Question

the point charges q and -q are fixed in space at separation a. the flux through infinite plane perpendicular to the line joining infinite plane perpendicular to the line joining the charges and at a distance od a/3 from +q

Answer 1

0

Answer 2

Solution:

1. Setup: Place +q at (0, 0, 0) and –q at (0, 0, a). The infinite plane (perpendicular to the z–axis) is at z = a/3 (i.e. a distance a/3 from +q).

2. Flux from a point charge through an infinite plane: For any point charge at a finite distance from an infinite plane, the plane intercepts a solid angle of 2π steradians, so it gets half of the total flux, i.e.,

   $$\Phi = \frac{q}{2\epsilon_0}.$$

3. Calculations for each charge:

   • For the charge +q (located below the plane), the flux crossing the plane upward is $$\Phi_{+} = \frac{q}{2\epsilon_0}.$$
   • For the charge –q (located above the plane), the flux crossing the plane downward is $$\Phi_{-} = \frac{-q}{2\epsilon_0}.$$

4. Superposition: The net flux through the plane is the sum:

   $$\Phi_{\text{net}} = \Phi_{+} + \Phi_{-} = \frac{q}{2\epsilon_0} + \frac{-q}{2\epsilon_0} = 0.$$

Explanation (minimal): Place charges with +q at (0,0,0) and –q at (0,0,a) and the plane at z = a/3. An infinite plane intercepts half the flux of a point charge, so net flux = q/(2ε₀) + (–q/(2ε₀)) = 0.