Electric Potential Hemisphere Problem

Click For Summary
SUMMARY

The discussion centers on calculating the electric potential and electric field at the center of a hemisphere with radius R and surface charge density σ. The initial approach incorrectly used dA = R^2 dθ dα instead of the correct dA = R^2 sin(θ) dθ dα, leading to erroneous results. The correct electric potential is V = σR / (2ε₀) and the electric field magnitude is E = σ / (4ε₀). The participants clarified the integration process and the significance of sine and cosine in the calculations.

PREREQUISITES
  • Understanding of electric potential and electric field concepts
  • Familiarity with surface charge density and its implications
  • Knowledge of spherical coordinates and their integration
  • Basic proficiency in calculus, particularly multiple integrals
NEXT STEPS
  • Study spherical coordinate integration techniques
  • Learn about electric fields generated by surface charge distributions
  • Explore advanced problems involving three-dimensional electric potential
  • Review the derivation of electric potential and field equations in electrostatics
USEFUL FOR

Students and professionals in physics, particularly those focusing on electromagnetism, as well as educators seeking to enhance their understanding of electric fields and potentials in three-dimensional geometries.

Saketh
Messages
258
Reaction score
2
There is a hemisphere of radius R and surface charge density \sigma. Find the electric potential and the magnitude of the electric field at the center of the hemisphere.​
I started by saying V = \int dV = \int \frac{\sigma dA}{4\pi \epsilon_0 R}. This, at least, I am confident is correct.

Then, I changed dA into Rd\theta and Rd\alpha, where the first one goes across the hemisphere's surface, and the second one goes around the hemisphere's circular edge.

Continuing:
<br /> V = \frac{\sigma R}{4\pi \epsilon_0}\int_{0}^{\pi/2}d\theta \int_{0}^{2\pi}d\alpha
Which, when evaluated, gives me the wrong answer.

Oblivious to my error at the time, I continued to the second part.

E = \int dE = \int \frac{\sigma dA}{4\pi \epsilon_0 R^2}\hat{r}.
I said that \hat{r} = \cos{\theta}\hat{i} + \sin{\theta}\hat{j}. Using similar logic as I did for the potential part of the problem, I continued:

<br /> E = \frac{\sigma R}{4\pi \epsilon_0}\int_{0}^{\pi/2}\cos{\theta}\hat{i} + \sin{\theta}\hat{j} d\theta \int_{0}^{2\pi}d\alpha

Which, when evaluated, also gives me the wrong answer. The correct answers for the parts are, respectively, V = \frac{\sigma R}{2 \epsilon_0} and E = \frac{\sigma}{4\epsilon_0}.

I have no idea what I'm doing incorrectly. I've never done a three-dimensional integration before, but this is the way that I thought it should be done. What have I done wrong?
 
Physics news on Phys.org
Saketh said:
There is a hemisphere of radius R and surface charge density \sigma. Find the electric potential and the magnitude of the electric field at the center of the hemisphere.​
I started by saying V = \int dV = \int \frac{\sigma dA}{4\pi \epsilon_0 R}. This, at least, I am confident is correct.

Then, I changed dA into Rd\theta and Rd\alpha, where the first one goes across the hemisphere's surface, and the second one goes around the hemisphere's circular edge.
I am not sure what you mean by this but from the rest of your calculation (which I did not check), you used dA = R^2 d\theta d\alpha. But the correct relation is dA = R^2 sin \theta d\theta d\alpha. You can see that your equation was not right since the total integral of da over a sphere should give 4 \pi R^2. Your expression would give 2 \pi^2 R^2.
 
nrqed said:
I am not sure what you mean by this but from the rest of your calculation (which I did not check), you used dA = R^2 d\theta d\alpha. But the correct relation is dA = R^2 sin \theta d\theta d\alpha. You can see that your equation was not right since the total integral of da over a sphere should give 4 \pi R^2. Your expression would give 2 \pi^2 R^2.
Yes - I don't understand why it's supposed to be dA = R^2 \sin{\theta} d\theta d\alpha. Why is there a sine? I realize that dA = R^2 d\theta d\alpha doesn't give the right answer, but conceptually I don't understand why there needs to be a \sin{\theta}.

I still don't get the second part. The solution puts in a cosine as well as a sine. If I understand why there's a sine in the equation for potential, I'll understand why it's here. However, I don't understand why there should be a cosine. The solution says that dE_z = dE \cos{\theta}, but I thought it would be dE_z = dE \sin{\theta}.
 
The reason the sin is there is because of the way the integral is done. You are integrating over two variables, the azimuthal and the zenith coordinates. The sin is introduced when integrating the azimuthal variable. The radius is actually Rsin(zenith angle). You then add up these strings into a hemisphere through the second integration.

http://img156.imageshack.us/img156/870/hemiyl3.gif
 
Last edited by a moderator:
mezarashi said:
The reason the sin is there is because of the way the integral is done. You are integrating over two variables, the azimuthal and the zenith coordinates. The sin is introduced when integrating the azimuthal variable. The radius is actually Rsin(zenith angle). You then add up these strings into a hemisphere through the second integration.
Ok, I think I get it now. The only way I'll see if I get it is by doing more problems, but your explanation makes me understand what's going on. Thanks!
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
Replies
6
Views
2K
Replies
12
Views
2K
Replies
2
Views
681
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
Replies
11
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 11 ·
Replies
11
Views
1K