39well, this is easy...
work done is change in potential energy if the system.
10 Answers
39
1i m not getting how to relate the distance...[2]plz give a hint 4 dat...
1take the potential at final position and that at initial position subtract final potential from initial potential and multiply it with the test charge....
11Uinitial = k Qq/r1 [ r1 = dis. b/w origin & A (use dist. formula ) ]
{ Uintermadiate = k Qq/r2 [ r2 = dis. b/w origin & (0,0.06,0.09) (use dist. formula ) ] } <--- (unnecessary)
Ufinal = k Qq/r3 [ r3 = dis. b/w origin & (0,0.04,0) (use dist. formula ) ]
Work done = | Ufinal -- Uinitial |
1yaar work done is independent of path it depends on the final and initial position
11@prateek
ya.... dats why that "intermediate" part is " unnecessary " !!!