# Find the magnitude and direction of the electric force

1. Dec 9, 2006

### metalmagik

1. The problem statement, all variables and given/known data
A particle has a charge of +1.5 µC and moves from point A to point B, a distance of 0.15 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 is EPEA - 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.

2. Dec 9, 2006

### vanesch

Staff Emeritus
Your solution seems right, except for one silly application, namely q/V = C. The q is the charge of a particle, and V is the potential difference between two points on its trajectory, so this has nothing to do with any charge on a capacitor.

But, yes, you have correctly calculated the potential difference needed for a certain potential energy to correspond to a charged particle ;

given that potential difference, over a certain distance, that gives you the electric field, also ok, and knowing the electric field and the charge, you know the mechanical force acting. Also ok.

Now, the field and the force are in the same direction (if q is positive).
Let's just reason about mechanical potential energy: if the potential energy is HIGHER at point A than at point B, then that means that it is "easier" for the particle to be at point B than at point A. So the force points then how ?
And given that the force is in the same direction as the field ...

(btw I didn't check the arithmetic...)

3. Dec 10, 2006

### metalmagik

Thanks vanesch, You really helped me out a lot with that last paragraph of your response, its like 1 AM right now, im going to bed but Im going to take your information and fully reason it out tomorrow, when im not tired...thanks a bunch