Field lines help us to visualize the electric field. The density of the field lines is proportional to the strength of the field. If field lines are drawn correctly according to the rules, then counting the number of lines that pass through a unit area perpendicular to the direction of the field yields a number proportional to the strength of the electric field.
The number of field lines passing through a geometrical surface of given area depends on three factors.
 | the strength of the field |
| the surface area |
| the orientation of the surface |
The diagram above shows a locally uniform electric field E. The lines are parallel and have constant density. The same surface is inserted in three different orientations. The maximum number of field lines is intercepted when the normal to the surface n is parallel to the field E, while no field lines pass through the surface when the normal is perpendicular to the field. In general, the number of field lines passing through an area A is directly proportional to Acosq, where q is the angle between the electric field and the normal to the surface. This leads to the definition of the electric flux.
DF
e = E DA cosq.DF
e is the electric flux through some small area DA, whose normal make an angle q with the direction of the electric field. E is the magnitude of the field. The SI unit of flux is Nm2/C.E is a vector quantity. It is useful also to represent the area A by a vector A. The length of this vector is the size of the area, while its orientation is perpendicular to the area. It is in the direction of the normal n. We have A = An. The normal to the surface can point into two different directions. For a closed surface, by convention, the normal points outward.
With our definition of A we can write the flux as the dot product of E and DA.
DF
e = E×DA.DF
e is the flux through a small are DA, which may be part of a larger area A. The total electric flux Fe through A can be evaluated by integrating the differential flux over the surface A,F
e = òE×dA = òEndA,where the integral is taken over the entire surface. En is the component of the electric field normal to the surface.
For the flux through a closed surface we write
.
The flux through a given surface can be positive of negative, since the cosine can be positive or negative. Consider the electrical flux passing through a cubical surface with two of its faces perpendicular to a uniform electrical field.
The flux passing through the sides of the cube is zero since these sides are parallel to the field lines and thus do not intercept any of them. As drawn, the field lines are parallel to the normal vector n for the right side, so the flux through this side is Fe = EA. The field lines are anti-parallel to the normal vector n for the left side, so the flux through this side is Fe = -EA. The total flux through the surface of the cube is the sum of the fluxes through all sides, and it is zero.
Let us calculate the flux passing through a spherical surface that surrounds a point charge q at the center of the sphere.
First let us consider the field lines. The field lines radiate outward from the charge. The density of field lines reflects the magnitude of the electrical field at any point. The magnitude of the field and the density of the field lines decrease proportional to the inverse square of the distance r from the charge. The area of the spherical surface increases proportional r2. The total number of field lines passing through the spherical surface, which is perpendicular to the direction of the field at any point, is the product of their density and the area of the surface. It is therefore independent of the size of the sphere. The number of field lines depends only on the magnitude of the charge at the center of the sphere.
Because the flux is proportional to the number of field lines passing through the area, we expect the flux to be proportional to the charge inside the surface. The field of the point charge at the center is radial, pointing towards or away from the point charge. It is everywhere normal to the surface area of the sphere. Its magnitude is E = q/(4pe0r2). En is therefore constant over the entire surface area and can be take out of the integral. We then have
.
The flux is be proportional to the charge inside the surface and the proportional constant is 1/e0.
Link:
| Tutorials on Vector Fields |
Newton's laws (1st, 2nd, and 3rd law) describe the dynamics of physical systems. They describe how objects behave when forces act on them. They do not address the origin of the forces. Newton's law of gravitation states that any objects with mass exert forces on each other and it lets us calculate the direction and strength of the gravitational forces.
Maxwell's equations
let us calculate the forces charged particles exert on each other. While Newton's laws only hold for particles moving with speeds much less than the speed of light, Maxwell's equations hold for particles moving with any speed. We say that Maxwell's equations are relativistically correct.Maxwell's equations are a set of four equations. The first of these equations is Gauss's Law. It states that the electric flux through any closed surface is equal to the total charge inside divided by e0.
Using Coulomb's law, we have already deduced that this is the case for a point charge at the center of a spherical surface. But Gauss' law holds for a closed surface of any shape and for any charge distribution inside that surface.
Maxwell's equations are more general than Coulomb's law. Coulomb's law only holds for charges at rest, (or, approximately, for charges whose accelerations are small.) Coulomb's law can be derived from the first two of Maxwell's equations for static situations, i.e. when all charges are at rest or moving with uniform velocity.
The first two of Maxwell's equations let us calculate the electric field due to any static charge distribution. Once we know the field then we know the force that this charge distribution exerts on another charge.
If a charge distribution has a high degree of symmetry, then Gauss' law alone (i.e. the first of Maxwell's equations alone,) can be used to determine the magnitude of the electric field. The direction must be deduced from the symmetry of the situation.
| Electric field for a uniform sphere of charge Imagine a sphere of radius R with charge Q uniformly distributed inside. The symmetry of the charge distribution requires a spherically symmetric electric field. The field must either point radially inward toward the center or outward from the center of the sphere. (If we have a spherically symmetric charge distribution, then, no matter how we orient our coordinate system, the distribution always looks the same. The field must therefore also look the same, no matter how we orient our coordinate system. A field, that is not radial, will look different, if we rotate our coordinate system, i.e. if we look at it from another angle.) For a spherical symmetric charge distribution, the magnitude of E can therefore only depend on the radial coordinate r and on the charge Q. To determine E as a function of r, we use Gauss' law. We draw a spherical Gaussian surface of radius r centered at the center of the spherical charge distribution. The radius r of the surface can be larger or smaller than the radius R of the distribution.
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| Electric field near an infinite plane of charge For an infinite plane of charge lying in the x-y plane, with its normal parallel to the z-axis, the field cannot depend on x or y. (If we shift the origin of our coordinate system along the x- or y-axis, or rotate our coordinate system about the z-axis, the charge distribution looks the same. The field must therefore also look the same.) The field must therefore be parallel to the z-axis. The field must also be symmetric under reflection about the x-y plane. This means that the field must have the same magnitude at +z as it has at -z, but that it must point in opposite directions at +z and at -z.
For our Gaussian surface we can choose a simple right circular cylinder with faces parallel to the plane of charge. The field lines are parallel to the sides of the cylinder, so the sides do not contribute to Fe = òE×dA. The flux through the top surface is
and the flux through the bottom surface is
(On the top, both E and n point upward,
and on the bottom both E and n point downward.) Fnet = 2EA = Qinside/e0. If the surface charge density is s, the Qinside = sA, and Fnet = 2EA = sA /e0, and The field of an infinite planar charge distribution is uniform. It does not decrease with distance. Of course, there are no infinite sheets of charge. But our result is a good approximation to the field near a finite-size plane of charge, as long as the dimensions of the plane are much larger than the distance away from the plane where we want to know the field.
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The most important application of the above result is the superposition of the fields from two planar charge distributions which are separated by some distance d. Place a uniformly charged plane with charge density s at x = +d/2 and a similar plane with charge density -s at x = -d/2.
The field due to the upper plane of charge is
E1 = s/2e0, x > d/2, E1 = -s/2e0, x < d/2.
The field due to the lower plane of charge is
E2 = s/2e0, x < -d/2, E2 = -s/2e0, x > -d/2.
The total field in the region x < -d/2 is E = E1+E2 = -s/2e0+s/2e0 = 0. Similarly, the total field in the region x > d/2 is zero.
In the region between -d/2 and +d/2 the total field is is
E = E1+E2 = -s/2e0-s/2e0 = -s/e0.
The fields add to yield a uniform field between the planes, but they precisely cancel outside the planes to give zero net field outside. Between the planes the field points from the positive towards the negative plane. This is the common configuration of a parallel plate capacitor.
Problems:
| A solid sphere of radius R = 40cm has a total positive charge of 26 mC
uniformly distributed throughout its volume. Calculate the magnitude
of the electric field at (a) 0 cm, (b) 30 cm, and (c) 60 cm from the center of the sphere.
| ||
| A sphere of radius R = 40cm has a total positive charge of 26mC
uniformly distributed over its surface. Calculate the magnitude of the
electric field at (a) 0 cm, (b) 30 cm, and (c) 60 cm from the center of the sphere.
|
Charged objects exert forces on each other. But even objects that have no net charge can be acted on by electrical forces. Consider the case of a neutral insulator placed close to a positively charged rod, as shown in the figure below.
The electrons in the insulating material are bound to their atomic nuclei. They are not free to move throughout the material. But the electric field produced by the positive charges on the rod attracts the electrons and repels the nuclei. The electrons are therefore no longer symmetrically distributed around their nuclei, but are pulled over to the side closer to the rod. The net effect is that the entire negative charge is displaced by a small amount with respect to the positive charge in the material. This results in an effective surface charge density, which is negative on the side close to the rod and positive on the side farther away from the rod. This is the phenomenon of polarization. The field of the rod decreases with distance, so the attractive force on the negative surface charge is stronger than the repulsive force on the positive surface charge. The material is attracted towards the rod due to its polarization.
If, instead of an insulator, we have an insulated conductor, then an excess of free electrons gathers on the side close to the rod, and the side farthest away from the rod is depleted of free electron and therefore has a net positive surface charge. Again the material is attracted towards the rod.
Link to other Web material:
| Electric field |
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