Identical 50 μC charges are fixed on an x axis at x = ±3.0 m. A particle of charge q = -15 μC is then released from rest at a point on the positive part of the y axis. Due to the symmetry of the situation, the particle moves along the y axis and has kinetic energy 1.2 J as it passes through the point x = 0, y = 4.0 m.
(a) What is the kinetic energy of the particle as it passes through the origin?
(b) At what negative value of y will the particle momentarily stop?

14 J

Explanation:

Let the angle of charge be θ

Then the distance from the x-axis = 3 m

The angle is given as

Similarly, the charge is on the left side, so the distance will be - 3 m

The angle therefore will be  = 45°

So we need to find the net force given by the law:

The distance between the points will be

= 3

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.

it becomes positively charged by charging it