Question

a) A particle with a charge of -4.0 μC and a mass of 4.9 x
10^{-6} kg is released from rest at point A and accelerates
toward point B, arriving there with a speed of 83 m/s. The only
force acting on the particle is the electric force. What is the
potential difference *V*_{B} -
*V*_{A} between A and B? If *V*_{B}
is greater than *V*_{A}, then give the answer as a
positive number. If *V*_{B} is less than
*V*_{A}, then give the answer as a negative
number.

b) A particle has a charge of +2.6 μC and moves from point
*A* to point *B*, a distance of 0.10 m. The particle
experiences a constant electric force, and its motion is along the
line of action of the force. The difference between the particle's
electric potential energy at *A* and *B* is
EPE* _{A}* - EPE

Thank you!

Answer #1

a)

Ans:

Solution:

Given,

mass of the particle

charge of the particle

The particle is at rest at point A. It has a velocity of on reaching point B. The work done by the electric field is stored as the kinetic energy of the particle.

Thus, workdone by the field is

The workdone by the particle is negative because particle is pulled by the field. So,

The Potential difference between A and B () is given by,

The point B is at higher potential than point A.

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