A caterpillar tries to climb straight up a wall a meter high, but for every 2 cm up it climbs, it slides down 1 cm. Eventually,
2 answers:
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Answer:
W = 0×(kq2L)
Explanation:
We know that the work to assemble a charge configuration of two charges a distance r from each other is simply W = kq2/r
If we want to assemble three charges A, B, and C. It's necessary to consider the distances between them
WABC = kq2/(rAB + rAC + rBC)
So, to assemble four charges A, B, C, & D....
WABCD = kq2/(rAB + rAC + rAD + rBC + rBD + rCD)
Considering a square charge configuration with sides L, such as in figure attached A, B, & C are positive & D is negative
rAB = L
rAC = L√2
rAD = L (-)
rBC = L
rBD = L√2 (-)
rCD = L (-)
⇒ W = kq2/(L + L√2 + (-L) + L + (-L√2) + (-L)
⇒ ∴ W = 0 × (kq2/L)
This way, working through each option...
(a)
The positive charges are equidistant from each other at a distance of L.
rAB = L
rAC = L
rAD = ½L⋅sin(60) (-)
rBC = L
rBD = ½L⋅sin(60) (-)
rCD = ½L⋅sin(60) (-)
Wa = kq2/(3L - (3/2)L⋅(0.866))
⇒ ∴ Wa = (1/1.7) × (kq2/L) = (0.5879)× (kq2/L)
(b)
rAB = L
rAC = 2L
rAD = 3L (-)
rBC = L
rBD = 2L (-)
rCD = L (-)
Wb = kq2/(4L - 6L)
⇒ ∴ Wb = (-1/2) × (kq2/L) = (-0.5)× (kq2/L)
(c)
The factor doesn't matter, so Wc = 0 × (kq2/L)
In this case, the greater work is actually the less work. Therefore, the positive work represents the amount of work the system actually exhibits, that we don't have to do. If there is negative work, we have to make up that work in order to place the charges as desired.
This way, charge configuration (a) requires the least amount of work.
