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 potential energy change of a charge moving in an electric field, you can use the equation:
ΔPE = q * ΔV
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, you need to know the charge of the particle. In this case, the charge is +12.5 μC (microcoulombs), which is equivalent to 12.5 * 10^-6 C.
Since the electric field is uniform and directed in the negative y direction, the potential difference ΔV can be calculated using the formula:
ΔV = -E * d
where E is the magnitude of the electric field and d is the displacement in the direction of the field.
Given that the charge moves from the origin to the point (x, y) = (19.5 cm, 51.8 cm), the displacement in the y-direction, d, is 51.8 cm.
Substituting the given values:
ΔV = -(-203 V/m) * 0.518 m
Simplifying:
ΔV = 105.254 V
Now you can calculate the change in potential energy:
ΔPE = (12.5 * 10^-6 C) * (105.254 V)
Simplifying:
ΔPE = 1.3157 * 10^-3 J
Therefore, the change in potential energy of the charge is 1.3157 * 10^-3 Joules.
2) The potential difference, ΔV, is found to be 105.254 V. Therefore, the charge moved through a potential difference of 105.254 V.