JEE Main 20244 Apr, Evening shiftPhysicsHardElectrostatics167× since 2002A charge qqq is placed at the center of one of the surface of a cube. The flux linked with the cube is:AOption A. q2ϵ0\frac{q}{2 \epsilon_0}2ϵ0qBOption B. ZeroCOption C. q4ϵ0\frac{q}{4 \epsilon_0}4ϵ0qDOption D. q8ϵ0\frac{q}{8 \epsilon_0}8ϵ0qCheckSkip · show solution
