Calling all electrical engineers

Question

Problem: a uniform electric field of magnitude 250 V per meter is directed in the positive x direction. A 12 uC (micro-coulomb) charge moves from the origin to the point (x,y) = (20 cm, 50 cm).

Question: what was the change in potential energy of this charge?

Answer 1

Distance the charge travelled = sqrt (20^2 + 50^2)
= sqrt (2900)cm = sqrt (0.29) m.
Now if the charge is moving against the field it's potential energy decreases else it will increase.
In both cases the change is
(Charge)*(potential difference)
(12*10^-6)*(250*sqrt (0.29)) = 0.001616 joules of potential difference.

Comments

Yes, you are correct.

Now for the next question:
Through what potential difference (expressed in volts) did the charge move?

Actually, I see you've already answered that as part of your equation:  250 volts*sqrt (0.29) = 134.63 volts

---

Thank you !