1110 Phys Chapter3
Electric Flux and Gauss’s Law
Electric Flux
Figure (1)
Consider an electric field that is uniform in both magnitude and direction, as shown in Figure 1. The total number of lines penetrating the surface is proportional to the product EA. This product of the magnitude of the electric field Eand surface area Aperpendicular to the field is called the electric flux E.
E = E A
From the SI units of Eand A, we see that Ehas units of newton-meters squared per coulomb (N. m2/C.) Electric flux is proportional to the number of electric field lines penetrating some surface.
Example 1(Electric Flux Through a Sphere)
What is the electric flux through a sphere that has a radius of 1.0 m and carries a charge of +1.0 µC at itscenter?
Solution
The magnitude of the electric field is giving by
The field points radially outward and is therefore everywhere perpendicular to the surface of the sphere. The flux through the sphere (whose surface area A = 4 r2= 12.6 m2) is thus
Figure (2)
If the surface under consideration is not perpendicular to the field, as show in Figure 2, the electric flux E given by
E = E A Cos
From this equation, we see that the flux through a surface of fixed area Ahas a maximum value EA when the surface is perpendicular to the field (when the normal to the surface is parallel to the field, that is, = 0° in Figure 2); the flux is zero when the surface is parallel to the field (when the normal to the surface is perpendicular to the field, that is, = 90°).
Figure (3)
Gauss’s Law
In this section we describe a general relationship between the net electric flux through a closed surface (often called a gaussian surface) and the charge enclosed by the surface. This relationship, known as Gauss’s law, is of fundamental importance in the study of electric fields.
Let us again consider a positive point charge qlocated at the center of a sphere of radius r, as shown in Figure 3. we know that the magnitude of the electric field everywhere on the surface of the sphere is
The field lines are directed radially outward and hence are perpendicular to the surface at every point on the surface. That is, at each surface point, E is parallel to the vector Airepresenting a local element of area Aisurrounding the surface point. Therefore,
the net flux through the Gaussian surface is
where we have moved Eoutside of the integral because Eis constant over the surface and given byFurthermore, because the surface is spherical,
Hence, the net flux through the Gaussian surface is
According to the superposition principle, which states that the electric field due to many charges is the vector sum of the electric fields produced by the individual charges. Therefore, we can express the flux through any closed surface as
Figure (4)
Figure 4 shows that the number of lines through S1 is equal to the number of lines through the non-spherical surfaces S2 and S3. Therefore, we conclude that the net flux through any closed surface surrounding a point charge q is given by and is independent of the shape of that surface.
Figure (5)
Consider the system of charges shown in Figure 5. The surface Ssurrounds only one charge, q1;hence, the net flux through S is . The flux through Sdue to charges q2, q3, and q4 outside it is zero because each electric field line that enters S at one point leaves it at another. The surface S\ surrounds charges q2 and q3; hence, the net flux through it is . Finally, the net flux through surface S\\ is zero because there is no charge inside this surface. That is, all the electric field lines that enter S\\ at one point leave at another. Notice that charge q4 does not contribute to the net flux through any of the surfaces because it is outside all of the surfaces.
Gauss’s law, which is a generalization of what we have just described, states that the net flux through any closed surface is
where qin represents the net charge inside the surface and E represents the electric field at any point on the surface.
Figure (6)
Example 2 (The Electric Field Due to a Point Charge)
Starting with Gauss’s law, calculate the electric field due to an isolated point charge q.
Solution
We choose a spherical gaussian surface of radius r centered on the point charge, as shown in Figure 6. The electric field due to a positive point charge is directed radially outward by symmetry and is therefore normal to the surface at every point. Thus, E is parallel to dA at each point. Therefore,
Gauss’s law gives
where we have used the fact that the surface area of a sphere is 4r2. Now, we solve for the electric field:
References
This lecture is a part of chapter 23 from the following book
Physics for Scientists and Engineers (with PhysicsNOW and InfoTrac),
Raymond A. Serway - Emeritus, James Madison University , Thomson Brooks/Cole © 2004, 6th Edition, 1296 pages
Problems
(1)An electric field with a magnitude of 3.50 kN/C is appliedalong the x axis. Calculate the electric flux through a rectangularplane 0.350 m wide and 0.700 m long assumingthat (a) the plane is parallel to the yz plane; (b) the planeis parallel to the xy plane; (c) the plane contains the y axis,and its normal makes an angle of 40.0° with the x axis.
(2)A 40.0-cm-diameter loop is rotated in a uniform electric fielduntil the position of maximum electric flux is found. Theflux in this position is measured to be 5.20 x 105 N.m2/C.What is the magnitude of the electric field?
(3)The following charges are located inside a submarine:5.00 µC, -9.00 µC, 27.0 µC, and -84.0 µC. (a) Calculatethe net electric flux through the hull of the submarine.(b) Is the number of electric field lines leaving thesubmarine greater than, equal to, or less than the numberentering it?
(4)Four closed surfaces, S1 through S4, together with thecharges -2Q , +Q , and -Q are sketched in Figure 8.(The colored lines are the intersections of the surfaceswith the page.) Find the electric flux through eachsurface.
Figure (8)
(5)A charge of 170 µC is at the center of a cube of edge80.0 cm. (a) Find the total flux through each face of thecube. (b) Find the flux through the whole surface of thecube.
Problems solutions
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