F = k Q1 Q2/d^2
I am ignoring powers of ten and k
distance of Q1 from Q2 = 5 (hypotenuse)
cos angle = 3/5
sin angle = 4/5
in x direction:
2.3 * 2 * (3/5)/25 + 2.3 * 3.7 /25
note that 25 is cm^2, multiply by 10^-4
for m^2
in y direction:
2.3 * 2 * (4/5)/25
magnitude = (Fx^2+Fy^2)^.5
tan angle = Fy/Fx
Three point charges are placed on the x-y plane as follows: Q1=-1.2nC at ( 3,4)cm, Q2=2.3nC at (0,0)cm and Q3=-3.7nC at (5,0)cm. What is the net force exerted by the two point charges on Q2?
2 answers
8.75x10^-5,65.3 degrees above +x axis