Question

a) A particle with a charge of -4.0 μC and a mass of 4.9 x 10-6...

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 VB - VA between A and B? If VB is greater than VA, then give the answer as a positive number. If VB is less than VA, 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 EPEA - EPEB = +7.5 x 10-4 J. (a) Find the magnitude of the electric force that acts on the particle. (b) Find the magnitude of the electric field that the particle experiences.

Thank you!

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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