1

In all these 3 cases, the net electric flux is found to be $q/2ε_0$.

I think this has something to do with the integral of $\left(\mathbf{E}\boldsymbol{\cdot}\mathrm d\mathbf{S}\right)$ which appears on the left side of Gauss's Law. We are somehow exploiting the symmetry to arrive at the result where the $\mathbf{E}$ (or electric field) in the formula comes out of the integral. But I don't know how?

Secondly, does the electric flux remain the same when we place the point charge on the face of an unsymmetrical 3d object instead of these 3 objects (cube, cone and hemisphere) ?

If anyone would be kind enough to help me out in the simplest way possible, that would make my understanding clear!

Frobenius
  • 15,613
Arman Ansary
  • 15

1 Answers1

1

By gauss law, we have that $\displaystyle \int \vec{E} \cdot \mathrm{d} \vec{S}=\dfrac{Q_{\text{enc}}}{\epsilon_0}$, so if for example, when a charge is kept on the face of a hemisphere, the charge is not enclosed. We want to enclose it fully in a body such that electric field still remains uniform so that gauss law can be applied. So, we complete the rest half by taking another hemisphere, so now flux through the whole sphere would become $\dfrac{Q_\text{enc}}{\epsilon_0}$, so the flux through the hemisphere would be just half of it due to symmetry. Same goes with cone and cube.

V.G
  • 352