Gauss’ law states that the net electric flux through a closed Gaussian surface is equal to the total charge q inside the surface divided by ε 0. Gauss’ law for a point charge can be finally expressed as Φ E = q / □ 0………………… (3) Using equations (1) and (2) we now can write the equation of electric flux for a point charge: Φ E = q / □ 0 This means that when the electric field is perpendicular to the surface, it becomes parallel to the area vector. Note that the area vector is normal to the surface. Here, θ = angle between the electric field E and the area vector A. In other words, the electric flux Φ E is defined as the scalar product of A and E. In Gauss’ law, this product is especially important and is called the electric flux, Φ E and we can write Φ E = E. When the electric field is perpendicular to the surface, Gauss’ law for a point charge can be expressed primarily as EA = q / □ 0, where the left side of Equation 1 is the product of the magnitude E of the electric field at any point on the Gaussian surface and the area A of the surface. The surface area A of a sphere is A = 4□ r 2, and the magnitude of the electric field can be written in terms of this area as E = q/(A□ 0), or EA = q / □ 0 (when the electric field is perpendicular to the surface) ……… (1) figure 2: one example of a Gaussian surface. In figure 2, the electric field is perpendicular to the surface and has the same magnitude everywhere on it. Such a hypothetical closed surface is called a Gaussian surface, although in general, it need not be spherical. We now place this positive point charge at the center of an imaginary spherical surface of radius r, as Figure 2 shows. With this substitution, the magnitude of the electric field becomes E = q/(4□□ 0 r 2 ). The constant k can be expressed as k = 1/(4□□ 0), where □ 0 is the permittivity of free space. The magnitude E of the electric field at a distance r from the charge +q is E = kq/r 2. figure 1: The electric field lines for a positive point charge The electric field lines for a positive point charge radiate outward in all directions from the charge, as shown in figure 1. We begin by developing a version of Gauss’ law that applies only to a point charge, which we assume to be positive. A surface that includes the same amount of charge has the same number of field lines crossing it, regardless of the shape or size of the surface, as long as the surface encloses the same amount of charge (part (c)).Derivation of Gauss’ law that applies only to a point charge The same thing happens if charges of equal and opposite sign are included inside the closed surface, so that the total charge included is zero (part (b)). Hence the net “flow” of the field lines into or out of the surface is zero ( Figure 2.2.3(a)). Every line that enters the surface must also leave that surface. A typical field line enters the surface at and leaves at. You can see that if no charges are included within a closed surface, then the electric flux through it must be zero. This net number of electric field lines, which is obtained by subtracting the number of lines in the direction from outside to inside from the number of lines in the direction from inside to outside gives a visual measure of the electric flux through the surfaces. Therefore, the net number of electric field lines passing through the two surfaces from the inside to outside direction is equal. Recall that when we place the point charge at the origin of a coordinate system, the electric field at a point that is at a distance from the charge at the origin is given byįigure 2.2.2 Flux through spherical surfaces of radii and enclosing a charge are equal, independent of the size of the surface, since all -field lines that pierce one surface from the inside to outside direction also pierce the other surface in the same direction. To get a feel for what to expect, let’s calculate the electric flux through a spherical surface around a positive point charge q, since we already know the electric field in such a situation. Now, what happens to the electric flux if there are some charges inside the enclosed volume? Gauss’s law gives a quantitative answer to this question. Therefore, if a closed surface does not have any charges inside the enclosed volume, then the electric flux through the surface is zero. We found that if a closed surface does not have any charge inside where an electric field line can terminate, then any electric field line entering the surface at one point must necessarily exit at some other point of the surface. We can now determine the electric flux through an arbitrary closed surface due to an arbitrary charge distribution. Apply Gauss’s law in appropriate systems.Explain the conditions under which Gauss’s law may be used. By the end of this section, you will be able to:
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