Question

A charge q is placed at the centre of the open end of a cylindrical vessel (figure 30-Q3). The flux of the electric field through the surface of the vessel is
(a) zero
(b) q/εν
(c) q/2εν
(d) 2q/εν.
Figure

Answer 1

The flux of the electric field via the vessel’s surface is q/εν.

Explanation:

• Half of the electric flux’s charge q is shared by the given surface. Therefore, q = (1/2)(q/∈0) = (q/2∈0).
• At the center of the cylindrical vessel’s open end, when a charge Q is placed, then only half of the charge is donated to the flux. This is because half will life outside the surface and half will lie inside.
• So, the cylindrical vessel’s surface area = half of the electric flux’s total surface area. So, flux=Q/2ε0.