Question

question: A particle with charge 8.00×10⁻¹⁹ C is placed on the x axis in a region where the electric potential due to other charges increases in the +x direction but does not change in the y or z direction.

part A: The particle, initially at rest, is acted upon only by the electric force and moves from point a to point balong the x axis, increasing its kinetic energy by 1.60×10⁻¹⁸ J . In what direction and through what potential difference Vb−Va does the particle move?

A. The particle moves to the left through a potential difference of Vb−Va= 2.00 V .

B. The particle moves to the left through a potential difference of Vb−Va= -2.00 V

C. The particle moves to the right through a potential difference of Vb−Va= 2.00 V .

D. The particle moves to the right through a potential difference of Vb−Va= -2.00 V .

E. The particle moves to the left through a potential difference of Vb−Va= 20.0 V .

F. The particle moves to the right through a potential difference of Vb−Va= -20.0 V .

Part B :If the particle moves from point b to point c in the y direction, what is the change in its potential energy, Uc−Ub?

+ 1.60×10⁻¹⁸ J

− 1.60×10⁻¹⁸ J

0

Answer 1

Ans)

Part A)
change in electrical potential energy = -(change in kinetic energy)
q(ΔV) = -ΔK
q(Vb-Va) = -ΔK
given charge q=8.00x10^-19c
(8.00x10^-19 C)(Vb - Va) = -(1.60x10^-18J)
Vb-Va = [ -(1.60x10^-18J)] /(8.00x10^-19 C)

Vb-Va=-2.00V
This is because if no forces other than the electric force acts on a positively charged particle, the particle

always moves towards a point a lower potential

so the particle moves to the left through a potential difference of Vb-Va=-2.00V

Therefore the answer is option b

Part B)

Every time a charged particle moves along a line of constant potential

its potential energy remains constant and the electric field does not work on the particle.

so no potential energy

that is Uc-Ub=0J