
To find the electric flux through the surface \( S \) due to the given configuration of charges, we apply Gauss's Law. Gauss's Law states that the total electric flux \( \Phi_E \) through a closed surface is equal to the net charge enclosed \( Q_{\text{enc}} \) divided by the permittivity of free space \( \epsilon_0 \):
\(\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}\)
In this problem, you have five charges: +q, +5q, –2q, +3q, and –4q. It is given that the surface \( S \) encloses the charges +q, +5q, and –2q. Let's calculate the net charge within the surface:
So, the total enclosed charge \( Q_{\text{enc}} \) is:
\(Q_{\text{enc}} = q + 5q - 2q = 4q\)
Substitute this back into Gauss's Law:
\(\Phi_E = \frac{4q}{\epsilon_0}\)
Hence, the electric flux through the surface \( S \) is \(\frac{4q}{\epsilon_0}\).
The correct answer is the option: \(\frac{4q}{\epsilon_0}\).
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).
A line charge of length \( \frac{a}{2} \) is kept at the center of an edge BC of a cube ABCDEFGH having edge length \( a \). If the density of the line is \( \lambda C \) per unit length, then the total electric flux through all the faces of the cube will be : (Take \( \varepsilon_0 \) as the free space permittivity)
A metallic sphere of radius \( R \) carrying a charge \( q \) is kept at a certain distance from another metallic sphere of radius \( R_4 \) carrying a charge \( Q \). What is the electric flux at any point inside the metallic sphere of radius \( R \) due to the sphere of radius \( R_4 \)? 
