(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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Electric Field in a Hemisphere

Have something to add?

**Physics Forums | Science Articles, Homework Help, Discussion**