Coulomb's law — A negative charge balanced between 3 positive charges

AI Thread Summary
The discussion centers on understanding why the force F(3,Q) is expressed as 3kQ/r^2 in the context of three equal positive charges arranged in an equilateral triangle with a negative charge at the center. The confusion arises from calculating the force between one charge (q3) and the central charge (Q), leading to the question of why a factor of three is included. Clarifications indicate that the three refers to the contributions from all three positive charges acting on the negative charge at the center. Additionally, there are discussions about the correct interpretation of distances involved and the significance of using consistent units in calculations. The key takeaway is that the factor of three accounts for the cumulative effect of all three positive charges on the negative charge.
yesmale4
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Homework Statement
Three equal positive charges, q = 8.5 μC each, are arranged at the vertices of equilateral triangle (see figure below). A negative charge, Q, is placed at the center of the triangle such that all four charges are in equilibrium.
Relevant Equations
Ef=0
f=kq1q2/r^2
hello i would like to understand something, i found the right answer but there is still something i don't understand.
here is the figure
a.png


and here is my correct solution

aa.jpg

aaa.jpg


what i don't understand is why F(3,Q) is 3kQ/r^2
i mean why is the 3? i only calculat the force between q3 and Q so why the 3 before?
 
Last edited:
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You mean Coulomb's law. Not "colon law".
 
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Steve4Physics said:
You mean Coulomb's law. Not "colon law".
fixed thank you
 
yesmale4 said:
Homework Statement:: Three equal positive charges, q = 8.5 μC each, are arranged at the vertices of equilateral triangle (see figure below). A negative charge, Q, is placed at the center of the triangle such that all four charges are in equilibrium.
Relevant Equations:: Ef=0
f=kq1q2/r^2

hello i would like to understand something, i found the right answer but there is still something i don't understand.
here is the figure
View attachment 316259

and here is my correct solution

View attachment 316260
View attachment 316261

what i don't understand is why F(3,Q) is 3kQ/r^2
i mean why is the 3? i only calculat the force between q3 and Q so why the 3 before?
What does r represent in your solution? The distance from where to where?
 
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yesmale4 said:
i found the right answer
What is the question :smile: ? (i can guess...)

I suppose
1666953133805.png
means ##\sqrt{3}\over 2## ?

You may want to learn some ##\LaTeX## (button at lower left of edit window). Or write and draw more unambiguously (the 60##^\circ## at the lower left looks more like 45##^\circ##, smack between 30##^\circ## and 60##^\circ##...)​

You seem to think that the sides of the triangle are ##r##. Fine, but then the distance ##qQ## is not ##r## !

##\ ##
 
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EDIT. Apologies @yesmale4. I misread your working. So this post is wrong and I've struck it through.

By the way @yesmale4, you have the incorrect unit in your answer. So your answer is wrong by a factor of one million!

Also, the value of q in the question is given to 2 significant figures. For consistency, your calculated value for Q should also have 2 significant figures.
 
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Steve4Physics said:
Also, the value of q in the question is given to 2 significant figures. For consistency, your calculated value for Q should also have 2 significant figures.
I don't see a calculated value. All I see is ##Q=\dfrac{q}{\sqrt{3}}## C. The issue is that there should be no units of Coulombs in the expression since it is a relation between magnitudes of charges.
 
kuruman said:
I don't see a calculated value. All I see is ##Q=\dfrac{q}{\sqrt{3}}## C. The issue is that there should be no units of Coulombs in the expression since it is a relation between magnitudes of charges.
Whoops. In haste I misread ##\frac q {\sqrt 3}## as ##\frac 9 {\sqrt 3}## (though that would have been an incorrect answer in any case).
 
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yesmale4 said:
Homework Statement:: Three equal positive charges, q = 8.5 μC each, are arranged at the vertices of equilateral triangle (see figure below). A negative charge, Q, is placed at the center of the triangle such that all four charges are in equilibrium.
Relevant Equations:: Ef=0
f=kq1q2/r^2

hello i would like to understand something, i found the right answer but there is still something i don't understand.
here is the figure
View attachment 316259

and here is my correct solution

[ ATTACH type="full" alt="aa.jpg"]316260[/ATTACH]
[ ATTACH type="full" alt="aaa.jpg"]316261[/ATTACH]

what I don't understand is why F(3,Q) is 3kQ/r^2
i mean why is the 3? i only calculated the force between q3 and Q so why the 3 before?
This seems rather strange. It's your solution, but you do not know where the 3 comes from.

Notice that the distance from q1 to q2 , etc. is not the same as the distance from each q to the central Q.

If you let R be the distance from each q to the charge Q in the center you should find that, ##\dfrac{r/2}{R} =\cos(30^\circ)## .
 
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SammyS said:
It's your solution
Maybe not. Could be "found" as in "… in the back of the book".
SammyS said:
Notice that the distance from q1 to q2 , etc. is not the same as the distance from each q to the central Q.
… and this is where the 3 comes from. ##\frac 1{(\frac r{\sqrt 3})^2}=\frac 3{r^2}##
 
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