A particle has a charge of +1.5 A particle has a charge of +1.5 mu C and moves f

Question

A particle has a charge of +1.5 A particle has a charge of +1.5 mu C and moves from point A to point B, a distance of 0.25 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 = +9.00 times 10-4 J. (a) Find the magnitude and direction of the electric force that acts on the particle. (b) Find the magnitude and direction of the electric field that the particle experiences.

Explanation / Answer

a) Work is

W = F.d

Work is also equal to the difference of potential energy

EPEB - EPEA = F.d = -9x10^-4J

F = -9x10^-4J/0.25m = -3.6x10^-3N

for direction of force, consider that the particle lost potential energy, so it gained kinetic energy and thus accelerated. Since the particle is positive then the force is from point A toward point B.

b) The field is given by equation

F = qE

E = F/q = 3.6x10^-3N/1.5x10^-6C = 2400 V

The field is from A to B.

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