A particle with a charge of is in a uniform electric field 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: a. What work was done by the electric force? b. What is the potential of the starting point with respect to the endpoint? c. What is the magnitude of E?

Question is incomplete.

Complete Question :

A particle with a charge of is +4.2 nC in a uniform electric field 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: a. What work was done by the electric force? b. What is the potential of the starting point with respect to the endpoint? c. What is the magnitude of E?

a) Work done by the electric force = W = 2.20 x 10^-6

b) Potential of the starting point with respect to end point = 523.80 V

c) Magnitude of E = 8.73 x 10^3 N/C

Explanation:

a) The charge has moved from point a to point b, hence the work done basically equals the change in kinetic energy.

Hence ,

                           W(a to b)= ΔK = Kb-Ka

                           W(a to b) = Kb

                           W(a to b) = 2.20 x 10^-6

b)            W(a to b)/q = Va - Vb

                Va - Vb = W(a to b) / q

                              =2.20 x  10^-6 J / 4.20 x 10^-9

                              =523.80 V

c)              E = Va - Vb / l = 523.80 V / 0.060 = 8.73 x 10^3 N/C