Acceleration of a Charged Particle in 3D Vector Notation

[cactusblanket]

Homework Statement

Hi there!
Here is a problem from our 1st year course. We have covered the basics on charges and Coulomb's Law. However our prof said he designed the following question to be "deliberately obscure"!

Two charges of 4.00μC are fixed in space, Q1 at (0,0,0) and Q2 at (-1,4,4) with distances accurate to 3 significant figures. A third particle of mass 6.00 g and charge Q3=-3.50μC is placed at (2,3,2) and released. Express its acceleration at the moment of its release using vector notation.

Homework Equations

F=kQ₁Q₂ / r²₂₁

The Attempt at a Solution

My level of understanding is very basic.
A hint from our professor is:
1. find the F on the 3rd particle from charge one,
2. then from charge two.
3. Add them vectorially
4. divide by the mass.

So I was able to do the following:
In the denominator I used
√((0+2)+(0+3)+(0+2))²
And I got these answers
1. F = -7.40 x 10^-3 (-2i, -3j, -2k)N
2. F = -1.46 x 10^-3 (-1i, 1j, 2k)N
3. I added them together to get -8.86 x 10^-3 (-0.593i, -0.620j, -0.269k)N
4. I used F=ma to get final answer of:

-1.477 (-0.593i, -0.620j, -0.269k)m/s²

I would like to know if I am way off track here or possibly in the ball park.
Any help would be great, thank you!

 

[vela]

Homework Statement

Hi there!
Here is a problem from our 1st year course. We have covered the basics on charges and Coulomb's Law. However our prof said he designed the following question to be "deliberately obscure"!

Two charges of 4.00μC are fixed in space, Q1 at (0,0,0) and Q2 at (-1,4,4) with distances accurate to 3 significant figures. A third particle of mass 6.00 g and charge Q3=-3.50μC is placed at (2,3,2) and released. Express its acceleration at the moment of its release using vector notation.

Homework Equations

F=kQ₁Q₂ / r²₂₁

The Attempt at a Solution

My level of understanding is very basic.
A hint from our professor is:
1. find the F on the 3rd particle from charge one,
2. then from charge two.
3. Add them vectorially
4. divide by the mass.

So I was able to do the following:
In the denominator I used
√((0+2)+(0+3)+(0+2))²
And I got these answers
1. F = -7.40 x 10^-3 (-2i, -3j, -2k)N
2. F = -1.46 x 10^-3 (-1i, 1j, 2k)N
3. I added them together to get -8.86 x 10^-3 (-0.593i, -0.620j, -0.269k)N
4. I used F=ma to get final answer of:

-1.477 (-0.593i, -0.620j, -0.269k)m/s²

I would like to know if I am way off track here or possibly in the ball park.
Any help would be great, thank you!

You've made some basic errors.

The $r$ in the denominator is the distance between the two charges. The distance between Q3 and Q1 is different than the distance between Q2 and Q3, right? So you can't use the same denominator for calculating the two forces.

Also your expression for the distance should have minus signs in there somewhere. You shouldn't be adding the coordinates.

Why don't you start by figuring out what $\vec{r}_{31}$ and $\vec{r}_{32}$ are? Then use those results to calculate $\vec{F}_{31}$ and $\vec{F}_{32}$.

 

[cactusblanket]

Hi,
Actually, I did use different denominators.
Your advice to try to figure out what r⃗ 31 and r⃗ 32 is spot on.

Indeed that is my first road block (I wasn't able to clarify that in my original post.)
I don't know how to calculate that.

Can you please recommend how I go about solving for r⃗ 31 and r⃗ 32?
Thank you!

 

[vela]

How do you calculate the vector connecting the points (0, 0, 0) and (2, 3, 2), with the tail at (0, 0, 0) and the tip at (2, 3, 2)? Similarly, how do you calculate the vector connecting the points (-1, 4, 4) and (2, 3, 2), with the tail at (-1, 4, 4) and the tip at (2, 3, 2)? Note this is the stuff you learned way back when you first learned about displacement and vectors, so if you don't know off the top of your head how to do this, you should go back and review this material in your textbook.

 

[cactusblanket]

Actually, I thought that I added the vectors correctly:
The vector points (0, 0, 0) and (2, 3, 2) give (2, 3, 2).
The vector points (-1, 4, 4) and (2, 3, 2) give (1, 7, 6).

To find their magnitude I performed √(2² + 3²+ 2²)² and √(1² + 7²+ 7²)².

I used those as my denominators and I get
F⃗ 31=-7.40 x 10-3 N
F⃗ 32 = -1.46 x 10-3 N

This still gives me a final answer of -1.477 m/s².
Is my vectorial addition still off?

Just so you know, I didn't study physics in high school (30 years ago) and only recently passed an introductory physics course, so I acknowledge the gaps in my knowledge.
I hope you can still give me some pointers as I work through this.

 

[vela]

Say you were looking for the vector connecting the points (1, 0) and (1, 1). If you were to add those, you get (2,1). Try drawing a diagram of the situation. Is (2, 1) the right answer?

 

[cactusblanket]

I drew it tip to tail. Yes, (2, 1) is correct.

 

[vela]

No, it's not. You're not answering the right question.

 

[cactusblanket]

Sorry, but that is how I learned to add vectors.
I cannot seem to understand the right question.
What is the right question?

 

[vela]

Well, you are adding them correctly. One question is why are you adding them? Reread post #6. Does the vector (2,1) answer the question I asked?

 

[cactusblanket]

OK, so the answer (2, 1) gives the direction only.
Magnitude is √5.

It is 11:15 pm, time to put the kids to sleep and go to bed here in Nova Scotia.
Looking forward to replying to your next prompt and resuming the thread if you are willing and able.
Hope I don't appear TOO dumb!
Thanks for your patience, good night for now.

 

[cactusblanket]

Hi Vela, I see you are online now. Can you continue this thread?

 

[cactusblanket]

Okay, I've been working on it and here is what I have been able to do:

Step 1
F⃗₁₃ = (-35.9i, -53.9j, -35.9k) N

Step 2
F⃗ ₂₃ = (-72.1i, 24.0j, 48.1k) N

Step 3
Add vectorially= (-108.0i, -29.8j, 12.1k) N

Step 4
F⃗=ma yields

(-17,999i, -4974j, 2024k) m/s²