Posted by **matin** on Tuesday, May 15, 2012 at 7:34pm.

Three charges sit on the vertices of an equilateral triangle, the side of which are 30.0 cm long. If the charges are A = +4.0 C, B = +5.0 C, and C = +6.0 C (clockwise from the top vertex), find the force on each charge.

- physics -
**Elena**, Wednesday, May 16, 2012 at 7:33am
Charge A(q1)

F12 = k•q1•q2/a².

F13 = k•q1•q3/a².

F(A) = sqrt(F12² + F13² - 2•F12•F13•cos 120º).

Charge B(q2)

F21 = k•q1•q2/a².

F23 = k•q2•q3/a².

F(B) = sqrt(F21²+F23²-2•F21•F23•cos120º).

Charge C (g3),

F31 = k•q1•q3/a².

F32 = k•q2•q3/a².

F(C) = sqrt(F31² + F32² - 2•F31•F32•cos 120º).

- physics -
**sara**, Friday, August 17, 2012 at 10:45pm
How did you get F12 and F13 ??

- physics -
**Elena**, Sunday, August 19, 2012 at 2:49pm
physics - sara, Friday, August 17, 2012 at 10:45pm

How did you get F12 and F13 ??

F12 = k•q1•q2/a²=9•10^9•4•10^-6•5•10^-6/0.09=2 N

F13 = k•q1•q3/a² = 9•10^9•4•10^-6•6•10^-6/0.09=2.4 N

F(A) = sqrt(F12² + F13² - 2•F12•F13•cos 120º)=

=sqrt[4+5.76-2•2•2.4(-0.5)] =3.82 N

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