Asked by sara
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?
Answers
Answered by
bobpursley
Look at the top. distance to each charge is .3 m. I assume you want the magnitude of the force, not direction.
I will do A.
ForceonA=forcefromB+force from C.
Draw the duagran, Label the forces B, and C. Move force C to the end of A (as in adding verctors, the resultant is R
Law of cosines:
R^2=B^2+C^2-2BCcostheta.
with some figuring, you see the angle between B, C is 120. So the internal angle in the parallelogram is then 60 deg.
So what is B? k4mC*5mC/.3^2
What is C? k4mC*6mC/.3^2
There you go, solve it.
I will do A.
ForceonA=forcefromB+force from C.
Draw the duagran, Label the forces B, and C. Move force C to the end of A (as in adding verctors, the resultant is R
Law of cosines:
R^2=B^2+C^2-2BCcostheta.
with some figuring, you see the angle between B, C is 120. So the internal angle in the parallelogram is then 60 deg.
So what is B? k4mC*5mC/.3^2
What is C? k4mC*6mC/.3^2
There you go, solve it.
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