Question

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?

Answer 1

Answer:

14 J

Explanation:

Let the angle of charge be θ

Then the distance from the x-axis = 3 m

The angle is given as $tan (\theta) = \frac{3}{3}\\ \theta = tan^{-1} (1)\\ = 45$

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:

$F = \frac{kq_{1}q_{2} }{r^{2} }$

The distance between the points will be $\sqrt{(3)^{2} +(3)^{2} }$

= 3$\sqrt{2}$