Answer to Question #151614 in Physics for Muhammad hamza waseem

Question

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

Answer 1

1) $W=\Delta KE=-q\Delta \phi=-1\cdot(10-50)=40(J)$

$W=Fd=qEd$

$qEd=40\to d=40/(qE)=40/(1\cdot100)=0.4(m)$

or

2) $E=-\frac{\Delta\phi}{d}\to d=-\frac{\Delta\phi}{E}=-\frac{10-50}{100}=0.4(m)$