Three charged objects are located t the vertices of a right triangle. Charge A (5*10^-6) has Cartesian coordinates (0,4; Charge B (-5.0*10^-6) is at the origin; charge c (+4.0*10^-6) has coordinates (5,0), where the coordinates are in meters. What is the net force on each charge.

Question

Ans: I found the charge on B by finding the force F(a to b) and similarly the hypotenuse force and using the cosine rule and sine rule i found the direction and the net charge but when i try the same on point charges A and C i do not get the answer. Where am i going wrong?

Answer 1

Pick a charge: Say B. Find Fab and Fac. Note: Fba=-Fab and Fca=-Fac

so you have now only one other force, Fbc. Fcb=-Fbc

So you have had to calculate only three pairs of charges.

The for magnitude, consider NOT using the trig resolutions, do this

Fa^2= Fab^2+Fac^2
Fb^2=Fba^2+Fbc^2
Fc^2=Fca^2+Fcb^2

That is it.

Answer 2

The desired answers according to the textbook seem to be Fa=1.2 x 10^-2 N [W73 S]; Fa=1.6 x 10^-2 N [W63 S]; Fa=4.6 x 10^-3 N [W31 S];

Thing is, using those formulas I don't get the correct answers at all. From the triangle, we have a side with 4m, a side with 5m, and (Found using the Pythagorean theorem) a side with 6.4m. When using the formulas given I get Fa= 1.4x10^-2 N and the 73 degrees; then for the B i get the correct answers. For Fc I get 8.4x10^-3 and a 31 degree angle.... help