Electrostatics: 3 Charges In An Equilateral Triangle

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SUMMARY

The discussion revolves around calculating the speed of a charge in an electrostatic system involving three point charges arranged in an equilateral triangle. The charges, each with a value of Q = 10 µC and mass m = 25 g, are positioned at the corners of a triangle with a side length of 4 cm. The correct final speed of the charge when released is determined to be 56 m/s, achieved through the kinetic energy equation. The participants highlight the importance of using the correct electromagnetic force equations rather than breaking down energy into components, emphasizing that energy is a scalar quantity.

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hellojojo
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Hello, so this was a past assignment question that I attempted. I got the right answer but in an incorrect way (I guess my thinking was not based on the concept?)

1. Homework Statement

Three point charges of charge Q = 10 https://maple-ta.uwaterloo.ca/mapleta/tmp/am/dh/gm/finnniaffjjcdncgjfhcnmebcf.gif and mass m = 25 g are located at the corners of an equilateral triangle, with side length 4 cm. One of the charges is let go. Gravity should be ignored.

What will its speed be when it is at infinity?

The answer was 56 m/s

3. The Attempt at a Solution

My solution was:

E1 =E2= Kq/r = ((8.99x10^9)(1x10^-5)(1x10^-5))/(0.04) = 22.475J

Then the x component was:
E1(x)= 22.475J*sin 30=11.2375
E2 (x)=22.475J*-sin 30=-11.2375

E1(y)=22.475J*cos30=19.46
E2(y)=22.475J*cos30=19.46

THen using Ke=1/2mv^2
v=56m/s

I used components instead of the electromagnetic force equation between two charges.
This is wrong or an incorrect way of thinking and I should've used the equation:

(kQQ/r-3KQQ/r)

Can someone explain to me what is going on and how this equation explains what is happening between the three forces.

Thanks
 
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Energy is a scalar, it does not have components in different directions.
Setting kinetic energy equal to the sum of E1 and E2 works.
hellojojo said:
(kQQ/r-3KQQ/r)
I don't know where that comes from and what exactly it is supposed to mean.
 
Hello hellojojo. Welcome to PF!

hellojojo said:
My solution was:

E1 =E2= Kq/r = ((8.99x10^9)(1x10^-5)(1x10^-5))/(0.04) = 22.475J

You haven't told us what E1 and E2 represent. I'm guessing electric potential energy of two out of the three charges? What's the meaning of E1 as opposed to E2?
Did you mistype the formula for E1? You have only one charge in the expression Kq/r, but you calculated with two charges.

Then the x component was:
E1(x)= 22.475J*sin 30=11.2375
E2 (x)=22.475J*-sin 30=-11.2375

E1(y)=22.475J*cos30=19.46
E2(y)=22.475J*cos30=19.46

Is potential energy a vector quantity or is it a scalar quantity? Does it make sense to take components of a scalar quantity?

This is wrong or an incorrect way of thinking and I should've used the equation:

(kQQ/r-3KQQ/r)

Can someone explain to me what is going on and how this equation explains what is happening between the three forces.

Thanks

As one of the charges flies away, the potential energy of the system changes. What is an expression for the initial PE of the system of three charges? What is an expression for the final PE of the system of three charges?
 

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