A uniform electric field of 500. V/m is directed in the negative x direction. A +15ìC charge moves from the origin to the point (x,y) = (20.0 cm, 50.0 cm). (a) what was the change in the potential energy of this charge? (b) Through what potential difference did the charge move?
This is how i worked it out and it is incorrect.Please tell me what step is incorrect.
(15x10^6C)(500N/C)[(0.2)+0]
ÄP.E= .0015J
Is the sign right on the .2 meters? The charge went against the field. Otherwise, it is ok.
To determine the change in potential energy, we can use the formula:
ΔPE = qΔV
where ΔPE is the change in potential energy, q is the charge, and ΔV is the change in electric potential.
(a) Let's calculate the change in potential energy:
ΔPE = (15µC)(500V/m)(0.2m)
= 1.5mJ
Based on your calculation, it seems like you mistakenly used the incorrect unit for q (charge). The charge should be in microcoulombs (µC), not coulombs (C). Therefore, the correct calculation should be:
ΔPE = (15µC)(500V/m)(0.2m)
= 1.5mJ
So your calculation for the change in potential energy is correct.
(b) To find the potential difference, we can use the formula:
ΔV = -EΔx
where ΔV is the potential difference, E is the electric field strength, and Δx is the displacement in the x-direction.
In this case, the electric field is directed in the negative x-direction, so the potential difference will also be negative.
ΔV = -(500V/m)(0.2m)
= -100V
Your calculation for the potential difference is correct.
Therefore, both your answers for (a) and (b) are correct, and it seems like there was a misunderstanding regarding the units for the charge in your calculation.