Placing charges with coulomb's law problem

In summary, the third charge, q, should be placed at a distance d/3 from the origin with a negative sign and a charge of 3Q. This will result in all three charges being in equilibrium as the forces between Q & q and 4Q & q will cancel out. The specific value for the charge Q will determine the numerical value for q.
  • #1
kirby2
43
0
Problem: a charge +Q is located at the origin and a second charge, +4Q is at a distance d on the x-axis. where should a third charge, q, be placed, and what should be its sign and magnitude, so that all three charges will be in equilibrium.

attempt:

I don't know how to solve it fully, but my initial thought is to put the third charge in between the other two and give it a negative sign. The two other charges repel each other so i thought a negative one in the middle would put them in equilibrium. is this right?
 
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  • #2
kirby2 said:
Problem: a charge +Q is located at the origin and a second charge, +4Q is at a distance d on the x-axis. where should a third charge, q, be placed, and what should be its sign and magnitude, so that all three charges will be in equilibrium.

attempt:

I don't know how to solve it fully, but my initial thought is to put the third charge in between the other two and give it a negative sign. The two other charges repel each other so i thought a negative one in the middle would put them in equilibrium. is this right?
You are right about the sign of the third charge.

Considering, for a moment, only the positive charges:
1. Is there any location along the x-axis where the electric field is zero?

2. What is the magnitude of the force that the two positive charges exert on each other?​
Answering those should give you a good start.
 
  • #3
am i correct by saying that there is no point on the x-axis where the electric field is zero
 
  • #4
kirby2 said:
am i correct by saying that there is no point on the x-axis where the electric field is zero
No.

For any location between x = 0 and x = d, The direction of the electric field due to the +Q charge is in the opposite direction of the electric field due to the +4Q charge . Therefore, there is some location between x = 0 and x = d at which the electric field due to those two charges is zero. At that location, what is true of the magnitude of the field due to +Q compared to the magnitude of the field due to +4Q ?

In my experience, by the time you encounter a problem like the one in this thread, you would have already done some problems with two charges located on the x-axis and you are to find the location at which the electric field is zero.
.
 
Last edited:
  • #5
i think i got it. it should be placed right d/3 from origin. sign negative. and charge of 7.11 x 10^(-20).
 
  • #6
kirby2 said:
i think i got it. it should be placed right d/3 from origin. sign negative. and charge of 7.11 x 10^(-20).
That's the correct location.

You haven't given a specific value for the charge Q. How can you give a specific numerical value for the 3rd charge, q ?

Use the second hint I gave you.
 
  • #7
i assumed Q to be 1.6 x 10^(-19)
 
  • #8
kirby2 said:
i assumed Q to be 1.6 x 10^(-19)
Any reason for that assumption?

If you assume that, then the third charge, q, will be a fraction af one electron charge.
 
  • #9
the force that the two charges exert on each other is (k)((4Q^2)/d^2). but i don't know how that helps
 
  • #10
kirby2 said:
the force that the two charges exert on each other is (k)((4Q^2)/d^2). but i don't know how that helps
That is a force of repulsion.

The middle charge, q, must exert a force equal to that. As you have concluded previously, charge, q, must have a sign opposite to that of Q & 4Q, so that it attracts both of them.

What is the (magnitude of the) force exerted on each other by the charge, Q, at the origin and the charge, q, at x = d/3 ?

Set that force equal to [itex]\displaystyle \frac{4k\,Q^2}{d^2}\,.[/itex]
 
  • #11
OK. i think its -3Q.
 
  • #12
kirby2 said:
OK. i think its -3Q.

Magnitude of the force between Q & 4Q:
[itex]\displaystyle \frac{4k\,Q^2}{d^2}\,.[/itex]​

Magnitude of the force between Q & q: assuming q is at x=d/3 and q = -3Q:
[itex]\displaystyle \frac{4k\,(3Q)\cdot Q}{(d/3)^2}=27\frac{4k\,(Q^2)}{d^2}\,.[/itex]​
 

1. What is Coulomb's Law and how does it relate to placing charges?

Coulomb's Law is a fundamental principle in physics that describes the force between two charged particles. It states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This law is commonly used when calculating the force between two charges in a given system, which is important in determining the placement of charges.

2. How do you determine the direction of the force between two charged particles?

The direction of the force between two charged particles can be determined by the sign of the charges. Like charges repel each other, so the force will be in the direction away from each other. Unlike charges attract each other, so the force will be in the direction towards each other. This direction can also be determined by the vector components of the force equation.

3. What are the units used in Coulomb's Law?

The units used in Coulomb's Law are in Newtons (N) for force, Coulombs (C) for charge, and meters (m) for distance. This can also be written as N*m^2/C^2.

4. How can Coulomb's Law be used to determine the placement of charges in a system?

Coulomb's Law can be used to determine the placement of charges in a system by calculating the force between them and ensuring that the forces are balanced. If the forces are not balanced, the charges will experience a net force and will move. By adjusting the distances and charges, the forces can be balanced and the charges can be placed in a desired configuration.

5. Can Coulomb's Law be used for more than two charges?

Yes, Coulomb's Law can be used for more than two charges. The force between two charges can be calculated using the equation F = k*q1*q2/r^2, where q1 and q2 are the charges and r is the distance between them. To calculate the force between more than two charges, the forces between each pair of charges must be calculated and then added together using vector addition.

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