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 6.15(a)). Every line that enters the surface must also leave that surface. A typical field line enters the surface at d A 1 d A 1 and leaves at d A 2. 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 P that is at a distance r from the charge at the origin is given byįigure 6.14 Flux through spherical surfaces of radii R 1 R 1 and R 2 R 2 enclosing a charge q are equal, independent of the size of the surface, since all E-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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