Asked by Jan
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?
A). What is the change in the potential energy associated with this charge?
B).Through what potential difference did the charge move?
Answers
Answered by
drwls
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)
Get to know these important concepts.
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)
Get to know these important concepts.
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