A charge of 28.0 nC is placed in a uniform electric field that is directed verti

Question

A charge of 28.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 3.10×104 V/m .

Part A
What work is done by the electric force when the charge moves a distance of 0.420 m to the right?

W =                         J

Part B
What work is done by the electric force when the charge moves a distance of 0.720 m upward?

W =               J

Part C
What work is done by the electric force when the charge moves a distance of 2.60 m at an angle of 45.0 downward from the horizontal?

Explanation / Answer

Part A

Moving to the right, no work is done by the electric force since the force is perpendicular to distancce.

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Part B

Moving upwards, the work done by the electric force is:

W = Fd = qEd

= ( 28 x10-9) ( 3.10 x104) (0.720 m) = 6.25 x 10-4 joules

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Part C

distance , d = (2.60 m) sin(45°) = 1.84 m

W = Fd = qEd = ( 28 x10-9) ( 3.10 x104) (1.84) = 1.60 x 10-3 joules

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