# Electric flux

On the topic of the electric field, has been discussed the definition and equation of the electric field which can be used to calculate the electric field strength produced by an electric charge, some electric charge or by an electric charge distribution. The calculation of the electric field strength produced by an electric charge or two electric charges is easily solved using the formula of electric field strength. If what is calculated is the electric field strength generated by an electric charge distribution, the calculation is more complicated if the formula for electric field strength is used but it is easier to use Gauss’s law. Before studying Gauss law in depth, first understood that electric flux because of the concept of electric flux used in Gauss law.

**Definition of the electric flux**

The word flux is derived from the Latin word, fluere, which means to flow. Electrical flux can be interpreted as an electric field flow. The word flow here does not show an electric field flowing like flowing water but explains the existence of an electric field that leads to a particular direction. On the topic of electric field lines, it has been explained that the electric field is visualized or drawn using electric field lines hence electric fluxes are also described as electric field lines. So electric flux is electric field line that passes a specific surface area, as exemplified in the figure below.

**The equation of the electric flux**

Mathematically, electrical flux is the product of the electric field (E), surface area (A) and the cosine of the angle between the electric field line and the normal line perpendicular to the surface.

F = E A cos θ ……………. (Equation 1)

If the electric field lines are perpendicular to the surface area they pass as in the figure, then the angle between the electric field line and the normal line is 0^{o}, where cos 0^{o} = 1. Thus the formula for electric flux changes to:

F = E A cos 0^{o} = E A (1)

F = E A ……………. (Equation 2)

Based on the formula the electric flux above concluded several things. First, the electric flux is maximum when the electric field line is perpendicular to the surface area because at this condition the angle between the electric field line and the normal line is 0o, where the cosine 0o is 1. Second, the electric flux is minimum when the electric field line is parallel to the surface area because at this condition the angle between the electric field line and the normal line is 90^{o}, where the cosine 90^{o} is 0. Third, the electric flux depends on the electric field (E) and the surface area (A). In addition to the square-shaped surface area as in the example above, the surface area can also be spherical and others.

**Electric flux on the closed surface**

The electric charge described earlier uses an example of an open surface (square or rectangular surface area). How do electric fluxes on closed surfaces such as cubes, beams or balls? Suppose there are electric field lines that pass through the beam as shown below.

The electric field lines which are colored in blue coincide with the upper and lower surfaces of the beam so that they form an angle of 90^{o }with the normal line of the upper and lower surfaces. Thus the electric flux on the upper and lower surfaces of the beam is F = E A cos 90^{o }= E A (0) = 0.

The electric field lines which are given a yellow color coincide with the right and left side surfaces of the beam so that they form an angle of 90^{o }with the normal line of the left and right side surfaces. Thus the electric flux on the right and left side of the beam is F = E A cos 90^{o} = E A (0) = 0.

The electric field lines are given a red perpendicular to the front and back surfaces of the beam so that they form a 0^{o} angle with the normal line of the front and rear surfaces. Thus the electric flux is F = E A cos 0^{o }= E A (1) = E A.

In the figure above, visible red lines of the electric field move into the beam and then move out of the beam. When the electric field lines move into the beam as if there is a negative charge inside the beam, the electric flux is negative. Conversely, when the electric field lines move out of the beam as if there is a positive charge inside the beam, the electric flux is positive. Qualitatively, if the amount of electric field lines that enter the beam is equal to the number of electric field lines coming out of the beam, the resultant electric flux is zero. Quantitatively, the resultant electric flux passing through the beam is calculated in the following way: incoming electrical flux = F_{1} = – EA cos 0^{o} = – EA (1) = -EA and outgoing electric flux = F_{2} = + EA cos 0^{o} = + EA (1) = + E A. The total electric flux is F = – F_{1} + F_{2} = -EA + EA = 0.

Based on the above calculations it was concluded that the total electric flux passing through the beam as in the figure above is zero. It can be said that the total electric flux is zero because there is no electric charge in the beam. So if there is no electric charge in a closed surface such as beams, cubes, spheres, etc. the total electric flux is zero. What if there is an electric charge on a closed surface?

Suppose there is an electric charge on the center of the ball as shown in the figure on the side. The four lines of the electric field are described as representing the lines of other electric fields that move out from the center of the sphere perpendicular to the surface of the sphere. Each line is perpendicular to the surface of the ball through which it forms an angle of 0^{o} with a normal line perpendicular to the surface of the ball.

Electric flux on the ball: Φ = E A.

The formula of the electric field strength is E = k q / r^{2}, and the equation of the surface area of the sphere is A = 4 p r^{2} so that the formula of electric flux changes to:

If the charge at the center of the ball is + 2Q then the electric flux on the ball is

Based on the electric flux formula, it is concluded that if there is an electric charge in the closed spherical surface, the value of the electric flux on the ball does not depend on the diameter or radius of the ball. The magnitude of the electric flux is 4πk times the total electrical charge in the ball or 1/ε_{o} times the total electrical charge in the ball.

**Unit of electric flux **

The basic formula of electric flux is F = E A, where E is the electric field strength and A is the surface area. The electric field unit is Newton per Coulomb (N/C), and the unit of surface area is the square meter (m^{2}) so that the unit of electrical flux is Newton square meter per Coulomb (Nm^{2}/C).