Chapter 23: Click Here
Chapter 23 & 24
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CHAPTER # 24
Gauss's Law
ELECTRIC FLUX
What is an Electric flux?
Electric flux is proportional to the number of electric field lines
penetrating some surface.
Explanation:
Consider an electric field that is uniform in both magnitude and
direction, as shown in Figure 24.1.
The field lines penetrate a rectangular surface of area A, whose plane is
oriented perpendicular to the field.
Recall from Section 23.6 (Book page no.723, PDF page no. 732) that
the number of lines per unit area (in other words, the line density) is
proportional to the magnitude of the electric field. Therefore, the total
number of lines penetrating the surface is proportional to the product EA.
This product of the magnitude of the electric field E and surface area A
perpendicular to the field is called the electric flux! (uppercase
Greek phi):
Unit: From the SI units of E and A, we see that Electric
flux has units of newton meters squared per coulomb (N"m2/C.)
Example 24.1 Electric Flux Through a Sphere
Maximum and Minimum flux
If the surface under consideration is not perpendicular to the field, the flux
through it must be less than that given by Equation 24.1. We can understand this by
considering Figure 24.2, where the normal to the surface of area A is at an angle (theta) to
the uniform electric field. Note that the number of lines that cross this area A is equal
to the number that cross area A*, which is a projection of area A onto a plane oriented perpendicular to the field. From Figure 24.2 we see that the two areas are
related by A* = A cos (theta). Because the flux through A equals the flux through A*, we
conclude that the flux through A is
(24.2) :
From this result, we see that the flux through a surface of fixed area A has 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, Theta = 0° in Figure 24.2); 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, Theta = 90°).
The electric flux through a Vector area/General Area
We assumed a uniform electric field in the preceding discussion. In more general
situations, the electric field may vary over a surface. Therefore, our definition of flux
given by Equation 24.2 has meaning only over a small element of the area. Consider a general surface divided up into a large number of small elements, each of area delta A.
The
variation in the electric field over one element can be neglected if the element is sufficiently small. It is convenient to define a vector delta Ai whose magnitude represents the
area of the ith element of the surface and whose direction is defined to be perpendicular to the surface element, as shown in Figure 24.3.
The electric field Ei at the location
of this element makes an angle )i with the vector delta Ai . The electric flux delta FluxE through
this element is
where we have used the definition of the scalar product (or dot product; see Chapter
7) of two vectors (A.B = AB cos )). By summing the contributions of all elements, we
obtain the total flux through the surface.
If we let the area of each element approach
zero, then the number of elements approaches infinity and the sum is replaced by an
integral. Therefore, the general definition of electric flux is
Equation 24.3 is a surface integral, which means it must be evaluated over the surface
in question. In general, the value of FluxE depends both on the field pattern and on the
surface.
Electric Flux through a closed surface
Consider the closed surface in Figure 24.4.
The vectors delta Ai point in different
directions for the various surface elements, but at each point they are normal to the
surface and, by convention, always point outward. At the element labeled !, the field
lines are crossing the surface from the inside to the outside and Theta < 90°; hence, the
flux Delta FluxE = E. delta A1 through this element is positive. For element ", the field lines
graze the surface (perpendicular to the vector +A2); thus, Theta = 90° and the flux is zero.
For elements such as #, where the field lines are crossing the surface from outside to
inside, 180° > Theta > 90° and the flux is negative because cos ) is negative. The net flux
through the surface is proportional to the net number of lines leaving the surface,
where the net number means the number leaving the surface minus the number entering the
surface. If more lines are leaving than entering, the net flux is positive. If more lines are
entering than leaving, the net flux is negative. Using the integral symbol to represent an
integral over a closed surface, we can write the net flux FluxE through a closed surface as
where En represents the component of the electric field normal to the surface. If the
field is normal to the surface at each point and constant in magnitude, the calculation
is straightforward, as it was in Example 24.1. Example 24.2 also illustrates this point.
Flux through a cube
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 is known as Gauss’s law.
Explanation
Let us again consider a positive point charge q located at the center of a sphere of
radius r, as shown in Figure 24.6.
From Equation 23.9 we know that the magnitude
of the electric field everywhere on the surface of the sphere is E = (k*e*q)/r^2. As noted in
Example 24.1, 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 delta Ai representing a local element of area delta Ai surrounding the
surface point. Therefore,
Important Points in Gauss's Law
the net flux through any closed surface surrounding a point charge
q is given by q/epsile not and is independent of the shape of that surface.
the net electric flux through a closed surface that surrounds no
charge is zero.
the electric field due to many charges is
the vector sum of the electric fields produced by the individual charges.
Complete Explanation
and from Equation 24.4 we find that the net flux through the gaussian surface is
where we have moved E outside of the integral because, by symmetry, E is constant over
the surface and given by E = (k*e*q)/r^2. Furthermore, because the surface is spherical, integral dA = A = 4*Pi*r^2. Hence, the net flux through the gaussian surface is
Recalling from Section 23.3 that Ke = 1/4Pi*epsile not, we can write this equation in the form
We can verify that this expression for the net flux gives the same result as Example
24.1:
E = (1.00 * 10^-6 C)/(8.85 * 10^-12 C^2/N"m^2) = 1.13 * 105 N"m2/C.
Note from Equation 24.5 that the net flux through the spherical surface is proportional to the charge inside. The flux is independent of the radius r because the area of
the spherical surface is proportional to r^2, whereas the electric field is proportional to
1/r^ 2.
Thus, in the product of area and electric field, the dependence on r cancels.
Now consider several closed surfaces surrounding a charge q, as shown in Figure
24.7.
Surface S1 is spherical, but surfaces S2 and S3 are not. From Equation 24.5, the
flux that passes through S1 has the value q/epsile not. As we discussed in the preceding section,
flux is proportional to the number of electric field lines passing through a surface. The
construction shown in Figure 24.7 shows that the number of lines through S1 is equal to
the number of lines through the nonspherical surfaces S2 and S3. Therefore, we
conclude that the net flux through any closed surface surrounding a point charge
q is given by q/epsile not and is independent of the shape of that surface.
Now consider a point charge located outside a closed surface of arbitrary shape, as
shown in Figure 24.8. As you can see from this construction, any electric field line that
enters the surface leaves the surface at another point. The number of electric field
lines entering the surface equals the number leaving the surface.
Therefore, we
conclude that the net electric flux through a closed surface that surrounds no
charge is zero. If we apply this result to Example 24.2, we can easily see that the net
flux through the cube is zero because there is no charge inside the cube.
Let us extend these arguments to two generalized cases: (1) that of many point
charges and (2) that of a continuous distribution of charge. We once again use 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
where E is the total electric field at any point on the surface produced by the vector
addition of the electric fields at that point due to the individual charges. Consider the
system of charges shown in Figure 24.9. The surface S surrounds only one charge, q1;
hence, the net flux through S is q1/ epsile not.
The flux through S due to charges q2, q3, and
q 4 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
(q 2 + q 3)/ epsile not. Finally, the net flux through surface S^n is zero because there is no charge
inside this surface. That is all the electric field lines that enter S^n at one point and 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 q in represents the net charge inside the surface and E represents the electric
field at any point on the surface.
A formal proof of Gauss’s law is presented in Section 24.5. When using Equation
24.6, you should note that although the charge q in is the net charge inside the gaussian
surface, E represents the total electric field, which includes contributions from charges
both inside and outside the surface.
Application of Gauss’s Law
Applications of gauss's law are well explained in the book starting from page 746 and PDF's page no. 755
Thank you!
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