Work and potential in an electric field

[sebby]

any help would be appreciated


A particle with a charge of +5.80 nC is in a uniform electric field E directed to the left. It is released from rest and moves to the left. After it has moved 6.00 cm, its kinetic energy is found to be +1.00 x 10^-6 J.

(a) What work was done by the electric force?
got this right

(b) What is the potential of the starting point with respect to the endpoint?
_________ V

(c) What is the magnitude of E?
_________N/C

 

[ibc]

This might help you:
dV=E∙dx
for uniform electric field: ∆V=E∙∆x
W=∆V∙q
(W = word done, q = charge)

 

[sebby]

thank you very much that was exactly what i needed.