posted by sara on .
3 charges sit on vertical equilateral triangle the side of each is 30.0 cm. if the triangle are A= +4.0 uC B=+5.0 uC and C=+6.0 uC (clockwise from top) what is the force on each charge?
Can someone please explain me in detail i am really confuse :(((
OK. Lets walk through one. Consider the force on charge A. It has two components, which are vectors.
The force on A from C is kQaQc/distance^2, along the direction from C to A.
The force on A from B is
kQbQa/distance^2, along the direction from B to A
so you can calculate those two forces. Howver, they are not along the same direction, so have to be added as vectors, getting a resultant force.
So add these two forces by any of the following methods...
2. Using the law of cosines (magnitude only) then the law of sines to figure the angle if you need it. Draw the figure carefully, all you need fodr these will be in the figure ...
3. By breaking each of the forces into a vertical, and hoizontal component, then adding.
Make certain you draw a figure, you have to have this in mind when computing.
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
what is 5.76??