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 \( R_4 \)?
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).