Download Discussion Question 4C

Discussion Question 4C
P212, Week 4
Electric Potential due to a System of Point Charges
Three point charges are fixed at the positions shown
in the figure. We are interested in the electric
potential surrounding this system of charges ... we'll
set the potential at infinity to be zero in all of these
problems. Also, we'll work with algebraic answers in
this problem, so you can put your calculator away.
(a) What is the electric potential at the origin?
y
-q
d
-q
2d
2d
q
x
(i)
It is always a good idea to anticipate as much of the answer as possible before you
begin your calculations. In potential problems, the sign is particularly important. So,
using only your physical intuition, what will be the sign of the electric potential at the
origin? (Ask yourself: if I bring in a positive charge from infinity, where the
potential is set to zero, will I be pushing it ‘uphill’ or letting it roll ‘downhill’?)
(ii)
Now we’re ready to calculate. You could do this problem like the last one, using
integrals over the electric field. However, we already know an expression for the
potential due to a point charge. What is this expression? Remember, any formula for
potential is only good to within an additive constant ... check your formula: does it set
the potential to zero at the location we want?
(iii)
Superposition to the rescue! The total potential at the origin is just: the potential due
to charge q1, plus the potential due to charge q2, plus ... you get the idea. Write down
the potential V due to each of the three charges.
(iv)
Finally, calculate the net potential at the origin. Is the sign correct?
V=
−kq −kq kq −kq
+
+
=
d
2d 2d
d
y
-q
d
-q
2d
q
2d
x
(b) If you placed a negative charge –Q at the origin, what potential energy would it have?
What is the maximum speed it would achieve if it was initially at rest and you let it go?
⎛ −kq ⎞ kQq
U = −QV = −Q ⎜
⎟ = d = U ( 0)
⎝ d ⎠
U ( 0 ) + KE (0) = U ( ∞ ) + KE ( ∞ )
mv ∞2 kQq
=
→ v∞ =
d
2
2kQq
md
(c) What is the total potential energy of the system of 3 charges?
Let's think for a second ... what does this question mean? The potential energy of the system
of 3 charges is equal to the work it took to assemble this configuration. When all 3 charges
are at infinity, we define them to have zero potential energy. So ... bring in one charge at a
time from infinity, and calculate how much work was required at each step …
Zero work to place left charge. When right charge is brought in
from ∞, it would acquire KE so required work would be negative. WR =Work required to add top charge: WT =
K ( −q )2
( 2d )2 + d 2
+
K ( −q )( +q )
( 2d )2 + d 2
Kq2
Kq 2
=−
.
4d
4d
=0
Kq 2
. It is easy to see that it takes no work to place the
4d
top charge when the left and right charges are in place. Consider the direction of the
U = total work = WR + WT = -
force on the top charge and note it is perpendicular to the y-axis.
Did you check the sign of your result using your physical intuition?
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