# Homework Help: Two charges add to zero

1. Feb 5, 2009

### phrygian

1. The problem statement, all variables and given/known data

I am trying to find the point at which the electric field from two charges adds to zero and came to the equation (kq)/x^2-(kQ)/(x-d)^2 = 0

2. Relevant equations

(kq)/x^2-(kQ)/(x-d)^2 = 0

3. The attempt at a solution

I know that (kq)/x^2-(kQ)/(x-d)^2 = 0 is the correct equations but algebraically do not know what to do from here.

Thanks for the help

2. Feb 5, 2009

### LowlyPion

Are the charges both the same sign?

3. Feb 5, 2009

### phrygian

The charge at x = 0 is negative and has magnitude q = 2 mCoulombs. The charge at x = d, (d = +12 cm), is positive and has magnitude Q = 4 mCoulombs.

4. Feb 5, 2009

### LowlyPion

Then your equation is already set up wrong, because the field between a positive and a negative is always non-zero at every point in between.

So if it isn't 0 between them it must be 0 somewhere else.

5. Feb 6, 2009

### phrygian

I must have typed something wrong but in the help section of the homework it gives me this:

Ex = (1/4peo)*((q/x^2) - (Q/(x-d)^2)) (for x < 0)

Setting this expression equal to zero yields the equation:

x^2*(q-Q) + x*(-2qd) + qd^2 = 0

I am just wondering mathematically how you get from the first equation to the second one because I have no idea...

6. Feb 6, 2009

### LowlyPion

After canceling common factors, that step can be made by cross multiplying from

q/x2 = Q/(x-D)2

But you say they are choosing x<0, but I would note that for x<0 the distance to the other charge is x + D not x - D. This equation would be true for x>D however.

7. Feb 6, 2009

### phrygian

Ahaaa I get it now thanks a lot.

And then how do you get to x = d*((q/(q - Q))*(1 +/- sqrt(Q/q))

8. Feb 6, 2009

### LowlyPion

That's the solution for a quadratic equation?