Finding the Net Electrostatic Force on Particle 3 in a Coulomb's Law Problem

[Langerhorns]

Homework Statement

Particle 1 of charge q1 = +0.76 µC and particle 2 of charge q2 = -3.0 µC, are held at separation L = 13 cm on an x axis. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the x and coordinates of particle 3?

I'm really not sure how to proceed with this question. But I understand that the forces added together is zero.

 

[LowlyPion]

Homework Statement

Particle 1 of charge q1 = +0.76 µC and particle 2 of charge q2 = -3.0 µC, are held at separation L = 13 cm on an x axis. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the x and coordinates of particle 3?

I'm really not sure how to proceed with this question. But I understand that the forces added together is zero.

Welcome to PF.

OK. So what equation do you think you might use then?

 

[Langerhorns]

So I did:

c/d^2 + c/d^2 =0

put in the numbers

(0.76*10^-6)/d^2 = (3*10^-6)/(d+13)^2

simplify taking out the fractions

(0.76*10^-6)(d+13)^2 = d^2(3*10^-6)

expand the bracket

(0.76*10^-6)(d^2+26d+169) = d^2(3*10^-6)

eventual rearrangement

2.24*10^-6 - 1.976*10^-5 - 1.2844*10^-4 = 0

quadratic equation

13 and -4.3 but 13 is not the right answer. It might be my method or calculation error, I'm really not sure.

 

[LowlyPion]

I trust you didn't round away too much from the answer for your quadratic.

If you think about it, ≈ 13 is an expected result, because if 1 charge is 4 times larger and twice as far away ... you'd expect something about equal to the charge separation wouldn't you?

 

[Hannisch]

Hum, isn't q2 negative? I reckon you might've used it as a positive..

F1 - F2 = 0 .. so, q1/d^2 = q2/(d+x)^2, where q2 still is negative, and I only see +3 in your calculations :)

 

[LowlyPion]

Hum, isn't q2 negative? I reckon you might've used it as a positive..

F1 - F2 = 0 .. so, q1/d^2 = q2/(d+x)^2, where q2 still is negative, and I only see +3 in your calculations :)

Actually I think he's taken that into account.

∑F = 0 = q3*∑ E = 0 ⇒ E1 + E2 = 0

Taking account of the sign of the charge then |E1| = |E2| satisfies the condition.

(This of course is for x that does not lie between q1 and q2.)

 

[Hannisch]

I don't quite see it, but I reckon you're right 'cause you most definitely know better than I do and it's late & I'm tired. I think I'll come back and try to see it when it's not 20 to midnight ;)