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 find the magnitude and direction of the electric force that acts on the particle, we can use the formula for electric potential energy:

EPE = q * ΔV

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

In this case, we are given the values of EPEA - EPEB = +5.50 * 10^-4 J and the charge q = +1.5 µC = +1.5 * 10^-6 C. We need to find the value of ΔV to determine the electric potential energy.

Rearranging the equation, we have:

ΔV = (EPEA - EPEB) / q

Substituting the given values, we get:

ΔV = (+5.50 * 10^-4 J) / (+1.5 * 10^-6 C)
= 3.67 * 10^2 V

The electric potential difference between points A and B is 3.67 * 10^2 V.

To find the magnitude of the electric force, we can use the formula:

F = q * E

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

From the equation E = ΔV / d, we can calculate the electric field E.

E = ΔV / d
= (3.67 * 10^2 V) / (0.19 m)
= 1.93 * 10^3 V/m

The magnitude of the electric field is 1.93 * 10^3 V/m.

Now, substituting the known values of charge q = +1.5 µC = +1.5 * 10^-6 C and electric field E = 1.93 * 10^3 V/m, we can calculate the electric force F:

F = q * E
= (+1.5 * 10^-6 C) * (1.93 * 10^3 V/m)
= 2.90 * 10^-3 N

The magnitude of the electric force that acts on the particle is 2.90 * 10^-3 N.

Since the charge of the particle is positive (+1.5 µC), it experiences a force in the direction opposite to the electric field. Therefore, the direction of the electric force is opposite to the direction of the electric field.

The direction of the electric force is opposite to the direction of the electric field.