Thread: work, charge, electric field

It takes 10 joules of work to move a 2.0 coulomb charge in a certain electric field, and it takes 5 joules of work to move a 1.0 coulomb charge in the same field. Which charge gained more potential energy?

a. 1.0 coulomb charge
b. 2.0 coulomb charge
c. change in potential energy is same in both cases
d. not enough information to say

I'd say, (d). since, as it is not specified whether the field is conservative or not, you cannot say anything for certain about potential energy change, since, P.E is not defined for a non-conservative field

since i didn't learn about conservative vs nonconservative fields, i feel quite sketchy using that as an answer for why there is not enough information. does a,b, or c work somehow?

Potential difference (voltage, V), is given by V=E/Q...it is a measure of energy per Coulomb of charge (JC^-1).

So

V₁ = 10 / 2 = 5V
V₂ = 5 / 1 = 5V

So the potential gained is equal.

ok thanks. that makes sense. on a side note now though, what if we were dealing with electric potential instead of potential energy. which would have a greater increase.

Well, electrical potential is just one form of potential energy. Other forms are elastic potential energy, gravitational potential energy, etc.. so really it's one and the same thing.

Originally Posted by bit4bit

Well, electrical potential is just one form of potential energy. Other forms are elastic potential energy, gravitational potential energy, etc.. so really it's one and the same thing.

Hey, don't confuse potential with potential 'energy'. P.E.=Potential x charge.
And potential is integral of dot product of electric field and displacement , so, it depends on the field. For conservative fields, work done in a closed loop is zero. But this is not true for non- cons. fields, so, they don't have any potential defined.

You're right, my mistake.