Two charges of 5 Q and -2 Q are situated at the points (3 a, 0) and (-5 a, 0) respectively. The electric flux through a sphere of radius '4 a' having center at origin is :

Solution

Physics sir · 2 steps

Gauss's law counts only the charge inside the sphere, so first check which charge is actually enclosed.

1. Only enclosed charge. The flux through a closed surface depends only on the charge inside it. A charge outside sends in as many field lines as it takes out, so its net flux is zero.

   Charge	Distance from origin	Compared with 4a	Counts?
   5Q at (3a, 0)	3a	3a < 4a, inside	yes
   -2Q at (-5a, 0)	5a	5a > 4a, outside	no

   φ = q_(enc)/ε₀ = 5Q/ε₀

2. So. Only 5Q is enclosed, so the flux through the sphere is 5Q/ε₀, option (d).

The idea to keep. In any flux question, compare each charge's distance from the centre with the radius first. Charges outside can change the field on the surface, but never the net flux.

Answer from the NTA final answer key. Solution by PrepWiser.