Coulomb's Law: Determining the velocity of a charge when far apart

[lilpoopoo]

Homework Statement

"Three identical charges, each with charge 'Q' and a mass 'm', are arranged on the corners of an equilateral triangle of side length 'L'. The spheres are released simultaneously. What is the speed of each charge when they are very far apart?

Homework Equations

F = [(kq₁q₂)/r²] * r^{^}
F = ma

The Attempt at a Solution

Since all of the charges are the same, and they are equidistant from one another, then the force on each should be identical.

I calculated the force on one charge (A) from the other two (B & C).

Therefore:

F_{on A} = (kQ²(root 3/4 L²))/L³

or (kQ²sin(60))/L²​

So for each individual charge, the force on A should be the same, as above, for B & C.

My issue is that I can't figure out how to calculate the velocity from this...

Or am I going completely in the wrong direction?

Edit: Forgive my equations, I haven't figured out how to make them nice. It's my first post.

 

[gneill]

Hi lilpoopoo. Welcome to Physics Forums.

Did you consider looking at the problem from a conservation of energy point of view?

 

[lilpoopoo]

Hi gneill. Thanks :)

Okay, so I did what you suggested, not sure if it's correct though.
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So we know the potential energy difference is:

V = (kq)/r​

And we know

PE = KE​

Therefore:

1/2 mv² + 1/2 mv² + 1/2 mv² = (kq)/r + (kq)/r + (kq)/r​

v = root [(2kQ)/(mr)]​

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I'm not sure about the 'r' part. Would that be the distance from the horizontal line drawn between the two charges at the base of the triangle up to the top charge at the apex, which is root (3/4 L²)?

 

[gneill]

1/2 mv² + 1/2 mv² + 1/2 mv² = (kq)/r + (kq)/r + (kq)/r​

v = root [(2kQ)/(mr)]​

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I'm not sure about the 'r' part. Would that be the distance from the horizontal line drawn between the two charges at the base of the triangle up to the top charge at the apex, which is root (3/4 L²)?

You're on the right track, but your electric potential calculation needs a tiny bit of massaging.

You're looking for the potential energy stored in the system when it's assembled in the triangle form. To calculate the potential energy you take the charges involved pairwise and calculate the individual potential energies, and sum them up. This appears to be what you were going for in your equation above. However, the potential energy for a pair of charges Q1 and Q2 separated by distance R (center to center distance) is:

$$PE = k\frac{Q1\;Q2}{R}$$

Note that the both Q values are required. Thanks to the particular triangular layout, the distance between all the charges is the same, L.

 

[lilpoopoo]

Okay, so doing the pairwise additions would yield:

(3/2) mv^2 = (kQ1Q2)/R + (kQ2Q3)/R + (kQ1Q3)/R

And since each charge is the same, +Q, and the charges are all 'L' distance apart:

(3/2) mv^2 = (kQ^2)/L + (kQ^2)/L + (kQ^2)/L

Therefore:

v = root [(2kQ^2)/(mL)]

:uhh:

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Edit: Also, is there a way to easily access our own posts? I just keep looking for it with search terms.

 

[gneill]

Edit: Also, is there a way to easily access our own posts? I just keep looking for it with search terms.

The primary forum page has the list of threads. If you've registered with a valid email address, then email notifications of additions to threads that you are subscribed to will be sent to it.

You can also select "My Pf" at the top of the page. If you don't see your thread i n the main section (perhaps it's no longer a "new" thread"), you can always select "List Subscriptions" under "My Control Panel". Your thread should show up there.

Personally, I tend to use the email notifications to find things. I just hit the link in the message and the browser opens to that message.

 

[lilpoopoo]

Oh, alright. Got it. Thanks :)

So was my solution correct?

 

[gneill]

Oh, alright. Got it. Thanks :)

So was my solution correct?

It looks okay to me. Why, aren't you happy with it?

 

[lilpoopoo]

Haha, no. Just unsure. Physics doesn't come easy for me. Thanks so much for your help.