solution For z>a, I found $$\vec{F}$$ = $$\frac{\normalsizeq\sigma\hat{z}}{\epsilon_{0}z^{2}}$$
Hmmm... isn't there a $q$ missing from this expression?...and $\sigma$ 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 $\theta$ to be $$\frac{1}{z}\left[\frac{z-r'}{\left(z^{2}-r'^{2}-2zr'\right)^{1/2}}+\frac{z+r'}{\left(z^{2}+r'^{2}+2zr' \right)^{1/2}}\right]$$ 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
(I fixed your $\LaTeX$...just click on the images to see the code I used)

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

Second,

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

Third, if you've learned Gauss's Law, a clever choice of Gaussian Surface would have allowed you to avoid integrating entirely!
P: 45
 Quote by gabbagabbahey Hmmm... isn't there a $q$ missing from this expression?...and $\sigma$ 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 $\LaTeX$...just click on the images to see the code I used) First, aren't you integrating over $r'$?...It shouldn't be present in your final result! Second, $$|z\pm a|=\left\{\begin{array}{lr}z\pm a &, z\geq \mp a\\a\pm z &, z<\mp a\end{array}\right.$$ 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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