It can be said that the total electric flux is zero because there is no electric charge in the beam. The total electric flux is F = – F 1 + F 2 = -EA + EA = 0.īased on the above calculations, it was concluded that the total electric flux passing through the beam as in the figure above 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. 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. 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. When the electric field lines move into the beam as if there is a negative charge inside the beam, the electric flux is negative. In the figure above, visible red lines of the electric field move into the beam and then move out of the beam. Thus, the electric flux is F = E A cos 0 o = E A (1) = E A. The electric field lines are given a red perpendicular to the front and back surfaces of the beam so that they create a 0 o angle with the normal line of the front and rear 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 which are given a yellow color coincide with the right and left side surfaces of the beam so that they create an angle of 90 o with the normal line of the left and right side 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 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. 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 charge described earlier uses an example of an open surface (square or rectangular surface area). In addition to the square-shaped surface area as in the example above, the surface area can also be spherical and others. Third, the electric flux depends on the electric field (E) and the surface area (A). 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. 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. Thus, the formula for electric flux changes to:īased on the formula, the electric flux above concluded several things. 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. 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. So electric flux is an electric field line that passes a specific surface area, as exemplified in the figure below. Regarding electric field lines, it has been explained that the electric field is visualized or drawn using electric field lines, and hence electric fluxes are also described as electric field lines. 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. Electrical flux can be interpreted as an electric field flow. The word flux is derived from the Latin word, fluere, which means to flow. Before studying Gauss law in depth, first understood that electric flux because of the concept of electric flux used in Gauss law. 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. 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. Regarding 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.
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