A particle with charge 674 times 106 coulombs is released at

A particle with charge -6.74 times 10^-6 coulombs is released at rest in a region of constant, uniform electric field. Assume that gravitational effects are negligible. At a time 3.36 seconds after it is released, the particle has a kinetic energy of 6.60 times 10^-9 joules. How long after the particle is released has it travelled through a potential difference of 0.601 volts?

Solution

kinetic energy = K = qV

Thus the potential diiference:

      V = K/q = (6.6x10^-9)/(6.74x10^-6) = 9.792e-4 V

The distance traveled is proportional to t² .

time taken to travel across 1.02 mV = 3.86 sec.

t² / 0.601 =3.36²/9.792e-4

t² = 6928.97489

   t = 83.24 s