For example;
The electric field between infinitely parallel plates is a uniform electric field, and its magnitude is
e =σdu/ε0 = & gt； ε0E=σ
For the six faces (Gaussian faces) of a cube, only the M and N faces are perpendicular to the electric field line, and the other four faces are parallel to the electric field line, which is determined by the Gauss law.
∑q/ε0=∫EdS=∑ESi
And the free charge in the cube is 0.
∴0=q(m)+q(n)=ε0[escosθ(m)+escosθ(n)]
Q (m) and q(N) are equal in size and opposite in sign.
θ is the included angle between the normal of Gaussian surface and the electric field line.
Because the advancing direction of Gaussian plane usually takes the outer normal direction.
∴θ(M)=π，θ(N)=0
∴q(M)=ε0EScosπ=-σS
q(N)=ε0EScos0=σS
That is, the amount of charge passing through the M plane is -σS, and the amount of charge passing through the N plane is σS..
Extended data:
Coulomb's law and Gauss's theorem can be used to solve the electric field strength. E. Coulomb's law and the principle of field strength superposition can be used to solve the field strength of point charge and point charge system. In principle, the field strength of a continuously distributed charged system can also be solved, but for specific problems, the continuous distribution function of charge must be known. There are some limitations in solving the field strength by Gauss theorem, and generally only the field strength with a certain symmetric distribution can be solved.
Using Gauss theorem to solve the field strength must follow two steps: firstly, the field strength produced by the charged system involved must be qualitatively analyzed, and the distribution law of field strength direction and size must be clear; Secondly, according to the law of field intensity distribution, it is judged whether it can be solved by Gaussian theorem, and if so, an appropriate Gaussian surface is constructed to solve it.
Baidu Encyclopedia-Gauss Law