Aparticle with a charge of +4.50 nc is in a uniform electric field e+ directed to the negative x direction. it is released from rest, and after it has moved 8.00 cm , its kinetic energy is found to be 1.50×10^−6 j . you may want to review (pages 567 - 572) . for related problem-solving tips and strategies, you may want to view a video tutor solution of parallel plates and conservation of energy.
part a: what work was done by the electric force? part b: what was the change in electric potential over the distance that the charge moved? part c: what is the magnitude of e+ ? part d: what was the change in potential energy of the charge?

(a)

According to the work-energy theorem, the work done by the electric force is equal to the kinetic energy gained by the particle, so we can write:

where

W is the work done by the electric force

is the final kinetic energy of the particle

is the initial kinetic energy of the particle

Since the particle starts from rest, . Moreover,

Therefore, the work done by the electric force is

(b) -333.3 V

According to the law of conservation of energy, the gain in kinetic energy of the particle must correspond to a loss in electric potential energy, so we can write:

Where

is the gain in kinetic energy

is the loss in electric potential energy

So we have

The loss in electric potential energy can be rewritten as

where

is the charge of the particle

is the change in electric potential over the distance the charge has moved

Solving for ,

(c) 4166 V/m

The magnitude of the electric field is given by

where

is the magnitude of the change in electric potential

d is the distance through which the charge has moved

Since we have

The magnitude of the electric field is

(d)

The change in electric potential energy of the charge  has already been calculated in part (b), and it is

small = small current large mag large current