Physics 21
Fall, 2014
Solution to HW-3
21-66 A charge 5.05 nC is placed at the origin of an xycoordinate system, and a charge −1.96 nC is placed on the
positive x-axis at x = 4.05 cm. A third particle, of charge
5.97 nC is now placed at the point x = 4.05 cm, y = 2.98 cm.
(a) Find the x-component of the total force exerted on the
third charge by the other two. (b) Find the y-component of
the total force exerted on the third charge by the other two.
(c) Find the magnitude of the total force acting on the third
charge. (d) Find the direction of the total force acting on
the third charge.
and q2 . It can be seen that at the position of charge q3 ,
the electric field points down and to the right. Since q3
is positively charged and sits in this electric field, it will
experience a force in the same direction as the electric field,
as shown in the equation F = qE. This is in agreement with
the calculations we just did.
6
5
4
3
Let us label the three charges: the 5.05 nC charge is q1 , the
−1.96 nC charge is q2 , and the 5.97 nC charge is q3 . We will
apply the full vector version of the Coulomb force law to
find the force exerted on q3 by q2 (F2 on 3 ), then the force
exerted on q3 by q1 (F1 on 3 ), then apply the superposition
theorem to find the total force on q3 (F3 ). In order to find
the Coulomb forces, we must write out the position vectors
for the three charges, as r1 = 0, r2 = 4.05 cm ˆi, and r3 =
4.05 cm ˆi + 2.98 cm ˆj, respectively.
We are now ready to work out the force exerted on q3 by q2 :
F2 on 3
1 q2 q3 (r3 − r2 )
=
4πǫ0 |r3 − r2 |3
First, let’s do the subtraction of the position vectors and
find the magnitude. We will use the Pythagorean theorem
to calculate the magnitude of a vector from its components.
r3 − r2 = 0.0298 m ˆj and |r3 − r2 | = 0.0298 m
N · m2 (−1.96 nC) (5.97 nC) ˆj
F2 on 3 = 8.99 × 109
C2
(0.0298 m)2
= − 1.18 × 10−4 N ˆj
bc
q3
2
1
q1
q2
0
-1
-2
5
6
In a similar way, we work out the force exerted on q3 by q1 .
r3 − r1 = 0.0405 m ˆi + 0.0298 m ˆj and |r3 − r2 | = 0.0503 m
2
9 N·m
F1 on 3 = 8.99 × 10
C2
(5.05 nC) (5.97 nC) 0.0405 m ˆi + 0.0298 m ˆj
×
3
(0.0503 m)
= 8.63 × 10−5 N ˆi + 6.35 × 10−5 N ˆj
Finally, we apply superposition:
F3 = F1 on 3 + F2 on 3
= 8.63 × 10−5 N ˆi − 5.45 × 10−5 N ˆj
The answers to questions (a) and (b) are just these two components. To answer questions (c) and (d), we need to find
the magnitude and direction of this vector. Note that part
(d) on MasteringPhysics expects an answer in radians.
|F3 | = 1.02 × 10−4 N
θ = tan−1 −5.45 × 10−5 /8.63 × 10−5 = −0.564 rad
In the figure below, the positions of the three charges are
shown superposed over the electric field due to charges q1
September 9, 2014
21-73 A small m = 12.8 g plastic ball is tied to a very light
26.9 cm string that is attached to the vertical wall of a room.
A uniform horizontal electric field exists in this room. When
the ball has been given an excess charge of q = −1.10 µC, you
observe that it remains suspended, with the string making an
angle of 17.4◦ with the wall. (a) Find the magnitude of the
electric field in the room and (b) determine if the direction
of the electric field is to the right or to the left.
(b) The force due to the electric field in this problem is
pointing to the right, and since q is negative, the electric
field must be pointing to the left.
21-97 Negative charge −Q is distributed uniformly around
a quarter-circle of radius a that lies in the first quadrant,
with the center of curvature at the origin. Find the x- and
y-components of the net electric field at the origin.
y
dQ = λds
= λadθ
dθ
x
The electric field is given by the general formula
dE =
1 dQ(r − r′ )
.
4πǫ0 |r − r′ |3
(1)
We can write the x and y coordinates of a point on the
charged ring as a function of θ:
(a) The first thing to recognize is that the charged ball is
stationary; it experiences zero acceleration in the x and y
directions. Therefore the sum of all the forces on the ball is
zero, and we can write
X
F=0
As shown on the right hand side of the diagram above, the
ball experiences a force FG = −mg ˆj due to gravity, which
points down, a force due to the electric field FE = FE ˆi,
which points to the right, and a tension force T = Tx ˆi +
Ty ˆj that acts along the length of the string. Using some
trigonometry one finds that the components of the tension
force are
Tx = −T sin 17.4
Ty = T cos 17.4◦
◦
The total vector force acting on the ball must be zero, so one
can write separate equations for the x and y components of
the total force:
FE − T sin 17.4◦ = 0
−mg + T cos 17.4◦ = 0
One can eliminate T from these equations and solve for FE
to get:
FE = mg tan 17.4◦
Since the electric force and electric field are related by FE =
qE, and FE points in the +x direction, we only need to solve
for the x component of the electric field:
mg tan 17.4◦
FE
=
q
q
(.0128 kg)(9.81 m/s2 ) tan 17.4◦
= −3.58 × 104 N/C
=
−1.10 × 10−6 C
Ex =
The magnitude of the electric field is
E = |E| = |Ex | = 3.58 × 104 N/C
x = a cos θ
y = a sin θ
The field point r and the charge point r′ are
r = 0 (the origin)
r′ = a cos θ ˆi + a sin θ ˆj.
The linear charge density is the total charge divided by the
arc length of the quarter-circle, or
λ=
−Q
.
1
2 πa
Therefore
dQ = λa dθ =
2Q
−Q
a dθ = −
dθ.
1
π
2 πa
Now we substitute r, r′ , and dQ into Eq. (1), leading to
!
1
2Q
−a cos θˆi − a sin θ ˆj
dE =
−
dθ
4πǫ0
π
a3
Q
=
cos θ ˆi + sin θ ˆj dθ.
2
2
2π ǫ0 a
Now we can integrate to find E. Since the ring is only in the
first quadrant, θ will go from 0 to π/2. Remember that we
must do a separate integral for each component E.
Z π/2 Q
cos θ ˆi + sin θ ˆj dθ
E=
2
2
2π ǫ0 a 0
π/2
Q
ˆ
ˆ
sin θ i − cos θ j =
2π 2 ǫ0 a2
0
Q
ˆ
ˆ
=
i+j
2π 2 ǫ0 a2
Q
Q ˆ
i + 2 2 ˆj.
=
2π 2 ǫ0 a2
2π ǫ0 a
The negative charge on the ring exerts an attractive force on
a positive test charge at the origin, so the x and y components of E are positive.
21-99 Two 1.20 m nonconducting wires meet at a right
angle. One segment carries 2.00 µC of charge distributed
uniformly along its length, and the other carries −2.00 µC
distributed uniformly along it, as shown in the figure. Find
(a) the magnitude and (b) the direction of the electric field
these wires produce at point P , which is 60.0 cm from each
wire. If an electron is released at P , what is (c) the magnitude and (d) the direction of the net force that these wires
exert on it?
(a) In class we worked out the vector electric field due to a
finite line of charge at a field point located directly outward
from the midpoint of the line. (Also see Ex. 21.10 in the
text.) The magnitude of the field is
E=
a
1 λ
√
,
2
2πǫ0 d d + a2
where d is the distance of the point from the line of charge,
and a is half the length of the line. In this exercise, the two
wires have the same length, the same magnitude of charge,
and are the same distance from the point P . Thus, we can
adapt the formula above to the charge and spatial orientation
of each wire by putting in the correct direction for each field.
Recalling that λ = Q/L and a = L/2, we find
E=
1
1 Q
p
4πǫ0 d d2 + (L/2)2
= (9 × 109 Nm2 /C2 )
= 35360 N/C.
1
2 × 10−6 C
p
0.6m
(0.6 m)2 + (0.6 m)2
The field E− due to the negatively charged wire is directed
to the left, while the field E+ due to the positively charged
wire is straight down. We can use superposition to find the
net electric field
Enet = E− + E+ = −35360 N/C ˆi − 35360 N/C ˆj
q
2 + E 2 = 5.00 × 104 N/C
Enet = E−
+
(b) Since the field vectors are perpendicular and of equal
magnitude, the net electric field will be 45◦ from either vector, that is, the electric field will make an angle of −135◦
with the +x axis (the negative angle means 135◦ clockwise.)
(c) The force acting on a charge q is F = qE, so at P the
magnitude of the force on the electron is
Fnet = eEnet = (1.60 × 10−19 C)(5.00 × 104 N/C)
= 8.00 × 10−15 N.
(d) Since the electron has a negative charge, the force will
be directed opposite (180◦ ) from the electric field. The force
will be directed at an angle of 45◦ with respect to the +x
axis.
21-105 Three charges are placed as shown in the figure.
The magnitude of q1 is 2.00 µC, but its sign and the value of
the charge q2 are not known. Charge q3 is +4.00 µC, and the
net force F on q3 is entirely in the negative x-direction. a)
Calculate the magnitude of q2 . b) Determine the magnitude
of the net force F on q3 .
(a) We first determine the sign of the charge q1 . We can do
this by thinking about which direction the force will be in
for the different combinations of signs for charges q1 and q2 .
Since there is no y-component of the force on q3 we know
that q1 and q2 must have opposite signs. Since the force is
directed in the negative x-direction we can infer that q1 must
be negative and q2 must be positive.
To determine q2 we calculate the the total force F = F1 on 3 +
F2 on 3 on q3 . Expressions for F1 on 3 and F2 on 3 follow from
the general expression for F2 on 1 on the equation sheet:
F1 on 3 =
1 q3 q1 (r3 − r1 )
,
4πǫ0 |r3 − r1 |3
F2 on 3 =
1 q3 q2 (r3 − r2 )
.
4πǫ0 |r3 − r2 |3
We need the position vectors of each charge. Let q1 be at
the origin. Then the position vectors r1 and r2 of q1 and q2
are trivial. For r3 , we notice that cos θ = 4/5 = x/4 where
x is the x-component of the position vector of q3 . Along
with this and Pythagorean’s theorem we can find both x
and y-components of r3 . The result is
r1 = 0,
r2 = 5 cm ˆi,
r3 = 3.2 cm ˆi + 2.4 cm ˆj
Knowing these position vectors we can write F1 on 3 and
F2 on 3 , in terms of the unknown charge q2 :
F1 on 3 = −36 N ˆi − 27 N ˆj
F2 on 3 = (−24 × 106 N/C) q2 ˆi + (32 × 106 N/C) q2 ˆj
Since the net force on q3 is in the negative x-direction, the
sum of the y-components must be 0. From the sum of the
y-components we find that the charge q2 = +0.844 µC. The
charge is positive, as expected by the reasoning above.
(b) Using q2 , we can now determine the net force on q3
by adding the components together. We already know that
there is no net y-component, so the net force has only an xcomponent, which is the sum of the x components of F1 on 3
and F2 on 3 . The magnitude of the total force is the absolute
value of its x component, |F| = 56.26 N.

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