Homework #1 Solutions version 3
6 problems
Exercises
.
1. Charges A and B lie 5 cm apart, on a line. Charge A is qA =
+2 µC, and charge B is qB = −3 mC. (Be careful of the metrix
prefixes!)
(a) Before doing any calculation, determine the direction of the force
on both charges. Which will feel the larger force, A or B?
(b) Calculate the force (magnitude and direction) on both charges,
using Coulomb’s Law.
+2µC
–3mC
A
B
5cm
/ (a) The charges have opposite charges, so they attract one another; the force on A is to the right and the
force on B is to the left. The magnitudes of the forces are equal, due to Newton’s Third Law.
(b) We already determined the directions above. The magnitude of the force on either charge is given by
Coulomb’s Law:
|qA qB |
r2 2 (2 × 10−6 C)(−3 × 10−3 C)
9 Nm
= (9 × 10
)
C2
(0.05 m)2
Fon A = Fon B = k
= 2.2 × 104 N
Thus the force on A is 2.2 × 104 N to the right, and the force on B is 2.2 × 104 N to the left.
.
2.
Three charges are fixed on a line as shown. Find the force
(magnitude and direction) on charge B.
+1µC
–3µC
–1µC
A
B
C
0.01m
0.01m
/ B is the target; A and C are the sources. We find the force due to each source, and then add them
together.
The force on B due to A points to the left (attraction), and has magnitude
2
(1 × 10−6 C)(3 × 10−6 C)
9 Nm
FA = 9 × 10
= 270 N
C2
(0.01 m)2
The force on B due to C also points to the left (repulsion), and has magnitude
2
(1 × 10−6 C)(3 × 10−6 C)
9 Nm
FC = 9 × 10
= 270 N
C2
(0.01 m)2
And so the net force on B is 270 N to the left plus 270 N to the left, or 540 N to the left .
.
3. Coulomb’s Law is
q1 q2
F~ = k 2 r̂
r
where r̂ is the unit vector pointing from the source charge to the
target charge (that is, from the forcer to the forcee). The unit vector
is defined as the vector from charge to charge, divided by its length:
r̂ =
~r
|~r|
so we can alternatively write Coulomb’s Law as
q1 q2
q1 q2 ~r
= k 3 ~r
F~ = k 2
r r
r
B
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This is more useful calculationally, but remember that the force decreases with the square of the distance, not the cube of the distance.
Suppose a +1 µC charge (charge A) sits at coordinate (1,3), and a
+3 µC charge (charge B) sits at coordinate (4,5). (The coordinates
are in meters.)
(a) Find the vector ~r from A to B.
(b) Find the distance |~r| (or r) from A to B.
(c) Find the unit vector r̂ from A to B.
(d) Find the force on B due to A, in unit-vector notation.
(e) Find the force on A due to B, in unit-vector notation.
4
A
3
2
0
1
2
3
4
5
/ (a) The vector ~r from A to B is 3 meters to the right (call that x̂) and two meters up (call that ŷ), so
~r = 3x̂ + 2ŷ .
(b) We use the Pythagorean theorem here: |~r| =
√
32 + 22 =
√
9+4=
√
13 .
(c) The unit vector of ~r is
r̂ =
3
2
~r
3x̂ + 2ŷ
= √ x̂ + √ ŷ
= √
|~r|
13
13
13
(d) (Note: Unit-vector notation means something like 3x̂ − 4ŷ, or 3ı̂ − 4̂ if you prefer that notation.)
In this case, A is the source and B is the target, so ~r is the vector from source to target, as it needs to be in
Coulomb’s Law. Let’s do this in two slightly different ways. First, use F~ = k qArq2B r̂.
qA qB
F~ = k 2 r̂
r
N m2 (1 × 10−6 C)(3 × 10−6 C)
√
= (9 × 10
)
C2
( 13)2
9
= (2.08 × 10−3 N)(0.832x̂ + 0.555ŷ)
= (1.7 mN)x̂ + (1.2 mN)ŷ
Now let’s try the other form of Coulomb’s Law
qA qB
F~ = k 3 ~r
r
2
3
2
√ x̂ + √ ŷ
13
13
2
(1 × 10−6 C)(3 × 10−6 C)
9 Nm
√
(3x̂ + 2ŷ)
= 9 × 10
C2
( 13)3
= (5.76 × 10−4 N)(3x̂ + 2ŷ)
= (1.7 mN)x̂ + (1.2 mN)ŷ
√
Notice how the second way is a little less complicated-looking; you only have to deal with the 13 once, in
the denominator. This is what I meant when I said that this second form was slightly easier to use in real
calculations, particularly in cases when ~r does not point along a coordinate axis.
(e) Newton’s third law says that if B feels a force F~ due to A, then A feels a force −F~ due to B. Thus the
force on A is −(1.7 mN)x̂ − (1.2 mN)ŷ . (Don’t feel bad if you went through the whole exercise again with
the charges reversed; it was good practice. But don’t forget that Newton’s Third Law might save you some
time in the future, like on an exam.)
.
4.
In the figure, the four particles form a square of edge length
a = 5.00 cm and have charges q1 = +10.0 nC, q2 = −20.0 nC, q3 =
+20.0 nC, and q4 = −10.0 nC. In unit-vector notation, find the force
on charge q1 .
q1
q2
y
a
q4
x
q3
/ The charge q1 is the target; the other three are the sources. We need to consider each source in turn. The
trickiest part is coming up with ~r (the vector between source and target) for each.
q2 : The vector from q2 to q1 is ~r = −ax̂, so r = a and r̂ = −x̂. (I only calculate r̂ here because it’s obvious.)
Thus
qs qt
q2 q1
N m2 (−20 × 10−9 C)(10 × 10−9 nC)
F~q2 = k 2 r̂ = k 2 (−x̂) = 9 × 109
(−x̂) = 7.2 × 10−4 N(x̂)
r
a
C2
(0.05)2
q3 : √
To get from √
q3 to q1 , we go a distance a up, and a distance a to the left, so ~r = aŷ − ax̂, and
r = a2 + a2 = a 2. Thus
q3 q1 (aŷ − ax̂)
kq3 q1
qs qt ~r
√
√ (ŷ − x̂)
=k √
=
F~q3 = k 2
r r
(a 2)2 a 2
(2a2 )( 2)
(It’s always good to simplify as much as possible before substituting numbers in.) With numbers,
−9
2
20
×
10
C
(10 × 10−9 C)
N
m
9
√
F~q3 = 9 × 10
(ŷ − x̂)
2
C
2(0.05)2 2
= 2.5 × 10−4 N(ŷ − x̂) = (2.5 × 10−4 N)ŷ − (2.5 × 10−4 N)x̂
q4 : The vector from q4 to q1 is ~r = aŷ, so r = a and r̂ = ŷ. Thus
qs qt
q4 q1
N m2 (−10 × 10−9 C)(10 × 10−9 nC)
F~q2 = k 2 r̂ = k 2 (ŷ) = 9 × 109
(ŷ) = 3.6 × 10−4 N(−ŷ)
r
a
C2
(0.05)2
3
The net force on the target is the sum of these three vectors:
F~ = 7.2 × 10−4 Nx̂
− 2.5 × 10−4 Nx̂
+2.5 × 10−4 Nŷ
−3.6 × 10−4 Nŷ
= 4.7 × 10−4 N(x̂)−1.1 × 10−4 N(ŷ)
Thus F~ = 4.7 × 10−4 N(x̂) − 1.1 × 10−4 N(ŷ)
Problems
.
5. Identical isolated conducting spheres 1 and 2 have equal charges
Q, and are separated by a distance which is large compared to their
diameters (Fig. a). The electrostatic force acting on sphere 2 due to
sphere 1 is F~ .
Suppose now that a third sphere 3, identical to the other 2 but initially
neutral and attached to an insulating handle, is touched first (b) to
sphere 1, then (c) to sphere 2, and then (d) finally removed. The
electrostatic force on sphere 2 is now F~ 0 . Find the ratio |F~ 0 |/|F~ |.
(a)
−F!
(b)
1
2
1
3
2
F!
1
2
3
1
2
(c)
!
(d) −F ’
F! ’
/ In picture (a), spheres 1 and 2 have charge Q, and sphere 3 has charge 0.
In picture (b), charge is allowed to flow from 1 to 3. Because the spheres are exactly the same, they should
both end up with the same charge. Therefore, after step b, sphere 1 and sphere 3 both have charge Q/2.
In picture (c), charge is allowed to flow between 2 and 3. The total charge in the two spheres is Q + Q/2 =
3Q/2. That charge, again, will distribute evenly between the two spheres. Therefore, after step (c), sphere
3 has charge 3Q/4, sphere 2 has charge 3Q/4, and sphere 1 still has charge Q/2.
If the distance between the two forces is R, then the initial force between the two is F = kQ2 /R2 . The final
force between the two is
(Q/2)(3Q/4)
3 Q2
3
F0 = k
=
k 2 = F
2
R
8 R
8
Therefore F 0 /F = 3/8 .
.
6. Consider the two fixed charges on an axis as shown. There is at
least one point on the axis where one could place an electron and have
it not feel a net force due to the other two charges. Find that point.
(Note that you will have to consider three separate cases, depending
on whether the electron is to the left of both charges, between them,
or to the right of both.)
+2C
–1
0
–3C
1
2
3
4
5m
/ Since the figure shows a number line, suppose we place the electron at position x, and the force on it is
zero. Our goal is to find the value of x. There are three possibilities to consider: either x < 0 (so the electron
is to the left of both charges), 0 < x < 4 (between the charges), or x > 4 (on the right of both).
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The magnitude of the force due to the 2 C charge is FL = k r2e2 , where r is the distance from x to 0; but this
distance is simply r = |x − 0| = |x|, so FL = k x2e2 .
The distance from the electron at x to the right charge at 4 is r = |x − 4|. The magnitude of the force due
3e
to the −3 C charge is thus FR = k (x−4)
2.
If the electron is between the two charges, it will be attracted to the left by the positive charge, and repelled
to the left by the negative charge. Thus the electron feels a net force to the left, and can’t be zero.
If the electron is to the left of both charges, then it is attracted to the right by the positive charge and repelled
to the left by the negative charge. Thus the force due to the left charge is F~L = FL (x̂) and F~R = FR (−x̂),
so the net force is
F~ = FL x̂ − FR x̂ = (FL − FR )x̂
The force on the electron is thus zero if FL = FR .
FL = FR
3e
2e
k 2 = k
x
(x − 4)2
2
3
=⇒ 2 =
x
(x − 4)2
2(x2 − 8x + 16) = 3x2
2x2 − 16x + 32 = 3x2
=⇒ x2 + 16x − 32 = 0
√
√
−16 ± 256 + 128
= −8 ± 4 6 = {1.80, −17.8}
=⇒ x =
2
The first value of x, 1.80, lies between the two charges, so that doesn’t make sense. But x = −17.8 m does
lie to the left of the charges, and so is indeed a point where the forces balance.
(Notice, by the way, that the charge of the electron cancelled out pretty early. If the electron feels no force
there, then no charge of either sign feels no force either; in fact, that is a point where the electric field of the
two charges is zero.)
What if the electron is to the right of the two charges? Then the electron is attracted to the left by the
positive charge and repelled to the right by the negative charge. Thus the net force on the right hand side is
F~ = FL (−x̂) + FR (x̂) = (FR − FL )x̂
This is only zero if FL = FR . However, we showed that FL = FR only when x = 1.80 or x = −17.8; in
neither case is the electron to the right of the two charges (i.e. x > 4). Therefore there is only one place on
the number line where the forces cancel: at x = −17.8 m .
(There is a way one could have guessed which side this point was on, without trial and error. If one moves
far away from these two charges, to the left or right, then the two charges seem to blur into one point with
net charge 2 − 3 = −1 C: this is a negative charge, so far away the electron would tend to be repelled away
from the center. If we put the electron just to the left of the positive charge, it will be attracted towards the
center. As the electron moves to the left, then, it must transition from being attracted to being repelled: at
the transition point, the force is zero.)
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