Originally Posted by **rcmango** a large positive charge +*Q* fixed at some location in otherwise empty space, far from all other charges. A positive test charge of smaller magnitude +*q* is launched directly towards the fixed charge. Of course, as the test charge gets closer, the repulsive force exerted on it by the fixed charge slows it down.
want an explanation for how the electrical potential encountered by the test charge changes as it gets closer to the fixed charge.
and then also how the EPE of the test charge changes as it gets closer to the fixed charge.
...Not sure exactly whats going on but I know its because the force of the fixed charge is bigger and the same positive charge as the test charge and thats why its slowing down, but need help for whats going on with the EPE. Thanks. |

It is not because the fixed charge applies a greater force than the moving charge that the moving charge is slowing down, it is solely because they are of the same polarity. In fact, the force felt on the small charge is exactly the same as the force felt on the larger charge (Newton's laws work here also). The fixed charge could be many times smaller than the moving charge, yet as that moving charge, that has the same polarity as the fixed charge, nears the fixed charge it will indeed slow down, stop and eventually move away from the fixed charge; unless something is applying a force to it such that it will eventually reach an equilibrium position with respect to the fixed charge.

The force felt by each charge is given by Coulombs law which is:

F = K(Q1Q2/r^2)

Now because the fixed charge has a greater amount of positive charge than does the moving charge, the force exerted on each charge by the other is greater than if they were of the same smaller charge amount, but nonetheless the moving charge is repelled.

Basically each positive charge has an electric field leaving it. The electric field will push on positive charges and try and move them in the direction of the E field seen. Since each positive charge gives away an E field that moves radially outward from each one, each charge sees an E field from the other that is trying to push it directly away from each charge (assuming nothing else is around but these two charges).

The electric field given by each charge can be found by:

E = kq/r^2

This is just the force per unit charge. Here q is the charge that is causing the electric force felt on the other charge. So the E field felt by the fixed charge is the force on the fixed charge divided by its own charge. This hints at what you were thinking. The E field emitted by each charge, felt at the other charge, is NOT the same, but the force on each charge is the same. The E field from the fixed charge that wishes to slow and push away the moving charge felt by the moving charge is larger than the E field on the fixed charge given from the moving charge.

The force felt on each charge is also simply: F = Eq, so let us say the fixed charge is Q and the moving charge is q. The force felt by each charge is the same so:

F = F

EQ = Eq, therefore the two E fields cannot be equal for the equation to be true. Furthermore let us say the E field given by the moving charge is E(q) and by the fixed charge is E(Q). Now rewriting we get:

E(q)Q = E(Q)q, and remember that E(q) = kq/r^2, so

k(qQ)/r^2 = E(Q)q = E(q)Q = F, and this is Coulombs law by definition. This is the mutual electric force felt by each charge due to the other charge having an electric field. Certainly if Q is made smaller, the rate at which the moving charge slows will decrease, just as if Q were to be made larger the moving charge would slow more quickly; either way it only slows, stops and reverses direction because the two charges have the same polarity.

As far as the potential difference felt by each charge, these values should also be the same, but opposite in sign (I will get to EPE in a minute). Voltage is a scalar and has no direction, but it is always taken with respect to something else. So if something has a positive voltage with respect to something else, that something else must have a negative voltage with respect to that something. This can be written as:

V = E"dot"d, where both E and d are vectors dotted together leaving a scalar. Also in this case one may ask, what is E? This is interesting for a couple of reasons.

If you have two like and equal amounts of charge, no matter where you place them apart, there will never be a voltage difference between them (always a zero volt difference). For instance, if you have a capacitor with no charge on the plates one plate will obviously measure at zero volts with respect to the other plate. Likewise, if you stuff a handful of electrons, of equal amounts, on each plate, the voltage measured will still be zero volts at one plate with respect to the other. The two plates filled with electrons each will have those electrons trying very hard to find a way to get off their plates as they do not like being close to each other, nor near a plate with an equal amount of like charges on them. There is indeed a force felt on each plate due to the other plate, but no voltage difference; why?

It is how voltage is defined and calculated/measured. The voltage difference between two ends of a path is the total potential energy change in moving a small electric charge along that path done by an electric field, divided by the magnitude of the charge.

In the case of two like and equal amounts of charge separated by any distance, any PE lost in moving an additional like charge from one pile to the other is exactly that gained when it reaches the second pile. That is the energy used to push it plus that to stop it are both the same, hence the net PE is zero. That means there is no voltage difference between the two plates.

Or just imagine connecting perfectly conducting wires between two piles of like charges that normally would conduct those charges, like electrons do upon and within copper, since the two piles have equal amounts of charge, neither pile will want to move charges to/from the other. No movement of charge means, by definition, no current flow. The only way no current would flow between two places where there exits free charges using perfectly conducting wire is if there were no voltage to move the charges. Basically with two piles of equal and like charges the net E field between them is zero, and without an E field to do work, no voltage can exist.

In your case there is an E field between the two charges and it is in the direction from the larger charge Q to the smaller, moving charge q. As the two charges get closer, the net E field actually increases because the charge Q is larger and its E field, like that of the smaller charge q, changes with the square of the separation distance. So while the ratio of the two E fields is always the same, the difference between them is growing larger. Like 1 to 2, 10 to 20, 100 to 200; same ratio but larger differences. It is the net E field that exists, able to do work on a particle changing its PE that creates the voltage.

So the voltage continuous to grow as the two charges get closer and eventually reaches a maximum when the moving charge stops. At which point if the one that was moving is not held in place it will start moving away in the direction in which it came decreasing the voltage between them. The EPE (electrostatic potential energy) is the change in energy a particle would obtain moving between two voltage points. The two points are your two charges, and moving from the "moving" charge to the fixed charge would mean a particle would have to take on more PE than what it had being next to the moving charge.

So when they are far apart the EPE between the two charges is small. As the two charges reach their closest positions next to each other, the EPE reaches its greatest level.

Take care,

Craig