(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A nonconducting hemispherical cup of inner Radius R has a total charge Q spread uniformly over its inner surface. Find the electric field at the center of curvature.

2. Relevant equations

[itex]\sigma[/itex] = [itex]\frac{Q}{2piR^{2}}[/itex]

dq = [itex]\sigma[/itex] dA = [itex]\sigma[/itex] R[itex]^{2}[/itex]Sin[itex]\theta[/itex]d[itex]\theta[/itex]d[itex]\phi[/itex]

k = [itex]\frac{1}{4\pi\epsilon_{}0}[/itex]

3. The attempt at a solution

dE = [itex]\frac{kdq}{R^{2}}[/itex]

dE = [itex]\frac{k \sigma R^{2} Sin\theta d\theta d\phi}{R^{2}}[/itex] = [itex]k \sigma Sin \theta d \theta d \phi[/itex]

E = [itex]k \sigma \int^{2 \pi}_{0} \int^{\frac{\pi}{2}}_{0} Sin \theta d \theta d \phi[/itex]

E = k [itex]\sigma 2 \pi [/itex] = k [itex]\frac{Q}{R^{2}}[/itex]

But for some reason, the answer is listed as k [itex]\frac{Q}{3R^{2}}[/itex]. I have no idea why my answer is factor of 3 larger. Any help would be appreciated.

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# Homework Help: Electric Field in a Hemisphere

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