A positive particle of charge 1.0 C accelerates in a uniform electric field of 100 V/m. The particle started from rest on an equipotential plane of 50 V. After t = 0.0002 seconds, the particle is on an equipotential plane of V = 10 volts. Determine the distance travelled by the particle.

Explanation

The work done in moving a charge in an equipotential surface is given by `W = -qΔV`
Substituting the values, we get
`W = (-1.0, C) (10V – 50V) = 40 J`
We know that, the work done in moving a charge in an electric field:
`W = qEd`
`40 = (1.0) (100)d`
`d = 0.4m`