A uniform electric field of magnitude 203 V/m is directed in the negative y direction. A +12.5 μC charge moves from the origin to the point (x, y) = (19.5 cm, 51.8 cm).
1) What was the change in the potential energy of this charge?
2) Through what potential difference did the charge move?
To find the change in potential energy and the potential difference, we need to use the formula:
ΔPE = qΔV
where ΔPE is the change in potential energy, q is the charge, and ΔV is the potential difference.
1) To find the change in potential energy, we need to calculate the potential difference first. The potential difference can be found using the formula:
ΔV = -EΔy
where Δy is the displacement in the y-direction and E is the magnitude of the electric field.
Given: E = 203 V/m, Δy = -51.8 cm = -0.518 m
ΔV = -203 V/m * (-0.518 m)
= 105.074 V
Now, we can calculate the change in potential energy:
ΔPE = qΔV
Given: q = +12.5 μC = 12.5 * 10^(-6) C, ΔV = 105.074 V
ΔPE = (12.5 * 10^(-6) C)(105.074 V)
= 1.3134 * 10^(-3) J
Therefore, the change in potential energy of the charge is 1.3134 * 10^(-3) J.
2) The potential difference through which the charge moved is given by ΔV. From the calculation above, the potential difference is found to be 105.074 V. Thus, the charge moved through a potential difference of 105.074 V.