This is a math problem that appeared in a physics one:
"Evaluate the density of electrical charge in the surface of a hollow sphere such that the electric field is constant (in direction and absolute value) inside of it."
I wonder if this can be solved by using Leibniz's Rule for integration.
First let's supose the field points towards the same direction of the $x$- axis in the usual $R^3$ ($\vec {i}$ direction) the plane $yz$ is perpendicular to the field so it is an equipotential surface. Therefore the solution of the problem does not depend on $\theta$ since the charge density must be equal on every point of the intersection of the plane with the sphere.
:
So:
$d\vec{E} =\frac{1}{4\pi\epsilon} \frac{(\vec r(\theta, \phi) - \vec r(P))dq}{(||\vec r(\theta, \phi) - \vec r(P)||)^3}$
$\vec{E} = \frac{1}{4\pi\epsilon} \int_{0}^{2\pi}\int_{0}^{\pi} \frac{\sigma (\phi) R^2 \sin(\phi)(\vec r(\theta, \phi) - \vec r(P)) d\theta d\phi}{||\vec r(\theta, \phi) - \vec r(P)||^3} $
$\vec{E} = \frac{R^2}{4\pi\epsilon} \int_{0}^{2\pi}\int_{0}^{\pi} \frac{\sigma (\phi) \sin(\phi)(\vec r(\theta, \phi) - \vec r(P)) d\theta d\phi}{||\vec r(\theta, \phi) - \vec r(P)||^3} $
where
$\vec r(\theta,\phi) = R(\cos(\theta)\sin(\phi), \sin(\theta)\sin(\phi), \cos(\phi))$
$\vec r(P) = (x_p,y_p,z_p)$ , $x_p^2 + y_p^2+z_p^2 < R^2$ is a constant vector
Is this solvable by using the Leibniz rule for integrals ?
$\frac{\partial \vec{E}}{\partial x_p} = 0$
$\frac{\partial \vec{E}}{\partial y_p} = 0$
$\frac{\partial \vec{E}}{\partial z_p} = 0$