Question

1. A charge of -80 µC is placed on the x axis at x = 0. A second charge of +50 µC is placed on the x axis at x = 0.5 m. What is the magnitude of the electrostatic force on a third charge of 4.0 µC placed on the x axis at x = 0.3 m?

Answer 1

Answer:

77 N

Explanation:

1st charge ( q1 ) = -80µC placed on x-axis at x = 0

second charge( q2) +50µC placed at x = 0.5 m

3rd charge(q3) 4.0µC placed at x = 0.3 m

Determine the magnitude of the electrostatic force on the third charge

calculate distance btw : 1st charge and 3rd charge = 0.3 m ( r1 )

distance btw : 2nd charge and 3rd charge = 0.5 - 0.3 = 0.2 m ( r2 )

hence the magnitude of electrostatic force on third charge

F = [  Kq1q3 / ( r1 )^2 ]  +  [ Kq2q3 / ( r2)^2 ]  ------ ( 1 )

where K = 9 * 10^9  N.m^2/c^2

Insert given values into equation 1

∴ F = 77.0N