1. ## electric force.

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 = +8.50 x 10^-4 J.

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

is the direction [along against or perpendicular]

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

is the direction [along against or perpendicular]

---> I know the magnitude of two charges is: http://img144.imageshack.us/img144/5986/39653156ri3.png
where k is Coulumbs constant 9 x 10^9

can i use this equation to help me find the force?

2. (a) The work done on the electric charge is $8.50\times10^{-4}\ \textrm{J}$; this is equal to the product of the magnitude of the constant force acting on it and the distance moved in the direction of the force. Hence the electric force is $\frac{8.50\times10^{-4}}{0.15}=5.67\times10^{-3}\ \textrm{N}$ in the direction from A to B.

(b) The electric field is the electric force per unit charge, so is $\frac{5.67\times10^{-3}}{1.5\times10^{-6}}=3.78\times10^3\ \textrm{N}\,\textrm{C}^{-1}$ in the direction from A to B.