Physics HW help please?
A particle of net charge, +Q, in a uniform, upwardly-directed electric field is moved from point (A) to point (B) along the path shown in the figure below. We want to calculate the voltage difference between the end point, (B), and the start point, (A), for the particle. The net vertical displacement of the particle is 2m.
1. Assume the electric field has a magnitude of |E|=44.2 V/m, and the particle has a net positive charge of Q = +5.40 C. Calculate the magnitude of the electric potential difference, |ΔV|=|VB −VA|.
2. We increase the magnitude of the electric field to |E(new)|=108 V/m, and repeat the experiment with the same Q = +5.40 C particle. Calculate the net change in the electric potential energy , ΔPE=PE (B) −PE(A). [Be sure the sign of your answer accurately reflects whether the particle gained or lost potential energy in moving from point (A) to point(B) in this experiment]
- NCSLv 71 month agoFavourite answer
1) |ΔV|=|VB −VA| = |E|*|Δd| = 44.2V/m * 2m = 88.4 V
2) ΔPE = q*E*Δd = 5.40C * 108V/m * (0 - 2)m = -1166 J
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