# In order to move a charge of 14.5 C in an electric field with an intensity of 350 N/C, 275 J of work was used. How far was the charge moved?

PHYSICS HOMEWORK HELP THANK YOU!!!!!

### 1 Answer

- 1 month ago
The correct answer is, we don't know. Any distance moved perpendicular to the electric field will contribute zero change in potential energy, so the answer depends entirely on that component of the motion, which is not specified.

That said, we can find how far "upstream" the object was moved by finding the change in potential energy and equating it to the amount of work used.

So if 275 Joules of work were used, and the object is at rest in both the initial and final states, and we neglect any losses due to friction, air resistance, et cetera, then the object must have gained 275 Joules of potential energy.

Potential energy equals charge times voltage (ΔU = qΔV) so the potential difference between the two points is ΔV = 275 J / 14.5 C = 18.966 V.

Technically here we should have to take a path integral of the field (which is part of the reason why this question doesn't work) but if we assume that we always move parallel to and against the field and that the field has a constant strength, then we can say that ΔV = E*d, where d is the distance moved and E is the field strength.

So 18.966 V = 350 V/m * d, and d = 18.966/350 m = 0.054187 m => 0.0542 m (3 s.f.)

Sorry about the wordiness, a lot of physics depends on your assumptions so I always like to be very specific in how I say things.

Here's a TL;DR:

W = ΔU = qΔV => ΔV = 275 J / 14.5 C = 18.966 V

ΔV = E*d => d = 18.966 V / 350 (V/m) = 0.0542 m

Hope this helps!