Prelecture Video: Electric Potential Part A Click Play to watch the video below.

Question

Prelecture Video: Electric Potential Part A Click Play to watch the video below. Answer the ungraded questions in the video and the graded follow-up questions at right. A negatively charged object is located in a region of space where the electric field is uniform and points due north. The object may move a set distance d to the north, east, or south. Rank the thre possible movements by the change in electric potential energy (Ue) of the object. Rank from greatest increase to decrease in U. To rank items as equivalent, overlap them Reset Help Figure 1012 10 v 10 v East South North Greatest increase in Ue Greatest decrease in Ue 20 v 20 v

Explanation / Answer

Part A:

potential difference = - scalar product of electric field and displacement

=- E . dx

Here electric field is pointing towards north

If the object of charge -q move towards east, then angle between Electric field and displacement is 90 degree so scalar product is zero as cos 90 =0 . in this case potential difference =0 as a result change in potential energy =0

If the object of charge -q moves towards north, then angle between Electric field and displacement is 0. so potential difference = - Ed. change in potential energy = charge x potential difference = qEd

If the object of charge -q moves towards south, then angle between Electric field and displacement is 180. so potential difference = - Ed(-1) as cos 180 =-1. change in potential energy = charge x potential difference = -qEd

so the increase in potential energy from the greatest to lowest = qEd, 0, -qEd i.e. north, east , south

PART B

Greater the potential difference, greater is the work done.

for path a-b, , potential difference = 0-20=-20 V

for path a-c, potential difference = 10-20=-10 V

for path a-d, potential difference = 20-10 =0 V

so magnitude of work done is more in path a-b.

PART C

Electric field is the negative gradient of potential. i.e. Electri field is in that direction in which potential decreases. so option is b.

PART D

Electric field = V/x

so at a, electric field = 80/1.2 = 66.66

so at b , electric field = 100/1.75= 57.14

so at c, electric field = 50/2 = 25

so a > b > c

all the best in the course work

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