Electric Fields- point charges

In summary, the problem is asking for the position, x=a, along the x-axis where a third charge, q, can be placed so that it does not experience any net force due to two other charges, 7.40 μC and -1.60 μC, placed at x=0.000 m and x=0.360 m respectively. The solution involves setting up the equation for the net force and solving for the position, taking into account the distances from the two charges. After simplifying the equation and solving algebraically, the position is found to be 0.217 m from the origin.
  • #1
n77ler
89
0
[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
 
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  • #2
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;

[tex]F_{7.40\mu C} + F_{-1.60\mu C} = 0[/tex]
 
  • #3
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[tex]_{7.40uc}[/tex]F[tex]_{-1.60uc}[/tex]= 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
 
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  • #4
n77ler said:
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[tex]_{7.40uc}[/tex]F[tex]_{-1.60uc}[/tex]= 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),

[tex]\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[/tex]

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 [itex]F=-\nabla\left(E\right)[/itex], 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.

n77ler said:
Edit: these superscripts should be subscripts I think they are called? Below the F
Yes, they are called sub-scripts.
 
  • #5
How will i solve for r[tex]_{}o[/tex] if I don't know q either?
 
  • #6
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.
 
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  • #7
I still don't understand though. I don't know the q charge. I don't know the radius, or E ?
 
  • #8
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 [tex]\frac{k*q1}{r^{2}}[/tex] = [tex]\frac{k*q2}{(0.36 - r)^{2}}[/tex] 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.
 
  • #9
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 :(
 
  • #10
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.
 
  • #11
So when I solve for r, do I have to fill it into 0.36-r?
 
  • #12
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.
 
  • #13
well I've tried 8 times and can't get it, I don't know what else to do
 
  • #14
n77ler said:
How will i solve for r[tex]_{}o[/tex] if I don't know q either?
Hootenanny said:
HINT: You may find it easier to solve if you does some cancellation first.
We start from the equation I posted previously,

[tex]\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[/tex]

We divide through by q and multiply through by 106,

[tex]\frac{7.40}{\left(r_0-r_1\right)^2} - \frac{1.60}{\left(r_0-r_2\right)^2}= 0[/tex]

We take the second quotient over to the RHS,

[tex]\frac{7.40}{\left(r_0-r_1\right)^2} = \frac{1.60}{\left(r_0-r_2\right)^2}[/tex]

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

[tex]\frac{\left(r_0-r_2\right)^2}{\left(r_0-r_1\right)^2} = \frac{16}{74}[/tex]

Can you take is from here?
 
  • #15
kbaumen said:
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.
 
  • #16
Thank-you so much, now only 11 more problems to go :( lol
 
  • #17
Hootenanny said:
r0 is the distance from the origin to the equilibrium point.

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

Related to Electric Fields- point charges

1. What is an electric field?

An electric field is a physical field that surrounds an electrically charged particle or object. It represents the force that would be exerted on another charged particle placed in the field.

2. How is an electric field created by a point charge?

An electric field is created by a point charge through its Coulomb force, which is the force of attraction or repulsion between two charged particles. The electric field lines point away from positive charges and towards negative charges.

3. What is the equation for the electric field created by a point charge?

The electric field created by a point charge can be calculated using the equation E = k*q/r^2, where E is the electric field strength, k is the Coulomb's constant, q is the charge of the point charge, and r is the distance from the point charge.

4. How does the distance from a point charge affect the electric field strength?

The electric field strength is inversely proportional to the square of the distance from the point charge. This means that as the distance increases, the electric field strength decreases.

5. What is the direction of the electric field at a point in space?

The direction of the electric field at a point in space is the direction of the force that would be exerted on a positive test charge placed at that point. This direction is indicated by the direction of the electric field lines, which always point towards negative charges and away from positive charges.

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