(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A particle has a charge of+1.5 µCand moves from point A to point B, a distance of0.15 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 isEPEA - EPEB = +9.0 10-4 J.

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

_______N the direction of motion (along, against, perpendicular)

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

_______N/C the direction of motion (along, against, perpendicular)

2. Relevant equations

PE = qV

C = q/V

E = F/q

V = Ed

3. The attempt at a solution

I have used the four formulas above to reach two answers. I'm no good with LaTeX so, i apologize but bear with me.

PE = qV

9e-4 J = (1.5e-6 C)V

V = 600 V

C = q/V

C = (1.5e-6)/600

C = 2.5e-9 F

V = Ed

600 = (.15)E

E = 4000 N/C

E = F/q

4000 = F/(1.5e-6)

F = .006 N

As for the direction...I believe the force is along the direction of motion...and the E field is perpendicular? I know the formulas and am doing the work right but if someone could clarify if I have I am understanding it right and using the right numbers I would appreciate it greatly. Thank you.

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# Homework Help: Find the magnitude and direction of the electric force

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