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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.

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  • $\begingroup$ 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. $\endgroup$
    – Ghoster
    Commented Sep 19, 2022 at 23:25

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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

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