A particle with charge 8.00×10 19 C is placed on the x axis in a region where th

Question

A particle with charge 8.00×1019 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 pointb along the x axis, increasing its kinetic energy by 1.60×1018 J . In what direction and through what potential difference VbVa does the particle move?

The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, increasing its kinetic energy by 1.60×1018  . In what direction and through what potential difference  does the particle move?

The particle moves to the right through a potential difference of VbVa= -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?

If the particle moves from point b to point c in the y direction, what is the change in its potential energy, ?

The particle moves to the left through a potential difference of VbVa= 2.00 V . The particle moves to the left through a potential difference of VbVa= -2.00 V The particle moves to the right through a potential difference of VbVa= 2.00 V . The particle moves to the right through a potential difference of VbVa= -2.00 V . The particle moves to the left through a potential difference of VbVa= 20.0 V .

The particle moves to the right through a potential difference of VbVa= -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?

If the particle moves from point b to point c in the y direction, what is the change in its potential energy, ?

+ 1.60×1018 J 1.60×1018 J 0

Explanation / Answer

part a ) using energy conservation

q(dV) = -dK
q(Vb - Va) = -dK
q(Va - Vb) = dK
(8x10^-19 C)(Va - Vb) = 1.6x10^-18 J
Va - Vb = 2 .00 = Vb - Va = -2.00 V

left side

The particle moves to the left through a potential difference of VbVa= -2.00 V

part b )

zero

because in y direction there is no change

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