Question

# question: A particle with charge 8.00×10−19 C is placed on the x axis in a region...

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 VbVa 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, UcUb?

+ 1.60×10−18 J

− 1.60×10−18 J

0

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

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