Question

A -60 nC charged particle is in a uniform electric field E = (30 v/m, east). an external force moves the particle 2.0 m north, then 6.0m east, then 4.0 m south, and finally 5.0 m west. the particle begins and ends its motion with zero velocity.

a) How much work is done on it by the external force?
b) What is the potential difference between the particle's final and initial positions?

Answer 1

a) The work done on the particle by the external force is zero since the particle begins and ends its motion with zero velocity.

b) The potential difference between the particle's final and initial positions can be found using the formula:

ΔV = -E•Δr

where E is the electric field and Δr is the displacement vector. The displacement vector can be found by summing the individual displacement vectors:

Δr = 2.0 m north + 6.0 m east + (-4.0 m south) + (-5.0 m west) Δr = -1.0 m east

Substituting into the formula gives:

ΔV = -(30 V/m) • (-1.0 m) ΔV = 30 V

Therefore, the potential difference between the particle's final and initial positions is 30 volts