MatinSAR said:
Thank you ... but what do you mean by relevant equation?BvU said:
could be interpreted as an invitation to show that for this case the gauss identity holds. In that case you could actually integrate the flux: you know the electric field (right?) and can benefit from symmetry.
I guess no it doesn't ... Thank you for your help ...BvU said:
Electrical flux is a measure of the amount of electric field passing through a given area. It is represented by the symbol Φ and is measured in units of volts per meter squared (V/m^2).
The electrical flux passing through a surface is calculated by multiplying the electric field strength by the surface area and the cosine of the angle between the electric field and the surface normal. This can be represented by the equation Φ = E x A x cos(θ).
The unit of measurement for electrical flux is volts per meter squared (V/m^2).
The electrical flux passing through a cube can be used to determine the strength and direction of an electric field. It is also important in understanding how electric charges interact with each other and their surroundings.
The shape of the cube does not affect the total electrical flux passing through it, as long as the surface area and angle between the electric field and surface normal remain constant. However, the distribution of the flux may vary depending on the shape of the cube.