So far, we have generally been working with charges occupying a volume within an insulator. We now study what happens when free charges are placed on a conductor. Generally, in the presence of a generally external electric field, the free charge in a conductor redistributes and very quickly reaches electrostatic equilibrium.
If an electric field is present inside a conductor, it exerts forces on the free electrons also called conduction electrons , which are electrons in the material that are not bound to an atom. These free electrons then accelerate. However, moving charges by definition means nonstatic conditions, contrary to our assumption. Therefore, when electrostatic equilibrium is reached, the charge is distributed in such a way that the electric field inside the conductor vanishes.
If you place a piece of a metal near a positive charge, the free electrons in the metal are attracted to the external positive charge and migrate freely toward that region.
The region the electrons move to then has an excess of electrons over the protons in the atoms and the region from where the electrons have migrated has more protons than electrons. Consequently, the metal develops a negative region near the charge and a positive region at the far end Figure. As we saw in the preceding chapter, this separation of equal magnitude and opposite type of electric charge is called polarization. If you remove the external charge, the electrons migrate back and neutralize the positive region.
The polarization of the metal happens only in the presence of external charges. You can think of this in terms of electric fields.
The external charge creates an external electric field. When the metal is placed in the region of this electric field, the electrons and protons of the metal experience electric forces due to this external electric field, but only the conduction electrons are free to move in the metal over macroscopic distances.
The movement of the conduction electrons leads to the polarization, which creates an induced electric field in addition to the external electric field Figure. The net electric field is a vector sum of the fields of and the surface charge densities and This means that the net field inside the conductor is different from the field outside the conductor. The redistribution of charges is such that the sum of the three contributions at any point P inside the conductor is.
That is, and hence. An interesting property of a conductor in static equilibrium is that extra charges on the conductor end up on the outer surface of the conductor, regardless of where they originate. Figure illustrates a system in which we bring an external positive charge inside the cavity of a metal and then touch it to the inside surface. Initially, the inside surface of the cavity is negatively charged and the outside surface of the conductor is positively charged.
When we touch the inside surface of the cavity, the induced charge is neutralized, leaving the outside surface and the whole metal charged with a net positive charge. To see why this happens, note that the Gaussian surface in Figure the dashed line follows the contour of the actual surface of the conductor and is located an infinitesimal distance within it.
Since everywhere inside a conductor,. But the Gaussian surface lies just below the actual surface of the conductor; consequently, there is no net charge inside the conductor.
Any excess charge must lie on its surface. A sketch of their apparatus is shown in Figure. Two spherical shells are connected to one another through an electrometer E, a device that can detect a very slight amount of charge flowing from one shell to the other.
When switch S is thrown to the left, charge is placed on the outer shell by the battery B. Will charge flow through the electrometer to the inner shell? The Electric Field at the Surface of a Conductor If the electric field had a component parallel to the surface of a conductor, free charges on the surface would move, a situation contrary to the assumption of electrostatic equilibrium.
Therefore, the electric field is always perpendicular to the surface of a conductor. At any point just above the surface of a conductor, the surface charge density and the magnitude of the electric field E are related by.
To see this, consider an infinitesimally small Gaussian cylinder that surrounds a point on the surface of the conductor, as in Figure. The cylinder has one end face inside and one end face outside the surface. The height and cross-sectional area of the cylinder are and , respectively. Because the cylinder is infinitesimally small, the charge density is essentially constant over the surface enclosed, so the total charge inside the Gaussian cylinder is.
Now E is perpendicular to the surface of the conductor outside the conductor and vanishes within it, because otherwise, the charges would accelerate, and we would not be in equilibrium. Electric flux therefore crosses only the outer end face of the Gaussian surface and may be written as , since the cylinder is assumed to be small enough that E is approximately constant over that area.
Electric Field of a Conducting Plate The infinite conducting plate in Figure has a uniform surface charge density. Compare this result with that previously calculated directly. Strategy For this case, we use a cylindrical Gaussian surface, a side view of which is shown.
Solution The flux calculation is similar to that for an infinite sheet of charge from the previous chapter with one major exception: The left face of the Gaussian surface is inside the conductor where so the total flux through the Gaussian surface is EA rather than 2 EA.
Significance This result is in agreement with the result from the previous section, and consistent with the rule stated above. Electric Field between Oppositely Charged Parallel Plates Two large conducting plates carry equal and opposite charges, with a surface charge density of magnitude as shown in Figure.
The separation between the plates is. What is the electric field between the plates? Strategy Note that the electric field at the surface of one plate only depends on the charge on that plate. Thus, apply with the given values. Solution The electric field is directed from the positive to the negative plate, as shown in the figure, and its magnitude is given by. Significance This formula is applicable to more than just a plate. Furthermore, two-plate systems will be important later.
What is the electric field both inside and outside the sphere? Strategy The sphere is isolated, so its surface change distribution and the electric field of that distribution are spherically symmetrical. We can therefore represent the field as. Solution Since r is constant and on the sphere,. Significance Notice that in the region , the electric field due to a charge q placed on an isolated conducting sphere of radius R is identical to the electric field of a point charge q located at the center of the sphere.
The difference between the charged metal and a point charge occurs only at the space points inside the conductor. For a point charge placed at the center of the sphere, the electric field is not zero at points of space occupied by the sphere, but a conductor with the same amount of charge has a zero electric field at those points Figure.
However, there is no distinction at the outside points in space where , and we can replace the isolated charged spherical conductor by a point charge at its center with impunity. Check Your Understanding How will the system above change if there are charged objects external to the sphere? If there are other charged objects around, then the charges on the surface of the sphere will not necessarily be spherically symmetrical; there will be more in certain direction than in other directions.
For a conductor with a cavity, if we put a charge inside the cavity, then the charge separation takes place in the conductor, with amount of charge on the inside surface and a amount of charge at the outside surface Figure a. For the same conductor with a charge outside it, there is no excess charge on the inside surface; both the positive and negative induced charges reside on the outside surface Figure b.
If a conductor has two cavities, one of them having a charge inside it and the other a charge the polarization of the conductor results in on the inside surface of the cavity a , on the inside surface of the cavity b , and on the outside surface Figure. The charges on the surfaces may not be uniformly spread out; their spread depends upon the geometry.
The only rule obeyed is that when the equilibrium has been reached, the charge distribution in a conductor is such that the electric field by the charge distribution in the conductor cancels the electric field of the external charges at all space points inside the body of the conductor. Summary The electric field inside a conductor vanishes.
Any excess charge placed on a conductor resides entirely on the surface of the conductor. That is, Figure is actually. Electric Field of a Line Segment Find the electric field a distance z above the midpoint of a straight line segment of length L that carries a uniform line charge density. Strategy Since this is a continuous charge distribution, we conceptually break the wire segment into differential pieces of length dl , each of which carries a differential amount of charge.
Then, we calculate the differential field created by two symmetrically placed pieces of the wire, using the symmetry of the setup to simplify the calculation Figure. Finally, we integrate this differential field expression over the length of the wire half of it, actually, as we explain below to obtain the complete electric field expression. Since it is a finite line segment, from far away, it should look like a point charge. We will check the expression we get to see if it meets this expectation.
The symmetry of the situation our choice of the two identical differential pieces of charge implies the horizontal x -components of the field cancel, so that the net field points in the z -direction. The total field is the vector sum of the fields from each of the two charge elements call them and , for now :.
Because the two charge elements are identical and are the same distance away from the point P where we want to calculate the field, so those components cancel. This leaves. The limits of integration are 0 to , not to , because we have constructed the net field from two differential pieces of charge dq. If we integrated along the entire length, we would pick up an erroneous factor of 2. In principle, this is complete. However, to actually calculate this integral, we need to eliminate all the variables that are not given.
In this case, both r and change as we integrate outward to the end of the line charge, so those are the variables to get rid of. We can do that the same way we did for the two point charges: by noticing that.
Significance Notice, once again, the use of symmetry to simplify the problem. This is a very common strategy for calculating electric fields. The fields of nonsymmetrical charge distributions have to be handled with multiple integrals and may need to be calculated numerically by a computer. Check Your Understanding How would the strategy used above change to calculate the electric field at a point a distance z above one end of the finite line segment? We will no longer be able to take advantage of symmetry.
Instead, we will need to calculate each of the two components of the electric field with their own integral. Electric Field of an Infinite Line of Charge Find the electric field a distance z above the midpoint of an infinite line of charge that carries a uniform line charge density.
Strategy This is exactly like the preceding example, except the limits of integration will be to. Solution Again, the horizontal components cancel out, so we wind up with. Significance Our strategy for working with continuous charge distributions also gives useful results for charges with infinite dimension.
In the case of a finite line of charge, note that for , dominates the L in the denominator, so that Figure simplifies to. If you recall that , the total charge on the wire, we have retrieved the expression for the field of a point charge, as expected. In the limit , on the other hand, we get the field of an infinite straight wire , which is a straight wire whose length is much, much greater than either of its other dimensions, and also much, much greater than the distance at which the field is to be calculated:.
An interesting artifact of this infinite limit is that we have lost the usual dependence that we are used to. This will become even more intriguing in the case of an infinite plane. Electric Field due to a Ring of Charge A ring has a uniform charge density , with units of coulomb per unit meter of arc. Find the electric potential at a point on the axis passing through the center of the ring.
Strategy We use the same procedure as for the charged wire. The difference here is that the charge is distributed on a circle. We divide the circle into infinitesimal elements shaped as arcs on the circle and use polar coordinates shown in Figure. Solution The electric field for a line charge is given by the general expression. A general element of the arc between and is of length and therefore contains a charge equal to The element is at a distance of from P , the angle is , and therefore the electric field is.
Significance As usual, symmetry simplified this problem, in this particular case resulting in a trivial integral. Also, when we take the limit of , we find that. The Field of a Disk Find the electric field of a circular thin disk of radius R and uniform charge density at a distance z above the center of the disk Figure. Strategy The electric field for a surface charge is given by.
Again, by symmetry, the horizontal components cancel and the field is entirely in the vertical direction. The vertical component of the electric field is extracted by multiplying by , so. As before, we need to rewrite the unknown factors in the integrand in terms of the given quantities.
In this case,. Also, we already performed the polar angle integral in writing down dA. Significance Again, it can be shown via a Taylor expansion that when , this reduces to. Active 3 years, 8 months ago. Viewed 1k times. Improve this question. William Ibarra William Ibarra 29 4 4 bronze badges.
In fact, it will distribute itself evenly over any surface with uniform curvature. Add a comment. Active Oldest Votes. Improve this answer. SmarthBansal SmarthBansal 11 11 silver badges 31 31 bronze badges. The OP agrees that the charge should be on the surface; the question is why this distribution is uniform on the surface of a sphere.
I think it could be reasoned the way I did. Think, e. I'll edit it to make it more clear. Unless I've missed something in your argument, all you've shown is that the charge is on the surface, not that it must be uniformly distributed.
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