Three equal positive charges of 2MicroC are arranged in the following way: charg
ID: 1964562 • Letter: T
Question
Three equal positive charges of 2MicroC are arranged in the following way: charge 1 is over Y, 2 cm away from the origin. Charge 2 is over the X axis 2cm from the origin and charge 3 is at the origin. Determine magnitude and direction of net force over charge 3. (Cartesian - 90 deg)2cm = .02 m
So I did the force diagram and I believe both F31 and F32 would be negative. Then I tried to find the forces by using this formula:
F = K |q3| |q1| / (r)^2
And
F = K |q3| |q1| / (r) ^2
However (maybe bcuz they're all 2 microcoulomb?) they turn out as 89.9 N which I think might be too big a number and by the time I try to find Fr I know the number is too ridiculously huge to be right. Help?
Explanation / Answer
F31=K*|q1|*|q3|/(2^2*10^-4) (-j) N where j is the unit vector along y- direction and this force acts on the -ve y-direction because q1 and q3 are both +ve and there is force of repulsion between the two charges.
F31=(9*10^9*(2*10^-6)^2)/4*10 (-j)N= -90(j)N
similarly, F32=(9*10^9*(2*10^-6)^2)/4(-i)N= 90(-i)N
so Fnet=F31+F32=90(-i-j)N
the answers are right. the distance of separation is extremely small (in cms) which causes the forces to come out relatively large.