Question

A charge q is placed at the centre of the open end of a cylindrical vessel. The flux of the electric field through the surface of the vessel is(a) zero(b) $\frac{q}{\epsilon_{0}}$(c) $\frac{q}{2\epsilon_{0}}$(d) $\frac{2q}{\epsilon_{0}}$

Answer 1

Hey

A charge q is placed at the centre of the open end of a cylindrical vessel. The flux of the electric field through the surface of the vessel is(a) zero(b) $\frac{q}{\epsilon_{0}}$(c) $\frac{q}{2\epsilon_{0}}$(d) $\frac{2q}{\epsilon_{0}}$

A correct

Answer 2

Answer: (c): q/2(epsilon nought)

Explanation: first consider an entirely closed cylinder, with the charge q, then residing on the inner surface of the now closed end. Then flux through the entire closed cylinder is, according to Gauss' Law, q/(epsilon nought). However, when you're opening the lid, the flux through the side surfaces of the cylinder would not change, but now half of the flux that should have normally passed through the open end, is not "actually passing" through the cylinder, so the flux reduces to half: q/2(epsilon nought)