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?
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Charge q = -41.8=-41.8
=55.2
The electric field E = 9.82 N/C
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