Finding zero net force between two charges

In summary, the question asked where a third point charge "q" would be placed between two charges (+4q and +2q) on a horizontal line of 2 m length to balance the forces acting on it. Using Coulomb's Law, the quadratic equation was used to solve for the distance "x" where "q" should be placed, resulting in the equation -2x^2 + 8x + 8 = 0. After fixing a mistake in the plugging in process, the correct answer of x = 1.17 m was obtained.
  • #1
kam761
4
0
1. Here is the question we were given: There are two charges on a horizontal line of 2 m length.

*-----------------------*
Qa = +4q Qb = +2q

If a third point of charge "q" is moved between these points, where will the forces acting on it be balanced? Our Professor gave us the answer to be 1.17m to the right of 4q.

2. Coulomb's Law : F = k[Qa][Qb]/(R^2)

k (Coulomb's constant) = 9.0E9
Qa = +4q
Qb = +2q
R = distance between two points (in this case 2 m) quadratic equation: -b +- sq.root((-b^2 - 4ac)/2a)


3. I know that "q" must be between the Qa and Qb for their forces to cancel (FQa is going to the right, FQb is going to the left.) I also know that you have to set the forces of Qa and Qb to each other. The problem I am having, and this is more basic algebra than anything else I think, is getting it into the quadratic equation to solve for "x" (the distance where "q" is.) I am also not sure what should go in the denominators of either side. This is what I have got so far:

k[Qa][q]/(2-x)^2 = k[Qb][q]/x^2

Getting rid of like terms and such, I get:

(Qa)(x^2) = (4 + 4x + x^2)(Qb)

and plugging in the charges for Qa and Qb:

4x^2 = 8 + 8x + 2x^2, or - 2x^2 + 8x + 8 = 0.

Yet, when I plug this into the quadratic equation, I am certainly not getting the correct answer. What am I doing wrong? Thank you for the help!
 
Physics news on Phys.org
  • #2
Welcome to PF :smile:

(2-x)^2 is not (4 + 4x + x^2)

Can you fix it now?
 
  • #3
Thanks!

I knew I was probably making a basic math problem. So this is the new reworked ending, which still doesn't seem to work out.

2x^2 + 8x - 8 = 0. Plugging this into the quadratic gives me 2.35 and 13.65, both of which are obviously not even close to 1.17, nor are they even in the x-axis length that I'm working in!
 
  • #4
You're right, something is still amiss.

Since x=1.17 is a solution to
k*4q*q / x^2 = k*2q*q / (2-x)^2,​
I would plug in x=1.17 at each step of your derivation, starting at the beginning, and see at which step it is no longer a solution. That will be where the mistake is.
 
  • #5
I'd check again when you plug into the quadratic formula..

Cause when I plug 2x2-8x+8=0, I get x1= -4.82 and x2=0.83.

What you have labelled as x is the distance from Qb to the new charge, so 2-x should give you the distance from Qa, right? 2-0.83=1.17. So, it's just your plugging in that's weird.

(Just a side note, I personally think it would be simpler to say:

k*Qa*q/x2=k*Qb*q/(2-x)2

because then your x is actually the distance from Qa to the new charge.)
 
  • #6
Ok,

I'm not sure what I'm doing wrong. Using 2x^2 - 8x + 8:

8 +- sq.root((8^2 - 4(2)(8))/2(2)) gives me +- 8. ?

I even tried to plug it into a quadratic formula calculator online, and that gave me 2 for both x's. I just don't understand. Would switching the denominators give me a different answer? I would think not, since you seemed to have gotten the right answer with the work I have done so far. Where the heck am I going wrong?
 
  • #7
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex]

This is the formula you're trying to use, right? The way you've written it confuses me a bit, but I think we're doing the same thing.

And that 0=ax2+bx+c .. But I know that you already know that.

No! I know what's wrong now! Oh dear, I'm not even reading my notes properly.

It's supposed to be 0= -2x2-8x+8. You've lost a minus somewhere, go and try to find where.

Edit: No, wait a second. the formula you came to was
2x^2 + 8x - 8 = 0 which does give me the correct answer too.

Mine's just yours times (-1). So, somewhere in your plugging in something's weird.
 
  • #8
Thank you so much! I had the variables right, I was putting the denominator underneath the sq. root sign, which of course would throw everything off. I knew it had to be a simple math mistake. Thank you for your help!
 

Related to Finding zero net force between two charges

Q: What is zero net force between two charges?

The concept of zero net force between two charges refers to the balance of attractive and repulsive forces between two charged particles. This means that the total force acting on each particle is equal and opposite, resulting in no overall movement.

Q: How is zero net force calculated between two charges?

To calculate zero net force between two charges, you need to use the Coulomb's law formula, which states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Q: What happens when there is zero net force between two charges?

When there is zero net force between two charges, the two particles will remain in a state of equilibrium, meaning they will not move towards or away from each other. This is because the attractive and repulsive forces are balanced, resulting in no overall movement.

Q: Can there be zero net force between two charges of the same polarity?

No, there cannot be zero net force between two charges of the same polarity. This is because charges of the same polarity repel each other, and there will always be a net force pushing them apart.

Q: How does the distance between two charges affect the zero net force?

The distance between two charges has a significant impact on the zero net force between them. As the distance between the charges increases, the force decreases, and at a certain distance, the forces become balanced, resulting in zero net force.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
241
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
5K
  • Introductory Physics Homework Help
Replies
9
Views
908
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
3K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
2K
Back
Top