Question

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 traveled by the particle.

Answer 1

Applying the law of conservation of energy, we get:

$$\Delta KE=KE_f-KE_i=q\Delta V.$$

From the other hand, from the work-kinetic energy theorem, we get:

$$\Delta KE=KE_f-KE_i=-W=-Fd=-qEd.$$

Equating this two formulas, we get:

$$q\Delta V=-qEd,$$ $$d=-\dfrac{\Delta V}{E}=-\dfrac{10\ V-50\ V}{100\ \dfrac{V}{m}}=0.4\ m.$$

Answer:

$d=0.4\ m.$