A uniform electric field with a magnitude of 5750 N/C points in the positive x d

Question

A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Part A Find the change in electric potential energy when a +19.5-C charge is moved 6.50 cm in the positive x direction. Part B Find the change in electric potential energy when a +19.5-C charge is moved 6.50 cm in the negative x direction. Part C Find the change in electric potential energy when a +19.5-C charge is moved 6.50 cm in the positive y direction. Express your answer using one significant figure. all answers should be in joules

Explanation / Answer

Electic field=5750 N/C

charge,q= +19.5 C

distance,r=6.50 cm

Electric field=Voltage/distance

So, 5750 V/m * 0.065 m = 373.75 volts

then

Voltage=Energy/charge

so

373.75 J/C * 19.5*10-6 C = 0.007288 J or 7.28 mJ

a) Here the charhe is moving in the direction of the field, so PE would decrease by that amount

b) opposite direction indicates that there is an increase in PE

c) motion perpendicular to field so no change in PE

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