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 a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
Resonance in X$_2$Y can be represented as
The enthalpy of formation of X$_2$Y is 80 kJ mol$^{-1}$, and the magnitude of resonance energy of X$_2$Y is: