Problem Set #4

Problem 30-11

This is a basic conservation of energy problem.

so

Similarly

where the last equality follows because energy is conserved. Therefore,

Problem 30-13

Again, a basic conservation of energy problem. The particle is moving in a stable circular orbit, so no work is being done on it, but to shift the particle from one radius, to another, we must do work. The work will be equivalent to the change in energy of the particle from one orbit to the other.

and

Because the orbits are stable we must have that the attractive force between the charges is equal to the centripital force for circular motion, i.e.,

Therefore,

and

Then

Problem 30-23

(a) For point charges

so

(b) The answer will be the same. We are adding scalars, not vectors, so it does not matter what direction the distance to the charge is, only how far. Put another way, the equipotential lines about point charges are circles of constant radius.

Problem 30-29

Assuming that the earth is a sphere (which is an approximation, since the earth bulges at the equator) the uniform electric field at the surface of the earth comes from a spherical charge distribution. As we have seen multiple times, a spherical charge distribution is equivalent to a point charge located at the center of the sphere, so

Now for a spherical charge distribution

and if we take the convention that then

Problem 30-32

(a) The system consists of two point charges, so we can write the total potential as

where distance to first charge and distance to the second charge. Remember that the electric potential is defined only up to an overall constant, hence the form of the equation above. Assuming that we set as is conventional, then we see that is the only point for which since we have a sum of two positive terms.

(b) is, of course, a different sort of beast, since the electric field is given by a vector sum instead of a scalar one. From the principal of superposition

It should be clear from the figure and the equation above that there are only two points where , (1) at and (2) on the line between the two charges, which is the only place where the two unit vectors will be equal and opposite.

Then we have

or

If we put the origin at the charge and call the distance from the origin then

Problem 30-45

Between the two parallel plates is constant, if we assume that the dimensions are large enough to ignore fringe effects (a good approximation here). Then in the region between the plates and we are given that

so

Problem 31­4

a. For a parallel plate capacitor

For this case, and m so

b. C.

Problem 31-9

This is another problem which shows how you can check your results by considering a useful limit. For a cylindrical capacitor we have

Now is the greater of the two radii, so we can set , and then we have

If the spacing is small then and , so

since is the total area for the internal cylinder.

Problem 31­12

and are in parallel, so they ADD. The result is in series with so they add reciprocals. Simply put

or

Problem 31-14

Need to find the equivalent capacitance. These capacitors are all in parallel, so the capacitances add.

F

and from the basic definition of capacitance we have

C

Problem 31-18

Neglecting the ``fringing fields'' near the edges of the semicircular plates,. we can think of this device varying from a minimum capacitance of zero when the plates do not overlap at all, to a maximum which will occur when the plates are completely overlapped. Note from the diagram that when the plates are all aligned the capacitors which they form are all in parallel so that the total capacitance is the sum of the individual ones. The area of each plate is and the separation between consecutive plates is . Again, we neglect fringing effects from the edges, and with these assumptions it is just a counting problem.

For there is no capacitance.
For there is one positive and one negative plate, so .
For there is one positive and two negative plates (or two positive and one negative) and either way we have two capacitors in parallel, so .

and so forth. We see that overall

Problem 31­22

Because the battery is disconnected, this is a case of constant charge in the system. The thing to keep in mind is that the capacitance does not change for the first capacitor, since that is purely a matter of geometry. So F. Volts, so

Afterwards, the total charge must be conserved, and the voltage drop across each capacitor must be equal. We are told that the voltage drop across the first capacitor falls to 35.8 Volts. Hence the charge on this capacitor is now

Finally, on the other capacitor the charge must make up the difference on the first one, and the voltage drop must be the same, so