[Electrical force] Electrical charge in equilateral triangle

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sea333
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Homework Statement
In each corner of equilateral triangle there is a charge with 5 nAs. What force works on the e3 charge (see image). Length of a side is 5 cm. e1 is charged with negative charge.
Relevant Equations
F = (1/(4*Pi*eps0))*(e*e)/r^2
Hand-drawn vector diagram of an equilateral triangle with charges e1, e2 and e3 at the vertices, 5 cm sides, 60° angles, with forces F1 and F2 and their resultant R drawn at the top charge.
Mathematica notebook cell with a comment reading Exercise 9 / NOT CORRECT, should be 9*10^-5, defining e = 5*10^-9, r = 0.05, eps0 = 8.85*10^-12, and force formulas Fe23 and Fe13 with their combined magnitude Fr, followed by outputs 0.000089918, -0.000089918 and 0.000127163.
 
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Not being a computer myself, I find that a bit hard to follow! Can you explain what you're doing?
 
F23 is a force on the 3rd charge from 2nd charge
F13 is a force on the 3rd charge from the 1st charge
Fr is a sum of both (Pythagorean theorem).
 
sea333 said:
F23 is a force on the 3rd charge from 2nd charge
F13 is a force on the 3rd charge from the 1st charge
Fr is a sum of both (Pythagorean theorem).
Doesn't Pythagoras only apply to right-angled triangles?
 
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PeroK said:
Doesn't Pythagoras only apply to right-angled triangles?
I will check if forces are right-angled
 
sea333 said:
I will check this
There is a generalisation called the "rule of cosines", which is very useful.
 
PeroK said:
There is a generalisation called the "rule of cosines", which is very useful.
I have used Cosines to calculate Fr and I still don't get the correct result:
Mathematica notebook cell with a comment reading Exercise 9 NOT CORRECT, should be 9*10^-5, defining e, r, epsilon-0, alpha, A = Cos[alpha Degree], Fe23, Fe13 and Fr = Fe13/Cos[alpha Degree], with output -0.000179836.

Handwritten force diagram on lined paper: charges q1, q2 and q3 at the corners of a triangle, with forces F12, F13 and resultant FR at q1, angles alpha = 60 and beta = 60 marked below.
 
Attachments
  • Mathematica notebook cell with commented code defining e, r, epsilon0, alpha, Fe23, Fe13 and Fr, followed by outputs 1/2, -0.000089918 and -0.000179836.
    Capture4.PNG
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That's not right! Did you look up the rule of cosines?
 
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Should I use the Law of cosines ? c^{2}=a^{2}+b^{2}-2ab*cos
Then I would need angle gamma?
I have updated the picture
Handwritten free-body diagram of three charges q1, q2 and q3 with force vectors F_{e13}, F_{e23} and F_R drawn at q1; angles labelled 60 and alpha, and below it alpha = 60, beta = 40.
 
I have used for gamma = 60 and got correct result but how do you know that gamma = 60 ?
I don't know why I thought that any of the angles is 90 degrees.
The correct result:
Mathematica notebook cell with a comment reading Exercise 9 NOT CORRECT, should be 9*10^-5, defining e = 5*10^-9, r = 0.05, epsilon0 = 8.85*10^-12, alpha = 60, y = 60, and computing the magnitude Fr from Fe13 and Fe23 with a cosine term, output 0.000089918, and EngineeringForm of Fe13*10^5 giving -8.9918.
 
If you want to use Pythagorean theorem or Cos/Sin function you need one angle to be 90 degrees.