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

In summary, the surface of a sphere of radius a is charged with a constant surface density \sigma. The total charge Q' on the sphere is given by: \vec{F} = \frac{\normalsizeq\sigma\hat{z}}{\epsilon_{0}z^{2}}. The force produced by this charge distribution on a point charge q located on the z axis is found to be \vec{F} = \frac{\normalsizeq\sigmaq\hat{z}}{\epsilon_{0}q^{2}}.
  • #1
Kizaru
45
0

Homework Statement


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.

Homework Equations


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)
 
Physics news on Phys.org
  • #2
Kizaru said:

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?:wink:...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!:wink:
 
  • #3
gabbagabbahey said:
Hmmm... isn't there a [itex]q[/itex] missing from this expression?:wink:...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!:wink:
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 :)
 

1. What is the formula for calculating the force on a point charge due to a constant sphere surface charge density?

The formula for calculating the force on a point charge (q) due to a constant sphere surface charge density (σ) is F = qσ/4πε0r², where ε0 is the permittivity of free space and r is the distance between the point charge and the center of the sphere.

2. How does the force on a point charge due to a constant sphere surface charge density change with distance?

The force on a point charge due to a constant sphere surface charge density follows an inverse square relationship with distance. This means that as the distance between the point charge and the center of the sphere increases, the force decreases.

3. Can the direction of the force on a point charge due to a constant sphere surface charge density change?

Yes, the direction of the force on a point charge due to a constant sphere surface charge density can change. If the point charge is located outside of the sphere, the force will always be directed towards the center of the sphere. However, if the point charge is located inside the sphere, the force will be directed away from the center.

4. How does the force on a point charge due to a constant sphere surface charge density compare to the force on a point charge due to a point charge?

The force on a point charge due to a constant sphere surface charge density is similar to the force on a point charge due to a point charge in terms of the inverse square relationship with distance. However, the force due to a constant sphere surface charge density is typically smaller than the force due to a point charge, as the charge is spread out over a larger surface area.

5. Can the force on a point charge due to a constant sphere surface charge density be repulsive?

Yes, the force on a point charge due to a constant sphere surface charge density can be repulsive if the point charge is located inside the sphere. This is because the point charge will have the same sign as the charges on the surface of the sphere, causing a repulsive force between them.

Similar threads

  • Advanced Physics Homework Help
Replies
13
Views
2K
  • Advanced Physics Homework Help
Replies
19
Views
701
  • Advanced Physics Homework Help
Replies
1
Views
301
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
7
Views
2K
  • Advanced Physics Homework Help
Replies
6
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
400
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
875
Back
Top