Where Should a Third Charge Be Placed to Experience No Net Force?

[n77ler]

[SOLVED] Electric Fields- point charges

Homework Statement

Two point charges of charge 7.40 μC and -1.60 μC are placed along the x-axis at x = 0.000 m and x = 0.360 m respectively. Where must a third charge, q, be placed along the x-axis so that it does not experience any net force because of the other two charges?

Homework Equations

E= k (q_1q_2) / (r^2)

The Attempt at a Solution

E = (8.99x10^9 Kgm^2/C^2) x (7.40x10^-6)(-1.60x10^-6) / (0.360m)^2

I calculated this value but I'm not really sure if I actually need it or what needs to be done in the problem

 

[Hootenanny]

No net force means that if you add the force acting on the charge from the two other charges, the result will be zero. So, you a looking for the point x=a such that;

$$F_{7.40\mu C} + F_{-1.60\mu C} = 0$$

 

[n77ler]

Ok so:

F=k (lq1l lq2l)/ r^2

=(8.99x10^9)x(7.40x106-6)x(1.60x10^-6) / (0.360)^2
= 0.821N
So, net force between these two point charges is 0.821N

F$$_{7.40uc}$$F$$_{-1.60uc}$$= 0

Can I just let force= 0.821 and using the above equation solve for r as the unknown? Thats all I can come up with but it's wrong. Any suggestions?

Edit: these superscripts should be subscripts I think they are called? Below the F

 

[Hootenanny]

Ok so:

F=k (lq1l lq2l)/ r^2

=(8.99x10^9)x(7.40x106-6)x(1.60x10^-6) / (0.360)^2
= 0.821N
So, net force between these two point charges is 0.821N

F$$_{7.40uc}$$F$$_{-1.60uc}$$= 0

Can I just let force= 0.821 and using the above equation solve for r as the unknown? Thats all I can come up with but it's wrong. Any suggestions?

Here you have calculated the force between the 7.40 μC and the -1.60 μC charges. You need the net force (from these to charges) acting on the third charge, q, to be zero. Do you understand why?

I'll help you set up the problem. Let r₁ and r₂ be the positions of the positive and negative charges respectfully. Further, let r₀ be the position of the third charge q such that the net force acting on q is zero. Then, by Coulomb's law (we can cancel the constants),

$$\frac{q\cdot 7.40}{\left(r_0-r_1\right)^2}\cdot10^{-6} + \frac{-q\cdot 1.60 }{\left(r_0-r_2\right)^2}\cdot 10^{-6} = 0$$

You now need to solve for r₀. Do you follow? HINT: You may find it easier to solve if you does some cancellation first.

As an aside, since the electrostatic force is conservative and hence defined as $F=-\nabla\left(E\right)$, we could equally approach the problem by find the position at which the electric field due to the two charges is zero, r₀. However, I feel that it make more intuitive sense (Newton's 3rd law) if we consider forces rather than fields.

Edit: these superscripts should be subscripts I think they are called? Below the F

Yes, they are called sub-scripts.

 

[n77ler]

How will i solve for r$$_{}o$$ if I don't know q either?

 

[kbaumen]

Another way to think is this: On the charge you place in, the intensity must be 0. E=k*q/r^2 where q is the charge that creates this electrical field. I think that's a bit easier way to solve, because there are less variables than in the Coulomb's law.

It doesn't matter whether you know the third charge or not. It can be 10^-20 as well as 10^20 C (though that's pretty much). The aim is to find a place where the other two charges don't interact with this third charge electrically.

P.s. Sorry for bad English.

 

[n77ler]

I still don't understand though. I don't know the q charge. I don't know the radius, or E ?

 

[kbaumen]

The third charge doesn't matter. Let's say that the third charge is r meters away from q1. In that case, it's 0.36 - r meters away from q2. So what you have to do, is solve $$\frac{k*q1}{r^{2}}$$ = $$\frac{k*q2}{(0.36 - r)^{2}}$$ in terms of r. k is a constant you probably know - 9 * 10^9 N*m^2/C^2 and both charges are given.

Remember that it's further from the biggest charge, so don't forget to compare r and 0.36 - r.

I hope that helps.

 

[n77ler]

I just can't get this, I worked out the equation you just wrote but it gets messy with all the big numbers. I come out with a quadratic and solve. r=2.59, r=-1.67 neither of them work and i compare them to 0.36-r and that doesn't work either :(

 

[kbaumen]

Hmm... weird. It should work. Try first solving it algebraically in terms of r and then put in the numbers. If it still doesn't work, try the same idea with forces, but I can't see why it shouldn't work.

 

[n77ler]

So when I solve for r, do I have to fill it into 0.36-r?

 

[kbaumen]

Yes. r will be distance from one of the charges, and 0.36-r from the other. The biggest of those two numbers is the distance from the biggest charge.

 

[n77ler]

well I've tried 8 times and can't get it, I don't know what else to do

 

[Hootenanny]

How will i solve for r$$_{}o$$ if I don't know q either?

HINT: You may find it easier to solve if you does some cancellation first.

We start from the equation I posted previously,

$$\frac{q\cdot 7.40}{\left(r_0-r_1\right)^2}\cdot10^{-6} + \frac{-q\cdot 1.60}{\left(r_0-r_2\right)^2}\cdot 10^{-6} = 0$$

We divide through by q and multiply through by 10⁶,

$$\frac{7.40}{\left(r_0-r_1\right)^2} - \frac{1.60}{\left(r_0-r_2\right)^2}= 0$$

We take the second quotient over to the RHS,

$$\frac{7.40}{\left(r_0-r_1\right)^2} = \frac{1.60}{\left(r_0-r_2\right)^2}$$

And now divide through by 7.40 & multiply though by $\left(r_0-r_2\right)^2$,

$$\frac{\left(r_0-r_2\right)^2}{\left(r_0-r_1\right)^2} = \frac{16}{74}$$

Can you take is from here?

 

[Hootenanny]

Yes. r will be distance from one of the charges, and 0.36-r from the other. The biggest of those two numbers is the distance from the biggest charge.

r₀ is the distance from the origin to the equilibrium point.

 

[n77ler]

Thank-you so much, now only 11 more problems to go :( lol

 

[kbaumen]

r₀ is the distance from the origin to the equilibrium point.

Yeah, that's what I said, just rephrased better.