Force between 2 point charges in VECTOR format (x i + y j + z k)

In summary: Coulomb's law:$$\vec F = k\frac{q_1q_2}{r^2} \hat r$$where ##\vec F## is the force of ##q_2## on ##q_1## and ##\hat r## is a unit vector from ##q_2## to ##q_1##. And ##r## is the distance between the two charges.How can I calculate a unit vector for force without its components though? Or am I supposed to calculate a unit vector for the distance and then multiply by that?
  • #1
nataelp
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Homework Statement
Point charges
q1 = 4.0 µC and q2 = 4.0 µCare located at r1 = (4.0î − 5.0ĵ + 2.0k) mand r2 = (9.0î + 5.0ĵ − 6.0k)m What is the force (in N) of q2 on q1? (Express your answer in vector form.)
Relevant Equations
F = k*q1q1/r^2
I tried using the distance between r2 and r1 and plugging them into the equation for i, j, k. >>
So for the force in the x direction it was k*(4E-6*4E-6)/(4-9)^2. The answer I got was wrong according to webassign. Can someone please tell me what I am missing?
 
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  • #2
And to be clear it was not the direction of the vector that was wrong, I can see that the magnitude is completely off, but the direction was correct.
 
  • #3
nataelp said:
And to be clear it was not the direction of the vector that was wrong, I can see that the magnitude is completely off, but the direction was correct.
What magnitude did you get?
 
  • #4
I got (-0.0058, -0.00144, 0.00225)
 
  • #5
nataelp said:
I got (-0.0058, -0.00144, 0.00225)
What's the vector from ##q_2## to ##q_1##?
 
  • #6
It should be 13.72 using the Pythagorean theorem.
 
  • #7
nataelp said:
It should be 13.72 using the Pythagorean theorem.
That is the magnitude of that vector
 
  • #8
malawi_glenn said:
That is the magnitude of that vector
If you mean in the vector form, it would be <-5^2, -10^2, 8^2>. And those are the values for r I used to find the force on each axis. The answer WebAssign gave me was this:
7e0a89c33665e8590c0152077175b1.gif

My problem is I don't know how to get to this answer.
 
  • #9
nataelp said:
If you mean in the vector form, it would be <-5^2, -10^2, 8^2>. And those are the values for r I used to find the force on each axis. The answer WebAssign gave me was this:
View attachment 320443
My problem is I don't know how to get to this answer.
I get that answer. How did you try to solve the problem?
 
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  • #10
So I did this:
For the force along each axis, I used F= k*q1*q2/r^2. I had the charges, and k is a known value, so I just needed the distance r.
For r along the x-axis, for example, I used 4-9 as r because it should be the distance in the x-direction. So I solved it with that value and got my answer for force in the x direction. I repeated it for x, y, and z.
I'm not sure where I am going wrong. And I did change it from microcoloumbs to coloumbs.
 
  • #11
nataelp said:
So I did this:
For the force along each axis, I used F= k*q1*q2/r^2. I had the charges, and k is a known value, so I just needed the distance r.
For r along the x-axis, for example, I used 4-9 as r because it should be the distance in the x-direction. So I solved it with that value and got my answer for force in the x direction. I repeated it for x, y, and z.
I'm not sure where I am going wrong. And I did change it from microcoloumbs to coloumbs.
Ah, this is not right. Once you have the magnitude of the force, you just need to calculate a unit vector in the right direction.
 
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  • #12
So I need to find the magnitude of the vector from q2 to q1 (in m), use that as r, and then to find the force along the i j k directions, I need to calculate a unit vector along those directions? Then multiply by that unit vector, which is each component divided by the magnitude?
 
  • #13
nataelp said:
So I need to find the magnitude of the vector from q2 to q1 (in m), use that as r, and then to find the force along the i j k directions, I need to calculate a unit vector along those directions? Then multiply by that unit vector, which is each component divided by the magnitude?
Yes, Coulomb's law is:$$\vec F = k\frac{q_1q_2}{r^2} \hat r$$where ##\vec F## is the force of ##q_2## on ##q_1## and ##\hat r## is a unit vector from ##q_2## to ##q_1##. And ##r## is the distance between the two charges.
 
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  • #14
How can I calculate a unit vector for force without its components though? Or am I supposed to calculate a unit vector for the distance and then multiply by that? I'm honestly not sure what you mean.
 
  • #15
Should the unit vector be <-5, -10, 8>/13.75 (magnitude of distance vector)? Then I multiply "F = k*q1*q2/13.75^2" by that unit vector, and I have my answer?
 
  • #16
nataelp said:
For r along the x-axis, for example, I used 4-9 as r because it should be the distance in the x-direction. So I solved it with that value and got my answer for force in the x direction.
No, you don't divide the q1*q2 in each direction by the square of the distance in that direction. You have to divide each by the square of the total distance between the charges.
 
Last edited:
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  • #17
I have figured out the problem now. Thank you all for your help!
 
  • #18
nataelp said:
How can I calculate a unit vector for force without its components though? Or am I supposed to calculate a unit vector for the distance and then multiply by that? I'm honestly not sure what you mean.
I suggest you revise your knowledge of vectors.
 

FAQ: Force between 2 point charges in VECTOR format (x i + y j + z k)

1. How do you calculate the force between two point charges in vector form?

The force \(\mathbf{F}\) between two point charges \(q_1\) and \(q_2\) located at positions \(\mathbf{r}_1 = x_1 \mathbf{i} + y_1 \mathbf{j} + z_1 \mathbf{k}\) and \(\mathbf{r}_2 = x_2 \mathbf{i} + y_2 \mathbf{j} + z_2 \mathbf{k}\) is given by Coulomb's law in vector form: \(\mathbf{F} = k_e \frac{q_1 q_2}{|\mathbf{r_{12}}|^3} \mathbf{r_{12}}\), where \(\mathbf{r_{12}} = (x_2 - x_1) \mathbf{i} + (y_2 - y_1) \mathbf{j} + (z_2 - z_1) \mathbf{k}\) and \(|\mathbf{r_{12}}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\).

2. What is the significance of the unit vector in the force equation?

The unit vector \(\hat{\mathbf{r_{12}}}\) in the force equation points from one charge to the other and indicates the direction of the force. It is defined as \(\hat{\mathbf{r_{12}}} = \frac{\mathbf{r_{12}}}{|\mathbf{r_{12}}|}\). The force vector \(\mathbf{F}\) is obtained by multiplying the magnitude of the force by this unit vector, ensuring that the direction of the force is correctly represented.

3. How does the distance between the charges affect the force in vector form?

The distance between the charges affects the magnitude of the force inversely as the square of the distance. In vector form, this is incorporated by the term \(|\mathbf{r_{12}}|^3\) in the denominator of the force equation. As the distance \(|\mathbf{r_{12}}|\) increases, the magnitude of the force decreases rapidly.

4. Can the force between two point charges be zero? Under what conditions?

The force between two point charges can be zero if either one of the charges is zero (\(q_1 = 0\) or \(q_2 = 0\)), or if the charges are infinitely far apart, making the distance \(|\mathbf{r_{12}}|\) infinitely large, which would result in the force approaching zero.

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