A particle has a charge of +1.5 µC and moves from point A to point B, a distance of 0.18 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 = +9.50 10-4 J.

(a) Find the magnitude and direction of the electric force that acts on the particle.
magnitude________ N
direction _________ (amongst, against, perpendicular to?)

(b) Find the magnitude and direction of the electric field that the particle experiences.
magnitude________N/C
direction __________ (amongst, against, perpendicular to?)

(a) The change in potential energy divided by the distance is the force.

F = 9.50*10^-4/0.18 Newtons
The force direction is from B to A, because the potential energy is greater at A, and the charge is positive.

(b) Divide the force by the charge to get the electric field strength. It will be in the same direction as the force.

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(a) Well, if we're talking about electric potential energy, we all know it's a shocking business! To find the magnitude of the electric force, we divide the difference in electric potential energy by the distance. So, using EPEA - EPEB = +9.50 * 10^(-4) J, and dividing it by 0.18 m, we got ourselves a force of 5.28 * 10^(-3) N.

Now, for the direction of the electric force, it really knows how to push things around. Since it's along the line of action of the force, it will be amongst the party and follow the particle's movement.

(b) As for the electric field, let's make sure it feels appreciated. To find its magnitude, we take the electric force we calculated earlier and divide it by the particle's charge. So, 5.28 * 10^(-3) N divided by 1.5 * 10^(-6) C gives us a magnitude of 3.52 * 10^3 N/C.

Now, the direction of the electric field can be quite shocking! It's always perpendicular to the electric force, showing off its electric personality. So, perpendicular it is!

To answer these questions, we can use the relationship between electric potential energy, charge, and electric field.

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

EPE = q * ΔV

Where EPE is the electric potential energy, q is the charge, and ΔV is the change in electric potential. In this case, we have:

ΔEPE = +9.50 * 10^(-4) J = q * ΔV

Plugging in the given charge q = +1.5 µC (= 1.5 * 10^(-6) C), we can solve for ΔV:

ΔV = ΔEPE / q

ΔV = (+9.50 * 10^(-4) J) / (1.5 * 10^(-6) C) = 633.33 V

Since the particle moves from point A to point B, and the electric potential energy increased, we know that the particle was moving against the electric field. Therefore, the electric force acts in the opposite direction, which is from B to A. Hence, the direction of the electric force is against the motion of the particle.

(b) To find the magnitude of the electric field, we can use the definition of electric potential:

ΔV = E * Δd

Where ΔV is the change in electric potential, E is the electric field, and Δd is the change in distance. In this case, we have:

ΔV = 633.33 V
Δd = 0.18 m

Plugging these values into the equation, we can solve for the electric field E:

E = ΔV / Δd

E = (633.33 V) / (0.18 m) = 3518.5 V/m

The magnitude of the electric field experienced by the particle is 3518.5 N/C.

Since the particle experiences an electric force towards point A, and the electric field is in the same direction, we can conclude that the direction of the electric field is in the same direction as the force, which is towards point A. Hence, the direction of the electric field is amongst the motion of the particle.