Question

Two identical positive point charges q are placed on y axis at y = -a and y = +a as shown. Electric flux due to these charges through xz plane passing through y = 0, is

• A. $\frac{q}{\epsilon_0}$

Answer 1

Answer: (A)

$\frac{q}{\epsilon_0}$

Answer 2

Two positive point charges q are placed at y = -a and y = +a. The xz plane is at y = 0.

1. For a single point charge q, the total electric flux emitted is $q/\epsilon_0$.
2. By symmetry, for a point charge placed at a distance from an infinite plane, exactly half of the electric field lines pass through the plane. So, the magnitude of the flux due to one charge through the xz plane is $\frac{1}{2} \frac{q}{\epsilon_0}$.
3. The charge at y = +a produces flux downwards through the xz plane.
4. The charge at y = -a produces flux upwards through the xz plane.
5. If "electric flux" refers to the net flux (signed quantity), then the fluxes are equal in magnitude and opposite in direction, resulting in a net flux of 0.
6. However, if the question asks for the total magnitude of flux crossing the plane (which is a common interpretation when the net flux is zero and a non-zero option is provided), then we sum the magnitudes of the fluxes from each charge. Total magnitude of flux = (Magnitude of flux from charge at +a) + (Magnitude of flux from charge at -a) Total magnitude of flux = $\frac{1}{2} \frac{q}{\epsilon_0} + \frac{1}{2} \frac{q}{\epsilon_0} = \frac{q}{\epsilon_0}$.

This interpretation leads to the answer $\frac{q}{\epsilon_0}$.