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?