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 :

<p>The electric flux through any closed surface is given by Gauss's law, which can be stated as:</p> $\Phi = \frac{Q_{\text{enc}}}{\varepsilon_0}$ <p>where:</p> <ul> <li>$\Phi$ = electric flux</li> <li>$Q_{\text{enc}}$ = total charge enclosed by the surface</li> <li>$\varepsilon_0$ = permittivity of free space (electric constant)</li> </ul> <p>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.</p> <p>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.</p> <p>The electric flux through the sphere is therefore given simply by the enclosed charge:</p> $\Phi = \frac{5Q}{\varepsilon_0}$