A particle with charge 6.40 times10^-19 C is placed on the x axis in a region wh
ID: 1528008 • Letter: A
Question
A particle with charge 6.40 times10^-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. 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 times 10^-18 J. In what direction and through what potential difference V_b - V_a does the particle move? The particle moves to the left through a potential difference of V_b - V_a = 2.50 V. The particle moves to the left through a potential difference of V_b - V_a = -2.50 V The particle moves to the right through a potential difference of V_b - V_a = 2.50 V. The particle moves to the right through a potential difference of V_b - V_a = -2.50 V. The particle moves to the left through a potential difference of V_b - V_a = 25.0 V. The particle moves to the right through a potential difference of V_b - V_a = -25.0 V. If the particle moves from point b to point c in the y direction, what is the change in its potential energy, U-c-U_b? + 1.60 times 10^-18 J -1.60 times 10^-8 J 0Explanation / Answer
A) Because no forces other than the electric force act on the particle, the positively charged particle must move in the direction parallel to the electric field, and the field must do positive work on the particle. Recall that when the electric field does positive work on a charged particle, the potential energy of the particle decreases. Thus, the particle must move in the direction in which its potential energy decreases (which is consistent with the fact that the particle's kinetic energy increases as it moves from a to b). Moreover, from the definition of potential and the energy conservation equation, you can directly calculate the potential difference Va - Vb.
The electric potential Vat any point in an electric field is the electric potential energy U per unit charge associated with a test charge q’ at that point:
V = U/q'
Ub - Ua = -K = 1.6*10^-18 J
In general, if no forces other than the electric force act on a positively charged particle, the particle always moves toward a point at lower potential.
The particle moves to the left through a potential difference of Vb – Va = - 2.5 V
b) Recall that the electric potential increases in the +x direction but does not change in the y or z direction.
Every time a charged particle moves along a line of constant potential, its potential energy remains constant and the electric field does no work on the particle.
ZERO
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