The electric flux through any closed surface is given by Gauss's law, which can be stated as:

where:

• $$\Phi$$ = electric flux
• $$Q_{\text{enc}}$$ = total charge enclosed by the surface
• $$\varepsilon_0$$ = permittivity of free space (electric constant)

In this scenario, we have a sphere of radius $$4a$$ with its center at the origin and two charges, $$5Q$$ at point (3a, 0) and $$-2Q$$ at point (-5a, 0). Since the sphere's radius is $$4a$$, the charge $$5Q$$, which is located at (3a, 0), lies inside the sphere, whereas the charge $$-2Q$$, located at (-5a, 0), lies outside the sphere. Gauss's law only considers charges that are enclosed within the surface.

Thus, the charge $$-2Q$$ has no impact on the electric flux through the sphere because it is not enclosed by the sphere. Only the charge $$5Q$$ contributes to the electric flux inside the sphere.

The electric flux through the sphere is therefore given simply by the enclosed charge: