# Does the number of field lines crossing an area depend upon angle between them?

Consider Electric Field Lines crossing a square area (for simplicity) such that all field lines are parallel and make an angle say $$\alpha$$ with the area vector of the square.

Let us vary the angle $$\alpha$$ such that the number of field lines crossing the area remains same.

If such a case, what conclusions can we draw if we compare the flux of Electric Field through the square area as the angle $$\alpha$$ between varies?

Note: I want to generalise these arguments for all cases but I have considered a square area for simplicity. If someone wants to generalise my question, please answer using a infinitesimal small area.

• The flux measures the "dot product" of the area vector and the electric field vector at the same point. If you increase $\alpha$ (which we assume means the electric field is now passing through the surface at more of an angle) the flux will decrease. Once the electric field passes through the surface at $90^o$ to the surface vector the flux is zero, as you would expect. Commented Jun 9, 2021 at 12:38
• I think the number of field lines is infinite and uncountable. The angle between the field lines is not defined. The angle between two specific field lines would be a concept making sense. (IMO) Commented Jun 9, 2021 at 13:01

As Charlie points out, the flux of the field depends on cos(α), (with the area vector defined as being normal to the surface). The number of field lines depends on how many you feel like drawing to represent the field. In a 3D sketch, the line density should be proportional to the field strength. If you change the flux, you change the number of lines. If you increase the angle, some of the field lines which formerly crossed the surface no longer do.

A typical physical problem where this sort of consideration appears is the reaction between charge and field:

Maxwell's equations state that:

$$\nabla \cdot E\ =\ \rho \ /\ \epsilon_{o}$$

Or equivalently:

$$\int \overrightarrow{E} .\overrightarrow{ds} \ =\ Q\ /\ \epsilon_{o}$$

In the simple case where E is constant (in module and direction) and the surface S is at an angle $$\alpha$$ respectively to E, then the scalar product (dot product) introduces a factor $$sin(\alpha)$$ (since $$.\overrightarrow{ds}$$ is normal to the surface).