A uniform electric field of magnitude 333 N/C is directed along the +y-axis. A 6.00 µC charge moves from the origin to the point (x, y) = (-10 cm, -37 cm).
A). What is the change in the potential energy associated with this charge?
B).Through what potential difference did the charge move?
physics - drwls, Monday, February 8, 2010 at 1:08pm
The product of the E-field (in N/C) and the charge (in Coulombs) equals the force, which will be in the +y direction since the charge is positive.
Multiply the force by the distance moved IN THE FIELD (y) DIRECTION for the P.E change, but add a "minus" sign. The distance moved must be in meters.
The potential difference between the two points is E * (change in y)
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