According to Gauss’s law, the electric flux Φ through a closed surface is given by:
\(Φ = \frac{q_{in}}{\epsilon_0}\),
where qin is the total charge enclosed by the surface.
In this problem, the charges enclosed by the surface are \(q_{in} = q + (-2q) + 5q = 4q\).
Thus, the electric flux is:
\(Φ = \frac{4q}{\epsilon_0}\).
The correct answer is Option (2).
Two charges of \(5Q\) and \(-2Q\) are situated at the points \((3a, 0)\) and \((-5a, 0)\) respectively. The electric flux through a sphere of radius \(4a\) having its center at the origin is:
Expression for an electric field is given by $\overrightarrow{ E }=4000 x^2 i \frac{ V }{ m }$ The electric flux through the cube of side $20 cm$ when placed in electric field (as shown in the figure) is ___$V cm$