A particle has a charge of +2.2 μC and moves from point A to point B, a distance of 0.16 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 = +7.8 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
4.8
To find the magnitude of the electric force that acts on the particle, we can use the formula:
Electric Force (F) = Change in Electric Potential Energy (ΔEPE) / Distance (d)
Given:
Charge of the particle (q) = +2.2 μC = 2.2 x 10^-6 C
Change in Electric Potential Energy (ΔEPE) = +7.8 x 10^-4 J
Distance (d) = 0.16 m
Substituting the given values into the formula:
F = (7.8 x 10^-4 J) / (0.16 m)
F = 4.875 x 10^-3 N
Therefore, the magnitude of the electric force that acts on the particle is 4.875 x 10^-3 N.
To find the magnitude of the electric field that the particle experiences, we can use the formula:
Electric Field (E) = Electric Force (F) / Charge (q)
Given:
Electric Force (F) = 4.875 x 10^-3 N
Charge of the particle (q) = +2.2 μC = 2.2 x 10^-6 C
Substituting the given values into the formula:
E = (4.875 x 10^-3 N) / (2.2 x 10^-6 C)
E = 2213.64 N/C
Therefore, the magnitude of the electric field that the particle experiences is 2213.64 N/C.