Electric Potential Hemisphere Problem

AI Thread Summary
The discussion revolves around 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 dθ dα instead of the correct dA = R² sin θ dθ dα, leading to incorrect results. The importance of the sine function arises from the geometry of the hemisphere, specifically when integrating over the azimuthal and zenith coordinates. The participants clarify that the cosine factor in the electric field calculation accounts for the vertical component of the electric field, which is essential for accurate integration. Ultimately, the discussion highlights the need for correct differential area elements in three-dimensional integrals to obtain the right answers.
Saketh
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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?
 
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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
 
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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!
 
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