A particle that carries a net charge of –41.8 µC is held in a constant electric field that is uniform over the entire region. The electric field vector is oriented 55.2° clockwise from the vertical axis, as shown in the figure. If the magnitude of the electric field is 9.82 N/C, how much work is done by the electric field as the particle is made to move a distance of d = 0.756 m straight up? work: J What is the potential difference AV between the particle's initial and final positions?

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter20: Electric Potential And Capacitance
Section: Chapter Questions
Problem 81P
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A particle that carries a net charge of -41.8 µC is held in a
constant electric field that is uniform over the entire region.
The electric field vector is oriented 55.2° clockwise from the
vertical axis, as shown in the figure. If the magnitude of the
electric field is 9.82 N/C, how much work is done by the
electric field as the particle is made to move a distance of
d = 0.756 m straight up?
d
55.2°
work:
What is the potential difference AV between the particle's
initial and final positions?
AV =
V
Transcribed Image Text:A particle that carries a net charge of -41.8 µC is held in a constant electric field that is uniform over the entire region. The electric field vector is oriented 55.2° clockwise from the vertical axis, as shown in the figure. If the magnitude of the electric field is 9.82 N/C, how much work is done by the electric field as the particle is made to move a distance of d = 0.756 m straight up? d 55.2° work: What is the potential difference AV between the particle's initial and final positions? AV = V
Expert Solution
Step 1 The given data

Charge q = -41.8μC=-41.8×10-6C

             θ=55.2°

The electric field E = 9.82 N/C

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