What work was done by the electric force on particle?

[eku_girl83]

Here's the problem:
A particle with a charge of 4.4nC 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 3E-6 Joules.
a) What work was done by the electric force?
.000003 J
b) What is the potential of the starting oint with respect to the endpoint?
681.818 J
I got these parts correct, but am having trouble with part c.
c) What is the magnitude of E?

Could someone give me a hint?? Thanks!

 

[HallsofIvy]

You already know the work done by the electric field.

Work= force* distance so you can calculate the force that the electric field applied for that distance of 6.00 cm.

force= magnitude of electric field * charge (which is 4.4nC) so you can calculate the magnitude from the force.

 

[M98Ranger]

To find the magnitude of E, you can use the equation for work done by the electric force: W = qEd, where W is the work done, q is the charge, E is the electric field, and d is the displacement. In this case, we know the work done (3E-6 J), the charge (4.4 nC = 4.4E-9 C), and the displacement (6.00 cm = 0.06 m). So, we can rearrange the equation to solve for E: E = W/(qd). Plugging in the values, we get E = (3E-6 J)/(4.4E-9 C * 0.06 m) = 681.818 N/C. This is the magnitude of the electric field.