# Two charges and electric field?

• timnswede
In summary, there are two possible values for the unknown charge located at x=+3a in an electric field produced by two charged particles located on the x axis. The values are -9Q and 27Q, based on the attraction of the unknown charge to the other particle. This is a valid way of finding the values, although it does not take into account the direction of the electric field. By correcting a typo in the equations, it can be seen that trying both positive and negative values for the resultant field is the same as trying both signs for the unknown charge.

## Homework Statement

Two charged particles are located on the x axis. The first is a charge +Q at x= -a. The second is an unknown charge located at x= +3a. The net electric field these charges produce at the origin has a magnitude of 2kQ/a^2. Explain how many values are possible for the unknown charge and find the possible values.

E=k(q/r^2)

## The Attempt at a Solution

I got the right answer, I'm just not sure if my thinking is right, because I kind of ignored the way the electric is facing. First of all I assumed q is negative and got 2kQ/a^2=kQ/a^2-k1/9a^2 based on the attraction of q to Q. That turns out to be q=-9Q. Then I did the same thing for when q is positive and got 2Qk/a^2=qk/9a^2 - kQ/a^2, again based on the attraction of q and Q. That turns out to q=27Q, which are the right answer. Is this a valid way of doing it or am I thinking of it wrong?

timnswede said:
Is this a valid way of doing it or am I thinking of it wrong?
Is there any particular doubt that stands out in your mind?

Bystander said:
Is there any particular doubt that stands out in your mind?
My main doubt is that I'm not taking into account the way that the electric field is facing due to the two charges, or would that just be a different way of thinking of it? Only my 4th day of E&M so I'm not too great at it yet :/

timnswede said:
My main doubt is that I'm not taking into account the way that the electric field is facing due to the two charges, or would that just be a different way of thinking of it? Only my 4th day of E&M so I'm not too great at it yet :/
Despite the way you worded your argument, trying both signs for the resultant field is precisely what you did.
timnswede said:
2kQ/a^2=kQ/a^2-k1/9a^2
2Qk/a^2=qk/9a^2 - kQ/a^2
Correcting the typo in the first equation:
2kQ/a^2=kQ/a^2-kq/9a^2​
the only difference from the second equation is the sign of the whole right hand side - or, equivalently, of the left hand side.

haruspex said:
Despite the way you worded your argument, trying both signs for the resultant field is precisely what you did.

Correcting the typo in the first equation:
2kQ/a^2=kQ/a^2-kq/9a^2​
the only difference from the second equation is the sign of the whole right hand side - or, equivalently, of the left hand side.
It's always the simplest parts that mess me up D:
But thank you for the help, makes sense then!