In the diagram at right, we see two negative charges and one positive charge, fixed in place and arranged in a equilateral triangle. They are isolated in the sense that all other electrical charges are very far away. So we represent very far away (or “infinity”) by a dotted line around the system. I have drawn flux lines for the system according to the rules we have discussed.

If you have two negative charges and one positive charge in a region of space, what does the field produced by these charges look like, when seen from a great distance away? a)Like 1 positive charge b)Like 1 negative charge c)Like 2 negative charges d)Like no charge (electrically neutral)

Correct Answer - B Since the field lines close to the infinity line have arrows pointing inward, they are like lines heading towards a negative charge. There are 8 of them, since we have 16-8 = 8 “extra” field lines from the interior which begin on the negative charges (2x8) but which cannot end on the positive charge (1x8). These field lines repel each other, so they will spread out into a radial pattern given enough space to do so, in the absence of any other field lines. A radial pattern of 8 field lines pointing inwards is like the field of a single isolated negative charge.

If we now place a small positive charge at the position labeled A on the diagram, where would you expect this charge, which is free to move, to end up? a)Near the positive charge b)Near the upper negative charge c)Near the lower negative charge d)Near infinity (far away) e)It depends on the precise starting point A

Correct answer B Test charges move along field lines when they are free to move. A positive test charge moves in the direction of the arrows on the field lines. Both field lines nearest to position A end on the upper negative charge, so we expect our test positive charge to end up there too.

C Suppose we put a negative charge at point C on the diagram. If it is free to move, where would be expect it to end up? a)Near the positive charge b)Near the upper negative charge c)Near the lower negative charge d)Near infinity (far away) e)It depends on the precise starting point

Correct Answer - D In this case the negative test charge, which moves against the flux arrow, is between two flux lines which end at infinity. It is clear then that it also should end up very far away from the system.

A If we put a negative charge at point A, where would you expect it to end up, if it was free to move? a)Near the positive charge b)Near the upper negative charge c)Near the lower negative charge d)Near infinity (far away) e)It depends on the precise starting point

Correct Answer - E In this case we know that the negative charge will initially move against the flux arrows in a direction which, not surprisingly, sends it away from the upper negative charge. But we do not have enough information on the diagram to decide where it goes after that. If its local flux line behaved like the next outer flux line it would follow that line towards infinity. If its local flux line behaved like the next inner flux line it would move towards the positive charge. In this case the answer depends on the precise details because when you are far away the field is like that of a negative charge, which would repel the negative test charge. But sufficiently close to the positive charge a negative test charge is attracted it regardless of the presence of the two negative charges which are further away. In the in-between region the result depends on the details of the flux lines.

If we now place a negative charge at the position labeled B on the diagram, where would we expect it to end up, if it was free to move? a)Near the positive charge b)Near the upper negative charge c)Near the lower negative charge d)Near infinity (far away) e)It depends on the precise starting point B

Correct Answer - A The way the diagram is described suggests that the test negative charge is placed in the triangle between the three fixed charges. In that case it pretty much has to end up near the positive charge. But the way the diagram is drawn means that E is not a bad answer. If the test charge is placed a little to the other side of the line connecting the two negative charges, then it should be forced off towards infinity. My sketch is not terribly accurate here, unfortunately, since it suggests the flux lines going towards infinity can penetrate the inner triangle, which is not the case.

Now let’s imagine that these 3 charges are the only charges in the universe. The boundary line is no longer dotted, because it doesn’t necessarily represent infinity, but instead represents the edge of the universe, which is electrically neutral, so no field lines can begin or end there. In this Universe which of the following is true? a) Individual positive charges are weaker than individual negative charges b) Individual negative charges are weaker than individual positive charges c) Individual charges of either sign have the same strength

Correct Answer B: If we take our flux line rules seriously, then the strength of fundamental charges is actually a function of how many of each kind there are in the Universe! Since all flux lines must begin and end on an actual charge, if there are twice as many negative charges as positive charges (which is the case in our imagined universe) then there must be twice as many flux lines coming out of positive charges than there are going into negative charges. Therefore the field around each negative charge is weaker. If the field around a charge is weaker, it makes sense to say that the charge is weaker. So it seems clear that flux lines are a measure of the strength or size of a charge.

Gauss’s Law: Gauss’s Law states that the total flux coming into or out of a region of space directly measures the total charge which is inside that region of space. Let us define the electric flux through a given area, , as the product of the average electric field passing through that area E times the area itself A.  = E A Only electric field actually passing THROUGH the area counts. If the electric field lines are not perpendicular to the area then you only count the components of the field lines which are perpendicular. In this case the flux depends on the angle  between the field lines and the line perpendicular to the area A.  = E A Cos 

Once we’ve defined electric flux, it is easy to state Guass’s Law: Coulomb’s Law gives the electric field at a distance r from a single charge as E = k q/r 2 Let’s draw a sphere around this charge. Since the field lines are radial, the field is perpendicular to the surface area of the sphere at all points, so  = E A = k q A/ r 2 The surface area of a sphere is A = 4  r 2 so this gives us  = E A = k q 4  = 4  k q = q/  o Where  o is a new quantity called the electrical permittivity of free space, and it is equal to 1/ 4  k.

Guass’s Law  = q/  o So the flux lines through any closed surface like a sphere measure the total charge inside the sphere. Note that Gauss’s Law is only valid for a closed surface, so you have to go the whole way around a system measuring its field to be sure of what is inside it. This is why a table, say, is perceived by us as a neutral object. In fact it seems to be full of uncountable numbers of charges inside its atoms. But these charges are so close together that even placing your hand on the surface of the table, the charges are very far away (“infinity”) from your hand, compared to the distance to each other. So the flux lines don’t reach your hand. They begin and end on the charges inside the table. If you cannot feel flux lines outside the table, then it is as electrically neutral, according to Guass’s Law, as if it had not a single charged particle inside it.

Electrical Induction: What about the case when we “induce” a charge on an object like a tin can by bringing another, electrically charged object close to it? In this case the tin can still has the same number of negative charges as positive charges, but the positive charges have moved to one side of the object and the negative charges to the other (attracted and repelled by the proximity of the charged rod). This means that they are no longer very close together, and some of the flux lines from the +ve charges between them loop outside the object to reach the -ve charges on the other side of the tin can. So now when we are close to the tin can we feel flux lines, and so it no longer behaves as if there were no charges inside it.

Let’s surround a electrically induced tin can with a sphere which is close enough to be penetrated by the flux lines emerging from it. What does Guass’s Law say now about the total charge within the sphere, which is to say the total charge within the can? a)The tin can is still electrically neutral b)The tin can now has a net positive charge c)The tin can now has a net negative charge

Correct Answer - A Although it seems as if the tin can is now “charged”, the fact that we don’t know whether it is positively or negatively charged is a clue to how Guass’s Law works. If you count them, there are as many flux lines going out through the surface as there are coming back in through it. This means some of the field vectors pointing through the surface are pointing in (they are negative) and some are pointing out (they are positive). If you add them all up they cancel each other out and the total charge inside is still zero. But in this case we can still feel some electrical effect, depending on whether we are closer to the negative or the positive end of the tin can.