Engineering Electric field of hemisphere with the given charge density

[YehiaMedhat]

Homework Statement

A hemispherical shell of radius $4.0 cm$ is located at the region $z > 0.0$ centered at the origin. It carries a surface charge distribution of density $\rho = 2.0\times 10^{-6} C/m^2$ in the region $y > 0.0$ and a surface charge distribution of density $\rho = -2.0\times 10^{-6} C/m^2$ in the region $y < 0.0$. Apply Coulomb's law to find the electric field intensity $E$ at the origin.

Relevant Equations

$\vec{E} = \frac{Q}{4\pi \epsilon r^2} \vec{a}$

The question is:
A hemispherical shell of radius $4.0 cm$ is located at the region $z > 0.0$ centered at the origin. It carries a surface charge distribution of density $\rho = 2.0\times 10^{-6} C/m^2$ in the region $y > 0.0$ and a surface charge distribution of density $\rho = -2.0\times 10^{-6} C/m^2$ in the region $y < 0.0$. Apply Coulomb's law to find the electric field intensity $E$ at the origin.
For me I considered slicing each half of the hemisphere into rings to get an intuition about how will the electric field will look like, so as in the picture, each vertical component is gonna be canceled with the other one from the opposing charge area, so we're left only with the horizontal components multiplied by 2.

Integrating with phi from 0 to pi over 2 and theta from 0 to 2pi
$\begin{aligned} & 2 \int_0^{2 \pi} \int_0^{\frac{\pi}{2}} \frac{(\rho d \phi)(f d \theta) \rho_s(\cos \phi)}{4 \pi \varepsilon_0 \rho^2} \\ & =\frac{2 \rho_s}{4 \pi \varepsilon_0} \int_0^{2 \pi} d \theta \int_0^{\frac{\pi}{2}} \cos \phi d \phi \\ & =\frac{(2 \pi) 2 \rho}{4 \pi \varepsilon_0}[\sin \phi]_0^{\frac{\pi}{2}}=\frac{\rho_s}{\varepsilon_0} \end{aligned}$
Is that correct?

 

[anuttarasammyak]

My strtegy is
1. Get E-field for uniformally charged hemisphere shell at its center.
2. Couple the hemisphere shell with another hemisphere shell of minus charge. The E-field is doubled.
3. Halve this sphere with plus-minus half and half latitude. The E-field is halved.
So your question is equivalent to 1 which is easier to treat. It is
\frac{\sigma}{4\pi\epsilon_0 r^2}\int_0^{\frac{\pi}{2}}rd\theta \ 2\pi r\sin\theta\cos\theta

 

[PhDeezNutz]

I’m inclined to agree with the above poster. You were right to project everything onto the $z-\text{axis}$ due to symmetry. However your expression for the differential surface element is missing a factor of $\sin \theta$. Rectify this and your answer should work out.

I assume you know spherical coordinates and I assume you’re using the physics convention.

For a differential element in spherical coordinates

$d \tau = r^2 \sin \theta dr d\theta d\phi$

Now restrict yourself to the surface of that sphere.

How does $d \tau$ change?