Question

the point charges q and -1 are fixed in space at seperation a. the flux through infinite plane perpendicular to line joining

Answer 1

(q-1)/(2ε₀)

Answer 2

Assume the two charges are placed on the z–axis at

$(0,0,\frac{a}{2})$ with charge +q, and $(0,0,-\frac{a}{2})$ with charge -1.

Let the infinite plane be the $xy$–plane (i.e. $z=0$), which is perpendicular to the line joining the charges and bisects the separation.

For a point charge, the total flux is $\Phi_{\text{total}}=\frac{q}{\epsilon_0}$. An infinite plane (when the point charge is at a finite distance from it) intercepts exactly half of the total flux (because the plane subtends a solid angle of $2\pi$ out of $4\pi$ steradians).

• For the $+q$ charge (above the plane):
  The flux crossing the plane (downward) is $$\Phi_+=\frac{q}{2\epsilon_0}.$$
• For the $-1$ charge (below the plane):
  The flux crossing upward from this charge is $$\Phi_-=\frac{-1}{2\epsilon_0}.$$

By superposition, the net flux through the plane is

$$\Phi = \Phi_+ + \Phi_- = \frac{q}{2\epsilon_0} + \frac{-1}{2\epsilon_0} = \frac{q-1}{2\epsilon_0}.$$