A particle has a charge of +1.5 µC and moves from point A to point B, a distance of 0.19 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 = +5.50 10-4 J.

(a) Find the magnitude and direction of the electric force that acts on the particle.
magnitude N
direction

(b) Find the magnitude and direction of the electric field that the particle experiences.
magnitude N/C
direction

To solve this problem, we will use the formula for electric potential energy:

EPE = q * ΔV,

where EPE represents the electric potential energy, q is the charge of the particle, and ΔV is the change in electric potential between point A and point B.

Given:
Charge of the particle (q) = +1.5 µC = 1.5 * 10^-6 C
Change in electric potential (ΔV) = +5.50 * 10^-4 J

(a) To find the magnitude of the electric force (F), we can use the formula:

F = -ΔEPE/Δx,

where ΔEPE is the change in electric potential energy and Δx is the distance traveled by the particle.

Given:
Change in electric potential energy (ΔEPE) = +5.50 * 10^-4 J
Distance traveled (Δx) = 0.19 m

Substituting the values into the formula, we have:

F = - (+5.50 * 10^-4 J) / (0.19 m)

Calculating this, we find:

F ≈ -2.89 N

Since the problem states that the motion of the particle is along the line of action of the force, the direction of the electric force is negative, indicating that it acts in the opposite direction of the particle's motion.

Therefore, the magnitude of the electric force is approximately 2.89 N, and its direction is opposite to the particle's motion.

(b) To find the magnitude and direction of the electric field (E), we can use the formula:

E = F / q,

where F is the electric force and q is the charge of the particle.

Given:
Electric force (F) = -2.89 N
Charge of the particle (q) = +1.5 µC = 1.5 * 10^-6 C

Substituting the values into the formula, we have:

E = (-2.89 N) / (1.5 * 10^-6 C)

Calculating this, we find:

E ≈ - 1.93 * 10^6 N/C

Since the electric field is a vector quantity, we need to specify its direction. In this case, the negative sign indicates that the electric field is directed opposite to the positive charge.

Therefore, the magnitude of the electric field is approximately 1.93 * 10^6 N/C, and its direction is opposite to the direction of the positive charge.