Phys 1120,
CAPA #6
solutions.
1) At the origin, the + on our left makes a "right" E field, the - on our right ALSO makes a
"right" E field ( - charges ATTRACT a positive test charge, + charges REPEL), and the + below
us makes an "up" E field. These all add up to an "up and right" E field, NOT zero, NOT in the y
direction!
The potential where Q3 is, due to the other pair, vanishes (+k q1/r + k(-q1)/r, where r is the
same to both, by symmetry) If the potential vanishes, it means it takes no work to get the charge
from infinity to there. And, similarly, the potential due solely to Q1 and Q2
will vanish
EVERYWHERE along the y axis (because you're equidistant from +q1 and -q1), so the work to
move Q3 anywhere along the y axis vanishes too.
The Potential at the origin is zero from Q1 and Q2 (we just argued this), but from Q3, you get
V=k Q3/d, which can also be written Q3/(4 pi epsilon_0 d).
The force where Q3 is is to the RIGHT (see Example 23-6) If Q3 is released from rest, it
therefore accelerates to the right.
Last but not least: to build up this configuration, start with all three charges at infinity. Bring Q1
in (that takes no work, 'cause there's no other charges there yet). Now bring Q2 in, that will take
k Q1 Q2 / (2a)
of work. (Do you see this? It's the potential energy of the pair, or if you prefer,
the potential at the location of Q2 times the charge of Q2)
Since Q1=-Q2, that's negative work!
(Makes sense: Q2 WANTS to come in, it's attracted to +Q1!
You do positive work when you
"fight" the field, you do negative work when the field is sucking the charge in by itself, as in this
situation) Now, as we argued above, the potential due to both Q1 and Q2 is zero all along the y
axis, even out to infinity. So bringing in Q3 doesn't take any additional work at all!
The total work is still just k Q1 Q2/(2a), which is negative.
2)
Work(by an external agent) = q*(delta V).
(There's no minus sign in this formula.
The work down by the ELECTRIC FIELD is -q*(delta
V) But we're not asking about work down by the field, but by you against the field. Think of
lifting a book: you do + work, gravity does - work. Or lowering a book:
You do - work, gravity
does + work)
Think about the signs until they make some physical sense to you.
In the given problem, from +500 V to -100 V the work done is