ragsdale (zdr82) – HW1 – ditmire – (58335)
This print-out should have 21 questions.
Multiple-choice questions may continue on
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before answering.
001 10.0 points
Two spheres, fastened to “pucks”, are riding on a frictionless airtrack. Sphere “1” is
charged with 3 nC, and sphere “2” is charged
with 6 nC. Both objects have the same mass.
1 nC is equal to 1 × 10−9 C.
As they repel,
1. they do not accelerate at all, but rather
separate at constant velocity.
2. sphere “2” accelerates 4 times as fast as
sphere “1”.
3. sphere “1” accelerates 4 times as fast as
sphere “2”.
4. sphere “1” accelerates 2 times as fast as
sphere “2”.
5. they have the same magnitude of acceleration. correct
6. sphere “2” accelerates 2 times as fast as
sphere “1”.
Explanation:
The force of repulsion exerted on each mass
is determined by
F =
1 Q1 Q2
= ma
4 π ǫ0 r 2
where r is the distance between the centers of
the two spheres.
~ 12 k = kF
~ 21 k
kF
Since both spheres have the same mass and
are subject to the same force, they have the
same acceleration.
002 10.0 points
A negatively charged balloon has 3.7 µC of
charge.
1
How many excess electrons are on this balloon?
Correct answer: 2.3125 × 1013 electrons.
Explanation:
Let : q = −3.7 µC and
ρ = −1.60 × 10−19 C/electron .
The charge density is
q
ρ=
N
q
N=
ρ
−3.7 × 10−6 C
=
−1.6 × 10−19 C/electron
= 2.3125 × 1013 electrons .
003 10.0 points
You have 2.6 kg of water. One mole of water
has a mass of 17.9 g/mol and each molecule
of water contains 10 electrons since water is
H2 O.
What is the total electron charge contained
in this volume of water?
Correct answer: −1.39956 × 108 C.
Explanation:
Let :
NA = 6.02214 × 1023 molec/mol ,
qe = −1.6 × 10−19 C/electron ,
m = 2.6 kg ,
M = 17.9 g/mol = 0.0179 kg/mol ,
Z = 10 electrons/molec .
and
The mass is proportional to the number of
molecules, so for m grams in n molecules and
M grams in NA molecules,
m
n
=
M
NA
m
n=
NA
M
Since 10 electrons are in each molecule of
water, then the total number of electrons ne
in the coin is
m
NA
ne = Z n = Z
M
ragsdale (zdr82) – HW1 – ditmire – (58335)
and the total charge q for the ne electrons is
q = ne qe
m NA qe
=Z
M
2.6 kg
= (10 electrons/molec)
0.0179 kg/mol
23
× 6.02214 × 10 molec/mol
× −1.6 × 10−19 C/electron
= −1.39956 × 108 C .
004 10.0 points
Three identical point charges, each of mass
160 g and charge +q, hang from three strings,
as in the figure.
The acceleration of gravity is 9.8 m/s2 ,
and the Coulomb constant is 8.98755 ×
109 N · m2 /C2 .
cm
cm
.5
.5
10
All three charges are in equilibrium, so for
each holds
X
F = 0.
Consider the forces acting on the charge on
the right. There must be an electrostatic force
F acting on this charge, keeping it balanced
against the force of gravity m g. The electrostatic force is due to the other two charges
and is therefore horizontal.
In the x-direction
F − T sin θ = 0 .
In the y-direction
T cos θ − m g = 0 .
These can be rewritten as
F = T sin θ
9.8 m/s2
10
40◦
160 g
160 g
160 g
+q
+q
+q
If the lengths of the left and right strings
are each 10.5 cm, and each forms an angle of
40◦ with the vertical, determine the value of
q.
Correct answer: 0.7304 µC.
and
m g = T cos θ .
Dividing the former by the latter, we find
F
= tan θ ,
mg
or
F = m g tan θ
= (0.16 kg) (9.8 m/s2 ) tan 40◦
= 1.31571 N .
Explanation:
Let : θ = 40◦ ,
m = 160 g = 0.16 kg ,
L = 10.5 cm = 0.105 m ,
g = 9.8 m/s2 , and
ke = 8.98755 × 109 N · m2 /C2 .
Newton’s 2nd law:
X
(1)
The distance between the right charge and
the middle charge is L sin θ, and the distance
to the left one is twice that. Since all charges
are of the same sign, both forces on the right
charge are repulsive (pointing to the right).
We can add the magnitudes
F = ma.
Electrostatic force between point charges q1
and q2 separated by a distance r
q1 q2
F = ke 2 .
r
2
F = ke
qq
qq
+ ke
2
(L sin θ)
(2 L sin θ)2
or
F =
5 ke q 2
.
4 L2 sin2 θ
(2)
ragsdale (zdr82) – HW1 – ditmire – (58335)
We have already found F , and the other quantities are given, so we solve for the squared
charge q 2
q2 =
4 F L2 sin2 θ
5 ke
(3)
or, after taking the square root (we know
q > 0) and substituting F from Eq. 1 into Eq.
3 and solving for q, we have
r
m g tan θ
q = 2 L sin θ
5 ke
= 2 (0.105 m) sin 40◦
s
(0.16 kg) (9.8 m/s2 ) tan 40◦
×
5 (8.98755 × 109 N · m2 /C2 )
= 7.304 × 10−7 C = 0.7304 µC .
|q1 | |q2 |
r2
By symmetry and the fact that force on charge
q by +Q is repulsive and by −Q is attractive,
~ k = ke
kF
Fy = 0 .
The x component of the forces on q by Q
and −Q are equal√in magnitude and direction.
2
. Hence,
Note: cos 45◦ =
2
√
q
Q
2
~ k = 2 ke
kF
2
L 2
= 2 (8.9875 × 109 N · m2 /C2 )
√
(3 µC) (10 µC) 2
×
(10 cm)2
2
= 38.1307 N .
10.0 points
3 µC
005
3
keywords:
10
cm
cm
10
90◦
−10 µC
10 µC
What is the magnitude of the electrostatic
~ k on the top charge?
force kF
The
Coulomb
constant
is
8.9875 × 109 N · m2 /C2 .
Correct answer: 38.1307 N.
Explanation:
006 (part 1 of 2) 10.0 points
Four charges −4 × 10−9 C at (0 m,0 m),
−2 × 10−9 C at (1 m,−4 m), 3 × 10−9 C at
(−3 m,−3 m), and 2 × 10−9 C at (5 m,−2 m),
are arranged in the (x, y) plane (as shown in
the figure below, where the scale is in meters).
y (m)
5
4
3
2
Let : Q = 10 µC ,
q = 3 µC , and
L = 10 cm .
1
0
−1
q
−2
−3
L
−4
−5
−5
θ
−Q
x
(m)
−4 nC
Q
2 nC
3 nC
−2 nC
−3
−1 0
1
2
3
4
5
Find the magnitude of the resulting force on
the −4 nC charge at the origin [coordinates,
ragsdale (zdr82) – HW1 – ditmire – (58335)
(0 m, 0 m)].
Correct answer: 3.14273 × 10−9 N.
Explanation:
Let : qo = −4 × 10−9 C ,
(xo , yo ) = (0 m, 0 m) ,
qa = −2 × 10−9 C ,
(xa , ya ) = (1 m, −4 m) ,
qb = 3 × 10−9 C ,
(xb , yb ) = (−3 m, −3 m) ,
qc = 2 × 10−9 C , and
(xc , yc ) = (5 m, −2 m) .
Coulomb’s Law for qo and qa is
qo qa
Fao = ke
(xa − xo )2 + (ya − yo )2
= 8.98755 × 109 N m2 /C2
−4 × 10−9 C −2 × 10−9 C
×
(1 m)2 + (−4 m)2
= 4.22944 × 10−9 N .
ya
θa = arctan
x
a
−4 m
= arctan
1m
◦
= 284.036 .
Fax = Fa cos θa
= 4.22944 × 10−9 N cos 284.036◦
= −1.02579 × 10−9 N .
Fay = Fa sin θa
= 4.22944 × 10−9 N sin 284.036◦
= 4.10316 × 10−9 N .
Coulomb’s Law for qo and qb is
qo qb
Fbo = ke
2
(xb − xo ) + (yb − yo )2
= 8.98755 × 109 N m2 /C2
−4 × 10−9 C 3 × 10−9 C
×
(−3 m)2 + (−3 m)2
= −5.9917 × 10−9 N .
4
yb
θb = arctan
x
b
−3 m
= arctan
−3 m
◦
= 225 .
Fbx = Fb cos θb
= −5.9917 × 10−9 N cos 225◦
= −4.23677 × 10−9 N .
Fby = Fb sin θb
= −5.9917 × 10−9 N sin 225◦
= −4.23677 × 10−9 N .
Coulomb’s Law for qo and qc is
qo qc
Fco = ke p
(xc − xo )2 + (yc − yo )2
= 8.98755 × 109 N m2 /C2
−4 × 10−9 C 2 × 10−9 C
p
×
(5 m)2 + (−2 m)2
= −2.47932 × 10−9 N .
yc
θc = arctan
x
c
−2 m
= arctan
5m
◦
= 338.199 .
Fcx = Fc cos θc
= −2.47932 × 10−9 N cos 338.199◦
= 2.302 × 10−9 N .
Fcy = Fc sin θc
= −2.47932 × 10−9 N sin 338.199◦
= −9.20798 × 10−10 N .
ragsdale (zdr82) – HW1 – ditmire – (58335)
The magnitude of the resultant force is
y θ = 191.286◦
5
4
3
2
1
3. 0
x
−1
−2
−3
−4
−5
−5 −3 −1 0 1 2 3 4 5
1/2
F = Fx2 + Fy2
h
= (Fax + Fbx + Fcx )2
2 i1/2
+ Fay + Fby + Fcy
= −1.02579 × 10−9 N
4. None of these figures is correct.
+ −4.23677 × 10−9 N
2
2.302 × 10−9 N
+ 4.10316 × 10−9 N
−4.23677 × 10−9 N
+−9.20798 × 10−10 N
= 3.14273 × 10−9 N .
2 i1/2
007 (part 2 of 2) 10.0 points
Select the figure showing the direction of the
resultant force on the −4 nC charge at the
origin.
1.
rect
y θ = 199.604◦
5
4
3
2
1
x
0
−1
−2
−3
−4
−5
−5 −3 −1 0 1 2 3 4 5
y θ = 76.1438
5
4
3
2
1
2. 0
x
−1
−2
−3
−4
−5
−5 −3 −1 0 1 2 3 4 5
◦
5
cor-
y
5
4
3
2
1
5. 0
−1
−2
−3
−4
−5
−5 −3 −1 0 1
y
5
4
3
2
1
6. 0
−1
−2
−3
−4
−5
−5 −3 −1 0 1
y
5
4
3
2
1
7. 0
−1
−2
−3
−4
−5
−5 −3 −1 0 1
y
5
4
3
2
1
8. 0
−1
−2
−3
−4
−5
−5 −3 −1 0 1
θ = 77.6119◦
x
2345
θ = 273.235◦
x
2345
θ = 227.129◦
x
2345
θ = 239.245◦
x
2345
ragsdale (zdr82) – HW1 – ditmire – (58335)
Explanation:
The direction θ as measured in a counterclockwise direction from the positive x axis
is
Fy
θ = arctan
Fx
−1.05441 × 10−9 N
= arctan
−2.96057 × 10−9 N
= 199.604◦ .
008 10.0 points
An object having a net charge of 23.4 µC is
placed in a uniform electric field of 430 N/C
directed vertically up.
The acceleration of gravity is 9.8 m/s2 .
What is the mass of this object if it “floats”
in the field?
Correct answer: 1.02673 g.
Explanation:
= 1.02673 g .
009 (part 1 of 2) 10.0 points
Three point charges are placed at the vertices
of an equilateral triangle.
The value of the Coulomb constant is
8.9875 × 109 N · m2 /C2
−2 C
−2 C
Find the magnitude of the electric field vec~ at P .
tor kEk
Correct answer: 2.39667 × 108 N/C.
Explanation:
Let : a = 10 m ,
q = −2 C , and
k = 8.9875 × 109 N · m2 /C2 .
q
̂
ı̂
a
QE − mg = 0
P
−2 C
g = 9.8 m/s2 .
For this to hold,
ı̂
60◦
Let : Q = 23.4 µC = 2.34 × 10−5 C ,
E = 430 N/C , and
Call the vertical direction the y-direction,
so the the unit vector ̂ points up. Then
force equilibrium in the vertical direction for
a charge Q of mass m yields
X
F = Q E ̂ + m g (−̂) = 0
̂
m
keywords:
QE
g
(2.34 × 10−5 C) (430 N/C) 1000 g
×
=
9.8 m/s2
1 kg
m=
10
y (m)
5
4
3
2
1
x
0
−4 nC
(m)
−1
−2
2 nC
−3
3
nC
−4
−2 nC
−5
−5 −3 −1 0 1 2 3 4 5
6
q
P
q
Basic Concepts: Electric field.
Electric field vectors due to bottom two
charges cancel out each other. The magnitude
ragsdale (zdr82) – HW1 – ditmire – (58335)
7
of the field vector due to charge at to top of
the triangle, which gives
14.9 cm
~ =
kEk
−5.53 µC
kq
√ !2
3
a
2
4 kq
3 a2
4 (8.9875 × 109 N · m2 /C2 ) (−2 C)
=
3
(10 m)2
O
=
= 2.39667 × 108 N/C ,
√
3
where h = a cos(30 ) =
a is the height of
2
the triangle.
◦
010 (part 2 of 2) 10.0 points
~ at P .
Find the direction of the field vector E
1
1. √ (ı̂ + ̂)
2
2. ̂ correct
3. −ı̂
4. ı̂
5. −̂
If the rod has a total charge of −5.53 µC,
find the horizontal component of the electric
field at O, the center of the semicircle. Define
right as positive.
Correct answer: −1.40661 × 107 N/C.
Explanation:
Let : L = 14.9 cm = 0.149 m and
q = −5.53 µC = −5.53 × 10−6 C .
Call the length of the rod L and its charge
q. Due to symmetry
Z
Ey = dEy = 0
and
Ex =
Z
dE sin θ = ke
dq sin θ
,
r2
where dq = λ dx = λ r dθ, so that
1
6. √ (ı̂ − ̂)
2
1
7. − √ (ı̂ − ̂)
2
1
8. − √ (ı̂ + ̂)
2
y
θ
b
x
Explanation:
~ at P due to q = −2 C is
By inspection, E
along ̂ direction (see figure above).
011 10.0 points
A uniformly charged insulating rod of length
14.9 cm is bent into the shape of a semicircle
as in the figure.
The value of the Coulomb constant is
8.98755 × 109 N m2 /C2 .
Z
ke λ
Ex = −
r
Z
3π/2
cos θ dθ
π/2
3π/2
ke λ
(sin θ)
=−
r
π/2
=2
ke λ
,
r
ragsdale (zdr82) – HW1 – ditmire – (58335)
where λ =
q
L
and r = . Therefore,
L
π
2 ke q π
L2
2 (8.98755 × 109 N m2 /C2 )
=
(0.149 m)2
× (−5.53 × 10−6 C) π
Ex =
= −1.40661 × 107 N/C .
Since the rod has negative charge, the field is
pointing to the left (towards the charge distribution). A positive test charge at O would
feel an attractive force from the semicircle,
pointing to the left.
8
~ = − ke λ ∆x ı̂ correct
4. ∆E
x2
~ = − ke λ ∆x ı̂
5. ∆E
x
~ = ke λ ∆x ı̂
6. ∆E
x
Explanation:
From the fact that λ > 0 and the diagram,
the field ∆E at O due to the segment on the
x-axis is in the negative x direction, and its
magnitude is given by
∆E = ke
∆q
λ ∆x
= ke 2 .
2
x
x
We can thus express the electric field as
012 (part 1 of 3) 10.0 points
A rod of length ℓ with uniform charge per unit
length λ, where λ > 0, is placed a distance d
from the origin along the x axis. A similar
rod with the same charge is placed along the
y axis as in the figure. Consider only a small
segment of length ∆x (it is of infinitesimal
length) on the horizontal rod, which is placed
along the x axis. The segment is at a distance
x from the origin.
y
λ
ℓ
d
d
O
ℓ
x
λ
~ at the oriDetermine the electric field ∆E
gin due to this segment.
~ = ke λ ∆x ı̂
1. ∆E
2x
~ = ke λ ∆x ı̂
2. ∆E
x2
~ = − ke λ ∆x ı̂
3. ∆E
2x
~ = −∆E ı̂ .
∆E
013 (part 2 of 3) 10.0 points
~ rod 1 at the origin
Determine the electric field E
due to the entire horizontal rod.
ke λ d
ı̂
ℓ2
~ rod 1 = ke λ d ı̂
2. E
ℓ
~ rod 1 = ke λ d ı̂
3. E
ℓ (ℓ + d)
~ rod 1 = ke λ ℓ ı̂
4. E
d
~ rod 1 = ke λ ℓ ı̂
5. E
d (d + ℓ)
~ rod 1 = − ke λ d ı̂
6. E
ℓ (ℓ + d)
k
~ rod 1 = − e λ ℓ ı̂
7. E
d
~ rod 1 = ke λ ℓ ı̂
8. E
d2
~ rod 1 = − ke λ ℓ ı̂ correct
9. E
d (d + ℓ)
~ rod 1 = − ke λ d ı̂
10. E
ℓ
Explanation:
For the entire rod, we must add all the
segments, i.e., integrate from x = d to x =
~ rod 1 =
1. E
ragsdale (zdr82) – HW1 – ditmire – (58335)
d + ℓ. The electric field due to the horizontal
rod is
Z
~ rod 1 =
E
Therefore the net field at the origin is
~ tot = E
~ rod 1 + E
~ rod 2
E
ke λ ℓ
=−
(ı̂ + ̂) .
d (d + ℓ)
~
dE
d+ℓ
dx
ı̂
x2
d
d+ℓ
1 ı̂
= −ke λ −
x d
1
1
= ke λ
ı̂
−
d+ℓ d
ke λ ℓ
ı̂ .
=−
d (d + ℓ)
= −ke λ
Z
014 (part 3 of 3) 10.0 points
~ tot at the origin
Determine the electric field E
due to both rods.
~ tot = − ke λ ℓ (ı̂ + ̂) correct
1. E
d (d + ℓ)
~ tot = ke λ d (ı̂ + ̂)
2. E
ℓ2
~ tot = ke λ d (ı̂ + ̂)
3. E
ℓ
~ tot = − ke λ d (ı̂ + ̂)
4. E
ℓ (ℓ + d)
k
~ tot = e λ d (ı̂ + ̂)
5. E
ℓ (ℓ + d)
~ tot = − ke λ d (ı̂ + ̂)
6. E
ℓ
~ tot = ke λ ℓ (ı̂ + ̂)
7. E
d2
~ tot = ke λ ℓ (ı̂ + ̂)
8. E
d (d + ℓ)
~ tot = ke λ ℓ (ı̂ + ̂)
9. E
d
~ tot = − ke λ ℓ (ı̂ + ̂)
10. E
d
Explanation:
The electric field due to the rod along the
y-axis is found in a similar fashion. The field
at the origin due to each rod has a magnitude
ke λ ℓ
. The field due to the rod
of E =
d (d + ℓ)
along the y-axis is in the “−̂” direction and
the field due to the rod along the x-axis in the
“−ı̂” direction.
9
015 10.0 points
Consider two charged plates with the same
net charge on each. Imagine a proton at rest
a certain distance from a negatively charged
plate; after being released it collides with the
plate. Then imagine an electron at rest the
same distance from a positively charged plate.
In which case will the moving particle have
the greater speed when the collision occurs?
1. The proton will have the greater speed on
impact.
2. The electron will have the greater speed
on impact. correct
3. It cannot be determined.
4. The proton and the electron will have the
same speed on impact.
Explanation:
The electron will have the greater speed on
impact. The force on each will be the same,
but the electron experiences more acceleration and therefore gains more speed because
of its smaller mass.
016 10.0 points
A charge of +1.90 × 10−9 C is placed at the
origin, and another charge of +6.30 × 10−9 C
is placed at x = 1.7 m.
The Coulomb constant is 8.98755 ×
109 N · m2 /C2 .
Find the point (coordinate) between these
two charges where a charge of +3.30 × 10−9 C
should be placed so that the net electric force
on it is zero.
Correct answer: 0.602638 m.
Explanation:
Let :
q1 = 1.90 × 10−9 C at the origin ,
ragsdale (zdr82) – HW1 – ditmire – (58335)
q2 = 6.30 × 10−9 C ,
q3 = 3.30 × 10−9 C ,
x2 = 1.7 m .
and
q3 is between q1 and q2 , so if r1,3 = P , then
r2,3 = x2 − P .
q1 q2
r2
The net electric force is zero, so
Felectric = kC
kC
F1,3
q1 q3
2
r1,3
q1
2
r1,3
q1
P2
= F2,3
q2 q3
= kC 2
r2,3
q2
= 2
r2,3
q2
=
(x2 − P )2
q2
(x2 − P )2
=
2
P
q
r1
x2 − P
q2
=
P
q1
r
q2
x2 − P = P
q1
r q2
= x2
P 1+
q1
Thus
x2
r
q2
1+
q1
1.7 m
r
=
6.3 × 10−9 C
1+
1.9 × 10−9 C
P =
10
3. E = 13438.9 N/C
4. E = 8868.32 N/C
5. E = 5675.34 N/C
6. E = 52327.8 N/C
7. E = 4666.68 N/C
8. E = 0 N/C correct
9. E = 22167.1 N/C
10. E = 4137.94 N/C
Explanation:
Let : q1 = q2 = 5.23 µC = 5.23 × 10−6 C ,
x = 344.5 cm = 3.445 m , and
k = 8.98755 × 109 N · m2 /C2 .
The electric field due to a point charge a
distance x away has a magnitude E given by
E=
k |q|
.
x2
The electric field is zero because the electric
field due to each charge is equal in magnitude
and opposite in direction to that due to the
other.
018 (part 2 of 2) 10.0 points
What is the magnitude of the electric field if
one charge is positive and the other negative,
both of magnitude 5.23 µC?
= 0.602638 m .
1. E = 22657.7 N/C
017 (part 1 of 2) 10.0 points
Two positive charges of 5.23 µC each are
689 cm apart.
Find the electric field midway between
them.
2. E = 14621.2 N/C
3. E = 9264.59 N/C
4. E = 19169.3 N/C
1. E = 4407.09 N/C
5. E = 5988.35 N/C
2. E = 7921.27 N/C
6. E = 7921.27 N/C correct
ragsdale (zdr82) – HW1 – ditmire – (58335)
Find the charge on the ball.
Correct answer: 11.4235 nC.
7. E = 38794.7 N/C
Explanation:
8. E = 16703 N/C
Let : Ex
Ey
θ
mb
9. E = 9641.66 N/C
10. E = 0 N/C
Explanation:
= 2.7 × 105 N/C ,
= 3.4 × 105 N/C ,
= 32◦ , and
= 0.9 g = 0.0009 kg .
In the ı̂ and ̂ directions, force equilibrium
tells us
Let : q1 = 5.23 µC ,
q2 = −5.23 µC , and
x = 344.5 cm = 3.445 m .
q Ex − T sin θ = 0
q Ey + T cos θ − m g = 0 ,
The field is twice the electric field due to either
charge separately, since the field due to each is
equal in magnitude and in the same direction
as that due to the other. It is given by
(1)
(2)
T sin θ = q Ex
T cos θ = m g − q Ey
T sin θ
q Ex
tan θ =
=
T cos θ
m g − q Ey
k |q|
x2
8.98755 × 109 N · m2 /C2 (5.23 µC)
=2
(3.445 m)2
E=2
(m g − q Ey ) tan θ = q Ex
q Ex
m g − q Ey =
tan θ
q Ex
tan θ
q Ex
mg
−
q=
Ey
Ey tan θ
m g tan θ
q=
Ey tan θ + Ex
= 7921.27 N/C .
q Ey = m g −
keywords:
019 (part 1 of 2) 10.0 points
A charged cork ball is suspended on a light
string in the presence of a uniform electric
field as in the figure. The ball is in equilibrium.
~ = (2.7 × 105 N/C)ı̂ + (3.4 × 105 N/C) ̂ .
E
The acceleration of gravity is 9.8 m/s2 .
̂
ı̂
0. 8
m
E
11
32◦
0.9 g
q=
(0.0009 kg) (9.8 m/s2 ) tan 32◦
(3.4 × 105 N/C) tan 32◦ + (2.7 × 105 N/C)
= 11.4235 nC .
020 (part 2 of 2) 10.0 points
Find the tension in the string.
Correct answer: 0.00582042 N.
Explanation:
The first equation for force equilibrium
gives
q Ex
T =
sin θ
(11.4235 nC) (2.7 × 105 N/C)
=
sin 32◦
= 0.00582042 N .
ragsdale (zdr82) – HW1 – ditmire – (58335)
021 10.0 points
A disk of radius 8 m lies in the yz plane with
its axis along the x axis and carries a uniform
surface charge density σ.
The permittivity of free space is 8.85 ×
10−12 C2 /N · m2 .
σ
.
Find the value of x for which Ex =
4 ǫ0
Correct answer: 4.6188 m.
Explanation:
Let :
R = 8 m and
ǫ0 = 8.85 × 10−12 C2 /N · m2 .
The electric field on the x axis is
σ
σ
x
=
Ex =
1− √
2ǫ
4 ǫ0
x2 + R 2
x
1
=1− √
2
x2 + R 2
1
x
=√
2
x2 + R 2
2
2
x + R = 4 x2
R
x= √
3
8m
= √
3
= 4.6188 m .
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