Question

A portion of the hemispherical surface is cut out by two planes passing through the same diameter, such that the angle between them at the base is $\alpha$. The surface carries a uniform charge density $\sigma$. Determine the electric field strength E at the centre O due to the charges on this isolated portion.

Answer 1

The electric field strength $\vec{E}$ at the center O is given by: $\vec{E} = \frac{\sigma}{8\epsilon_0} \sin\left(\frac{\alpha}{2}\right) \hat{u} + \frac{\sigma\alpha}{8\pi\epsilon_0} \hat{v}$ where $\hat{v}$ is the unit vector along the diameter, and $\hat{u}$ is a unit vector perpendicular to the diameter, lying in the plane that bisects the angle $\alpha$ between the two cutting planes. The magnitude of the electric field is: $E = \frac{\sigma}{8\epsilon_0} \sqrt{\sin^2\left(\frac{\alpha}{2}\right) + \left(\frac{\alpha}{\pi}\right)^2}$

Answer 2

The electric field at the center O due to a point charge $dq$ at position $\vec{r}$ is $d\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{dq}{|\vec{r}|^2} \hat{r}$. For a hemispherical surface of radius R with uniform surface charge density $\sigma$, the center O is at the origin. Any point on the surface is at a distance R from O, so $|\vec{r}| = R$. The charge element is $dq = \sigma dA$, where $dA$ is the surface area element. In spherical coordinates, $dA = R^2 \sin\theta d\theta d\phi$. The position vector is $\vec{r} = R(\sin\theta\cos\phi \hat{i} + \sin\theta\sin\phi \hat{j} + \cos\theta \hat{k})$. The unit vector is $\hat{r} = \sin\theta\cos\phi \hat{i} + \sin\theta\sin\phi \hat{j} + \cos\theta \hat{k}$.

The electric field element is $d\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{\sigma R^2 \sin\theta d\theta d\phi}{R^2} \hat{r} = \frac{\sigma}{4\pi\epsilon_0} \sin\theta d\theta d\phi \hat{r}$.

The portion of the hemisphere is cut by two planes passing through a diameter, with an angle $\alpha$ between them. Let this diameter be along the z-axis, and the hemisphere be in the region $z \ge 0$. We can set the integration limits for the azimuthal angle $\phi$ from $-\alpha/2$ to $\alpha/2$ and for the polar angle $\theta$ from $0$ to $\pi/2$. This choice aligns the x-axis to bisect the angle between the two planes.

The electric field at the center O is the integral of $d\vec{E}$ over this portion: $\vec{E} = \int_{-\alpha/2}^{\alpha/2} \int_0^{\pi/2} \frac{\sigma}{4\pi\epsilon_0} (\sin^2\theta\cos\phi \hat{i} + \sin^2\theta\sin\phi \hat{j} + \sin\theta\cos\theta \hat{k}) d\theta d\phi$

Evaluating the integrals: $\int_0^{\pi/2} \sin^2\theta d\theta = \frac{\pi}{4}$ $\int_0^{\pi/2} \sin\theta\cos\theta d\theta = \frac{1}{2}$ $\int_{-\alpha/2}^{\alpha/2} \cos\phi d\phi = [\sin\phi]_{-\alpha/2}^{\alpha/2} = \sin(\alpha/2) - \sin(-\alpha/2) = 2\sin(\alpha/2)$ $\int_{-\alpha/2}^{\alpha/2} \sin\phi d\phi = [-\cos\phi]_{-\alpha/2}^{\alpha/2} = -\cos(\alpha/2) - (-\cos(-\alpha/2)) = 0$ $\int_{-\alpha/2}^{\alpha/2} d\phi = \alpha$

The components of the electric field are: $E_x = \frac{\sigma}{4\pi\epsilon_0} \left(\frac{\pi}{4}\right) (2\sin(\alpha/2)) = \frac{\sigma}{8\epsilon_0} \sin(\alpha/2)$ $E_y = \frac{\sigma}{4\pi\epsilon_0} \left(\frac{\pi}{4}\right) (0) = 0$ $E_z = \frac{\sigma}{4\pi\epsilon_0} \left(\frac{1}{2}\right) (\alpha) = \frac{\sigma\alpha}{8\pi\epsilon_0}$

So, the electric field vector is $\vec{E} = \frac{\sigma}{8\epsilon_0} \sin(\alpha/2) \hat{i} + \frac{\sigma\alpha}{8\pi\epsilon_0} \hat{k}$. Here, $\hat{k}$ is the unit vector along the diameter, and $\hat{i}$ is a unit vector in the plane perpendicular to the diameter, bisecting the angle $\alpha$ between the two cutting planes. The magnitude of the electric field is $|\vec{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2} = \sqrt{\left(\frac{\sigma}{8\epsilon_0} \sin(\alpha/2)\right)^2 + 0^2 + \left(\frac{\sigma\alpha}{8\pi\epsilon_0}\right)^2}$. $|\vec{E}| = \frac{\sigma}{8\epsilon_0} \sqrt{\sin^2(\alpha/2) + \left(\frac{\alpha}{\pi}\right)^2}$

The electric field strength E at the center O is given by the vector: $\vec{E} = \frac{\sigma}{8\epsilon_0} \sin\left(\frac{\alpha}{2}\right) \hat{u} + \frac{\sigma\alpha}{8\pi\epsilon_0} \hat{v}$ where $\hat{v}$ is the unit vector along the diameter, and $\hat{u}$ is a unit vector perpendicular to the diameter, lying in the plane that bisects the angle $\alpha$ between the two cutting planes. The magnitude of the electric field is: $E = |\vec{E}| = \frac{\sigma}{8\epsilon_0} \sqrt{\sin^2\left(\frac{\alpha}{2}\right) + \left(\frac{\alpha}{\pi}\right)^2}$