Coulobs law and vectors?

Question

Consider two point charges of the same magnitude but opposite sign (+Q and -Q), which are a fixed distance d apart. Can you find a location where a thir positive charge Q could be placed so that the net electric force on this third charge is zero? What if the first two charges were both +Q?

Answer 1

Offhand, d/2? About the Other Question, I Think No.

Answer 2

Let's assume that charge +Q is placed on point A and charge -Q is placed on point B as in the following sketch:

Q(A)-----------d-----------Q(B)

If we place the third charge Q between points A and B it is impossible to have zero net force, since Q(A) would repel Q while Q(B) would attract it, i.e. the two forces would have the same direction.
The only possibility is to place the third charge either on the left of point A or on the right of point B.
Let's study the first case:

Q-----r-----Q(A)-----------d-----------Q(B)

Using Coulmb's law for the pair Q(A) - Q we get

F1=K*Q(A)*Q/r^2 (1)

Using Coulmb's law for the pair Q(A) - Q we get

F2=K*Q(B)*Q/(d+r)^2 (2)

In order to have equilibrium these two forces have to be of equal magnitude. Equating (1) and (2) we get
K*Q(A)*Q/r^2 = K*Q(B)*Q/(d+r)^2 ->
1/r^2 = 1/(d+r)^2 ->
r^2 = (d+r)^2 ->
r^2 = d^2 + r^2 + 2*r*d ->
0 = d^2 + 2*r*d ->
d*(d + 2*r) = 0

From this we get either
d=0 (meaning that A and B are the same point)
or
r = -d/2 which is obviously wrong since the quantities d and r are both positive.

So the only solution is if Q(A) and Q(B) are on the same point.

This result isn't as trivial as it sounds!

For example if a charge q is placed at a long distance from an atom then the negative charge of the electrons orbiting the nucleus is placed more or less on the same point as the positive charge of the nucleus itself. Therefore the atom appears to have negative charge.
But if q starts approaching then the atom at some point the aforementioned approximation does not stand so the charge "feels" a force from the atom.

Another thing worth mentioning is that the solution given above stands even if the third charge differs in magnitude from the first two. The only limitation is that Q(A) and Q(B) are of the same magnitude (but different sign).

Repeating the process if Q(A)=Q(B)=Q we conclude that the location of zero net force is in the middle of AB, i.e. r=d/2

Q(A)-----------Q-----------Q(B)

Again the magnitude (as well as the sign) of the third charge is irrelevant.

Answer 3

You have six force vectors with three charges. This can be seen from the table:

x 1 2 3
1 x o o
2 o x o
3 o o x; where o = force vector (e.g., Q1 <-- Q2, charge 1 attracts charge 2, which is the second element in the first row. If we say Q1>0 and Q2<0 are the first two charges, they are attracting each other according to Coulomb's Law, Fa = k(Q1)Q2/d^2. We can rewrite the table to show attraction and repulsion among the three charges.

x 1 2 3
1 x a o
2 a x a
3 o a x; where a shows attraction and o shows repulsion.

Since Q1 and Q2 are fixed at d(1,2), only Q3 can be moved to find a point where the sum of forces on it are zero. Thus F(3) = o(1,3) + a(2,3) = 0, which means o(1,3) = -a(2,3). Thus, Q3 would have net force on it where the repulsive force from Q1 = the attraction force from Q2. So we have kQ1Q3/r(1,3)^2 = kQ2Q3/r(3,2)^2 from which, we can see that r(1,3) = r(3,2) is the point where the net forces would be zero.

In other words, place Q3 equidistant from Q1 and Q2 anywhere, even directly between them, where r(1,3) = r(3,2) = d/2, and the net force will be zero on Q3.

If all three charges were positive, there would be no source of negative charge to offset the repulsive forces. So there would be no place where the Q3 charge could be placed with net zero force. It would accelerate off into limbo somewhere.

Answer 4

Let's take the two positive charges first. Say each is in the xyz plane on the x-axis at -d/2 and +d/2. Coulomb's Law is F = k(Q1*Q2)/r^2

So, the force from each point charge will be kQ^2/r^2 and collinear with the line joining the point charge to the third charge. Wherever these forces are equal and opposite, there will be no net electrical force on the third charge. In general, the forces will be equal in the yz-plane. And in particular, at the origin, the forces will be equal and opposite; and therefore, zero.

Now, let one of the point charges have a negative charge. Now, everywere in the yz-plane, as elsewhere, there will be an x component of the force.