# $E$-field on half charged sphere (only north hemisphere)

A sphere of radius $$R$$ is centered on the origin. The “northern” hemisphere defined by $$z > 0$$ carries a uniform surface charge $$\sigma$$, during the “southern” hemisphere carries no charge. Find the electric field at the origin. (Use symmetry to argue that it is enough to calculate $$ˆz \cdot E$$, and then compute this by integrating over the hemisphere.)

Using Gauss' Law, $$\text{electric flux} = \frac{\text{total charge}}{E_{0}} = E - \text{field} \int dA$$ the left side is $$\sigma 3 \pi \frac{r^{2}}{E_{0}}$$ if I'm correct. The right side should integrate the area as several rings from $$0$$ to $$z$$ but I need help setting up the integral, my calculus is pretty rusty.

• Using Gauss' Law… You can’t use Gauss’ Law; there isn’t enough symmetry to know that the field is going to be constant over some surface of integration. Commented Sep 19, 2022 at 23:25

Because of the symmetry, every projection of the electric field on the x-y plane has a corresponding opposite contribution on the other side of the circle, thus, the only contribution of the electric field emerge on the z-axis. It's possible to build the infinitesimal electric field of an infinitesimal portion of spherical surface $$dE=k\frac{dq}{R^2}=k\frac{\sigma dS}{R^2}=k\frac{\sigma R\sin\theta d\theta d\phi}{R^2}$$ Now, we must integrate $$E=k\sigma\int\int \frac{\sin\theta d\theta d\phi}{R}=\frac{k\sigma}{R}\int_0^\frac{\pi}{2}\sin\theta d\theta\int_0^{2\pi}d\phi$$ $$E=\frac{k\sigma}{R}[-\cos\theta]_0^{\frac{\pi}{2}}\cdot2\pi=\frac{2\pi k\sigma[0+1]}{R}=\frac{2\pi k\sigma}{R}=\frac{\sigma}{2\epsilon_0 R}$$ Where $$k=\frac{1}{4\pi \epsilon_0}$$ and the integration $$\int_0^{\frac{\pi}{2}} d\theta$$ instead of $$\int_0^{\pi} d\theta$$ takes into account just half sphere