Total flux passing through the close cube \(ϕ = \frac {q}{ε_0}\)
All six surfaces exhibit symmetry with respect to the charge, resulting in an equal contribution to the flux. Consequently, the flux through any single face is:
\(ϕ = \frac ϕ 6 = \frac {q}{6ε_0}\)
\(φ_{face} = \frac {q}{6ε_0} = \frac {4\pi q}{6(4\piε_0)}\)
So, the correct option is (B): \(\frac {4\pi q}{6(4πε_0)}\)
The current passing through the battery in the given circuit, is:
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :
A full wave rectifier circuit with diodes (\(D_1\)) and (\(D_2\)) is shown in the figure. If input supply voltage \(V_{in} = 220 \sin(100 \pi t)\) volt, then at \(t = 15\) msec:
Gauss law states that the total amount of electric flux passing through any closed surface is directly proportional to the enclosed electric charge.
According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.
For example, a point charge q is placed inside a cube of edge ‘a’. Now as per Gauss law, the flux through each face of the cube is q/6ε0.
As per the Gauss theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. Therefore, if ϕ is total flux and ϵ0 is electric constant, the total electric charge Q enclosed by the surface is;
Q = ϕ ϵ0
The Gauss law formula is expressed by;
ϕ = Q/ϵ0
Where,
Q = total charge within the given surface,
ε0 = the electric constant.