Asked by matin

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.
Answered by Elena
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).
Answered by sara
How did you get F12 and F13 ??
Answered by Elena
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
Answered by lester
same question above, but calculate the magnitude of the net force that each charge experiences.
There are no AI answers yet. The ability to request AI answers is coming soon!