# Static Equilibrium between 2 known points and 1 unknown

• Paul2011
In summary, in order for three charges to be in static equilibrium, the electric field must be zero, meaning the individual electric fields must cancel out. This can only occur if the third charge is located on the x-axis and has a charge value of -4q. The position of the third charge can be determined by solving a system of equations, with one possible solution being x=3a.

## Homework Statement

A charge -q and a charge 4/9q are located at coordinates x = 0 and x = a, respectively. If you place a third charge so that all three are in static equilibrium, justify that its position and its charge,respectively, can only be

x=3a Q=-4q

## The Attempt at a Solution

Since the three charges are in static equilibrium I know that the electric field is zero. Which means each individual electric field has to add up to zero as a whole. If F_ab = (q_1)(q_2)/r^2 is get F_ab= (-4q^2)/9a^2, so does that mean the third charge added in has to cancel that out?

The electric field would be zero only at the points where the charges are located.

The third charge c has to cancel F_ab by F_cb and and F_ba by F_ca. So it has to be F_ab=-F_cb and F_ba=-F_ca. So you have a system of two equations to solve with two unknown variables, the position and the charge value of the third charge.

You have to argue that the position of the third charge can only be somewhere in the x-axis, otherwise the forces that the other two charges would apply on it could not cancel out.

Ok, so if I find F_ab = -(4q^2)/9a^2 and -F_cb = (16/9)q^2/a^2. F_ba is the same as F_ab which is -(4q^2)/9a^2 and F_ca = 4q^2/a^2. Not quite sure if I should replace a^2 with the given position of x=3a or not

First, better to get rid of the minus signs (cause they depend on where we put the charge c).

Second we consider the position and charge of charge c unknown variables. Say x is its position and Q is its charge then
F_ab=Fcb means that 4q^2/9a^2=4qQ/9(x-a)^2 which is the first equation
F_ba=Fca means that 4q^2/9a^2=qQ/x^2 which is the second equation.

from these two equations we can get that 4x^2=9(x-a)^2 hence 2x=3(x-a) or 2x=-3(x-a) hence x=3a or x=3a/5. We keep only x=3a (because if x=3a/5 this means that the third charge is inbetween the other 2 charges and the overall force to it cannot be zero).

Now you can put x=3a to the second equation and solve for Q and you ll get Q=4q.

We have actually calculated the absolute value of Q as 4q. Hence Q=4q or Q=-4q. Both values seem valid. You sure the answer speaks only of Q=-4q?

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Thanks a lot. Really appreciate it, definitely understand it now.

Paul2011 said:
Thanks a lot. Really appreciate it, definitely understand it now.

The answer says that Q=-4q only?

Doesnt say that Q=-4q only, just to prove that Q=-4q and to verify the position.