Question

A particle with a charge of +5.90 nC is in a uniform electric field E directed to the left.It is released from rest and moves to the left; after it has moved 60.0mm,it's kinetic energy is found to be +3.50 × 10-⁶.What is the potential difference of the starting point with respect to the end? What is the magnitude of electric field?

Answer 1

Answer

Potential difference is

$V=\frac{KE}{q}\\=\frac{3.50 × 10^{-9}.}{5.9*10^{-6}}\\=0.60*10^{-9}Volts$

Electric field

$E=\frac{V}{d}\\=\frac{0.0006}{0.060}\\=0.01V/m$