Question

Find electric field due to uniformly charged hemispherical cap having charge density $\sigma$ at point P :-

Answer 1

$\mathbf{E} = \frac{\sigma}{4\epsilon_0}\,\hat{z}$

Answer 2

We wish to find the electric field at point $P$ (the top of the hemispherical cap) due to a uniformly charged hemispherical cap of radius $R$ and surface charge density $\sigma$.

1. Setup:
   Use spherical coordinates where a point on the cap is given by $(R, \theta, \phi)$ with:

   • $\theta$ from $0$ to $\pi/2$ (since the cap is a hemisphere)
   • $\phi$ from $0$ to $2\pi$

2. Choosing an Element of Charge:
   The surface element is

   $$dA = R^2 \sin\theta\, d\theta\, d\phi.$$

   Its charge is

   $$dq = \sigma\, dA = \sigma R^2 \sin\theta\, d\theta\, d\phi.$$

3. Contribution to the Electric Field:
   The field from $dq$ at $P$ (which is on the axis through the center of curvature) is given by Coulomb's law:

   $$d\mathbf{E} = \frac{1}{4\pi\epsilon_0}\frac{dq}{R^2}\hat{\mathbf{r}},$$

   where the distance from each element to point $P$ is $R$ (due to the symmetry of the spherical surface) and $\hat{\mathbf{r}}$ will have a component along the vertical (z-) direction given by $\cos\theta$. Hence, the vertical component is

   $$dE_z = \frac{1}{4\pi\epsilon_0}\frac{\sigma R^2 \sin\theta\, d\theta\, d\phi}{R^2}\cos\theta = \frac{\sigma}{4\pi\epsilon_0}\sin\theta \cos\theta\, d\theta\, d\phi.$$

4. Integrating:
   Integrate over the entire cap:

   $$E_z = \frac{\sigma}{4\pi\epsilon_0}\int_{\phi=0}^{2\pi}\,d\phi\int_{\theta=0}^{\pi/2} \sin\theta\cos\theta\, d\theta.$$

   The $\phi$-integral gives:

   $$\int_{0}^{2\pi} d\phi = 2\pi.$$

   The $\theta$-integral:

   $$\int_{0}^{\pi/2}\sin\theta\cos\theta\,d\theta.$$

   Use the substitution $u=\sin\theta$ so $du=\cos\theta\,d\theta$. When $\theta=0$, $u=0$; when $\theta=\pi/2$, $u=1$. Thus,

   $$\int_{0}^{1} u\,du = \frac{1}{2}.$$

   Putting it together:

   $$E_z = \frac{\sigma}{4\pi\epsilon_0}(2\pi)\left(\frac{1}{2}\right) = \frac{\sigma}{4\epsilon_0}.$$

5. Result:
   The net electric field at $P$ is:

   $$\mathbf{E} = \frac{\sigma}{4\epsilon_0}\,\hat{z}.$$