oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 18 2007, 11:00 pm
Question 1, chap 22, sect 2.
part 1 of 1
10 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 15 nC. Both objects have the same mass.
1 nC is equal to 1 × 10−9 C.
As they repel,
1. sphere “1” accelerates 25 times as fast as
sphere “2”.
2. they do not accelerate at all, but rather
separate at constant velocity.
1
placed at (0.0, 28 cm), and the last is placed
at (57 cm, 0.0).
Calculate the magnitude of the force on
the charge at the origin. The value of the
Coulomb constant is 9 × 109 N m2 /C2 .
Correct answer: 7.55781 N (tolerance ± 1
%).
Explanation:
Let : q = 8 µC = 8 × 10−6 C ,
(x1 , y1 ) = (0, 28 cm) = (0, 0.28 m) ,
(x2 , y2 ) = (57 cm, 0) = (0.57 m, 0) ,
(x0 , y0 ) = (0, 0) .
and
The electric field along the y-axis is
3. sphere “2” accelerates 5 times as fast as
sphere “1”.
4. sphere “1” accelerates 5 times as fast as
sphere “2”.
5. they have the same magnitude of acceleration. correct
6. sphere “2” accelerates 25 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
ke q
+ y12
9 × 109 N m2 /C2 8 × 10−6 C
=
0 + (0.28 m)2
= 918367 N/C ,
Ey =
x21
and along the x-axis
ke q
+ y22
9 × 109 N m2 /C2 8 × 10−6 C
=
(0.57 m)2 + 0
= 221607 N/C , so
Ex =
x22
E=
where r is the distance between the centers of
the two spheres.
~ 12 k = kF
~ 21 k
kF
m1 a1 = m2 a2 .
Since both spheres have the same mass and
are subject to the same force, they have the
same acceleration.
q
Ex2 + Ey2
and
q
Ex2 + Ey2
= 8 × 10−6 C
q
× (918367 N/C)2 + (221607 N/C)2
F = qE = q
= 7.55781 N .
Question 2, chap 23, sect 1.
part 1 of 1
10 points
Question 3, chap 22, sect 2.
part 1 of 1
10 points
Three equal charges of 8 µC are in the x-y
plane. One is placed at the origin, another is
Two identical small charged spheres hang
in equilibrium with equal masses as shown in
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 18 2007, 11:00 pm
the figure. The length of the strings are equal
and the angles with the vertical are identical.
The acceleration of gravity is 9.8 m/s2
and the value of Coulomb’s constant is
8.98755 × 109 N m2 /C2 .
T
T cos θ
Fe
2
θ
θ
T sin θ
mg
0. 2
Because the sphere is in equilibrium,
X
Fx = T sin θ − Fe = 0
m
4◦
0.03 kg
T sin θ = Fe
0.03 kg
Find the magnitude of the charge on each
sphere.
Correct answer: 4.22007 × 10−8 C (tolerance
± 1 %).
Explanation:
Let : L = 0.2 m ,
m = 0.03 kg ,
θ = 4◦ .
and
X
Fy = T cos θ − m g = 0
T cos θ = m g .
Dividing,
T sin θ
Fe
=
T cos θ
mg
Fe = m g tan θ
= (0.03 kg) 9.8 m/s2 tan(4◦ )
= 0.0205585 N
tan θ =
for the electric force.
From Coulomb’s law, the electric force between the charges has magnitude
|Fe | = ke
m
θ
a
m
|q|2
,
r2
where |q| is the magnitude of the charge on
each sphere. (The term |q|2 arises here because the charge is the same on both spheres.)
Thus
L
q
and
q
From the right triangle in the figure above,
we see that
a
L
a = L sin θ
= (0.2 m) sin(4◦ )
= 0.0139513 m .
sin θ =
The separation of the spheres is r = 2 a =
0.0279026 m . Consider the forces acting on
the left sphere:
|q| =
=
s
|Fe | r 2
ke
s
(0.0205585 N) (0.0279026 m)2
8.98755 × 109 N m2 /C2
= 4.22007 × 10−8 C .
Question 4, chap 23, sect 2.
part 1 of 1
10 points
A circular arc has a uniform linear charge
density of −6 nC/m.
The value of the Coulomb constant is
8.98755 × 109 N · m2 /C2 .
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 18 2007, 11:00 pm
y
In polar coordinates
135 ◦
2. 7
3
dq = λ (r dθ) ,
m
x
What is the magnitude of the electric field
at the center of the circle along which the arc
lies?
Correct answer: 36.9041 N/C (tolerance ± 1
%).
Explanation:
Let : λ = −6 nC/m = −6 × 10−9 C/m ,
∆θ = 135◦ , and
r = 2.7 m .
where λ is the linear charge density. The
positive y axis is θ = 90◦ , so the y component
of the electric field is given by
dEy = dE sin θ .
Note: By symmetry, each half of the arc
about the y axis contributes equally to the
electric field at the origin. Hence, we may just
consider the right-half of the arc (beginning
on the positive y axis and extending towards
the positive x axis) and double that result.
Note: The upper angular limit θ = 90◦ .
The lower angular limit θ = 90◦ − 67.5 ◦ =
22.5◦ , is the angle from the positive x axis to
the right-hand end of the arc.
!
Z ◦
λ 90
E = −2 ke
sin θ dθ ̂
r 22.5 ◦
= −2 ke
λ
[cos (22.5◦ ) − cos 90◦ ] ̂ .
r
Since
θ is defined as the angle in the counterclockwise direction from the positive x axis as
shown in the figure below.
67
◦
.5 ◦
r
.5
67
θ
~
E
First, position the arc symmetrically
around the y axis, centered at the origin. By
symmetry (in this rotated configuration) the
field in the x direction cancels due to charge
from opposites sides of the y-axis, so
Ex = 0 .
For a continuous linear charge distribution,
Z
dq
~
E = ke
r̂ .
r2
ke
λ
= (8.98755 × 109 N · m2 /C2 )
r
−6 × 10−9 C/m
×
2.7 m
= −19.9723 N/C ,
E = −2 (−19.9723 N/C)
× [cos(22.5◦ ) − cos 90◦ ] ̂
= (36.9041 N/C) ̂
~ = 36.9041 N/C .
kEk
~ in a
Alternate Solution: Solve for kEk
straightforward manner, positioning the beginning of the arc on the positive x axis (as
shown in the original figure in the question).
θ is still defined as the angle in the counterclockwise direction from the positive x axis.
!
◦
Z
ke λ 135
cos θ dθ ı̂
Ex = −
r 0◦
ke λ
(sin 135◦ − sin 0◦ )ı̂
r
= −(−19.9723 N/C) (sin 135◦ − sin 0◦ )ı̂
= (14.1226 N/C) ı̂ ,
=−
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 18 2007, 11:00 pm
4
!
◦
Z
ke λ 135
Ey = −
sin θ dθ ̂
Question 6, chap 23, sect 4.
r 0◦
part 1 of 1
10 points
ke λ
(cos 0◦ − cos 135◦ ) ̂
=−
A particle of mass 0.000138 g and charge
r
◦
◦
95 mC moves in a region of space where the
= −(−19.9723 N/C) (cos 0 − cos 135 ) ̂
electric field is uniform and is 4 N/C in the x
= (34.0949 N/C) ̂ ,
direction and zero in the y and z direction.
If the initial velocity of the particle is given
q
by vy = 2 × 106 m/s, vx = vz = 0, what is the
~ = E2 + E2
kEk
speed of the particle at 0.8 s?
x
y
h
Correct answer: 2.97536 ×106 m/s (tolerance
2
= (14.1226 N/C)
± 1 %).
i1/2
Explanation:
+ (34.0949 N/C)2
= 36.9041 N/C .
Let :
Question 5, chap 23, sect 1.
part 1 of 1
10 points
From the electric field vector at a point, one
can determine which of the following?
I) the direction of the electrostatic force on
a test charge of known sign at that point;
II) the magnitude of the electrostatic force
exerted per unit charge on a test charge
at that point;
III) the electrostatic charge at that point.
1. II and III only
2. I, II and III
3. I and II only correct
4. I only
5. III only
Explanation:
The definition of the electrostatic force is
~
~ = F or F
~ = q E.
~ This means F
~ is in the
E
q
~ depending on
same or opposite direction to E,
the sign of the charge. And if we only consider
the magnitude F = q E for a unit charge, the
F
= E, which is statement
force on it is just
q
II.
m = 0.000138 g = 1.38 × 10−7 kg ,
Ex = 4 N/C ,
Ey = Ez = 0 ,
vy = 2 × 106 m/s ,
vx = vz = 0 , and
t = 0.8 s .
According to Newton’s second law and the
definition of an electric field,
~ = m~a = q E
~.
F
Since the electric field has only an x component, the particle accelerates only in the x
direction:
q Ex
ax =
.
m
To determine the x component vxf of the
final velocity, use the kinematic relation
vxf = vxi + ax (tf − ti ) = ax tf
since ti = 0 and vxi = 0 .
q E x tf
m
(0.095 C) (4 N/C)(0.8 s)
=
1.38 × 10−7 kg
= 2.2029 × 106 m/s .
vxf =
No external force acts on the particle in the
y direction so vyi = vyf = 2 × 106 m/s and
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 18 2007, 11:00 pm
Basic Concepts: Field patterns of point
charge and parallel plates of infinite extent.
The force on a charge in the electric field is
given by
~ = qE
~
F
the final speed is given by
vf =
q
2 + v2
vyf
xf
= (2 × 106 m/s)2
+(2.2029 × 106 m/s)2
= 2.97536 × 106 m/s .
1/2
Note: This is analogous to a particle in a
gravitational field with the coordinates roπ
tated clockwise by (90◦ ).
2
Question 7, chap -1, sect -1.
part 1 of 2
10 points
A dipole (electrically neutral) is placed in
an external field.
(a)
−
+
−
+
(c)
(b)
+
−
−
+
(d)
For which situation(s) shown above is the
net force on the dipole equal to zero?
1. (a) only
2. (c) only
3. None of these
4. (b) and (d)
5. (a) and (c)
6. Another combination
7. (c) and (d) correct
Explanation:
5
and the torque is defined as
~ = ~r × F
~
T
~ = k ∆q r̂
∆E
r2
X
~ =
~i.
E
∆E
Symmetry of the configuration will cause
some component of the electric field to be
zero.
Gauss’ law states
I
~ · dA
~ = Q.
ΦS = E
ǫ0
Solutions: The electric dipole consists of
two equal and opposite charges separated by
a distance. In either situation (c) or (d), the
electric field is uniform everywhere between
the parallel infinite plates. Thus, the electric
force on one charge is equal but opposite to
that on another so that the net force on the
whole dipole is zero. By contrast, electric
fields are nonuniform for situations both (a)
and (b).
Question 8, chap -1, sect -1.
part 2 of 2
10 points
For which situation(s) shown above is the
net torque on the dipole equal to zero?
1. (c) and (d)
2. (a) and (b)
3. None of these
4. (b) and (d)
5. (a) and (c) correct
6. (c) only
7. other combination
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 18 2007, 11:00 pm
8. (a) only
Explanation:
A electric dipole can be regarded as a pair
of charges of opposite sign. Only in figures
(a) and (c), the electric fields are along the
direction of ~r , where ~r is the vector between
~ is
the pair of charges. Therefore the force F
also along ~r . This will lead to zero torque,
since
~ = ~r × F
~ ∝ ~r × ~r = 0 .
T
For figures (b) and (d), the torque on both
charges are nonzero and the resultant torques
are also nonzero.
Question 10, chap 24, sect 5.
part 1 of 1
10 points
A point charge q1 is concentric with two
spherical conducting thick shells, as shown
in the figure below. The smaller spherical
conducting shell has a net charge of q2 and
the larger spherical conducting shell has a net
charge of q3 .
r4
q3
r3
r2
r1
q2
Question 9, chap 24, sect 2.
part 1 of 1
10 points
Five charges are placed in a closed box.
Each charge (except the first) has a magnitude which is twice that of the previous
one placed in the box. All charges have the
same sign and (after all the charges have been
placed in the box) the net electric flux through
the box is 7.5 × 107 N · m2 /C.
What is the magnitude of the smallest
charge in the box?
Correct answer: 21.4214 µC (tolerance ± 1
%).
Explanation:
Let : Φ = 7.5 × 107 N · m2 /C .
Let the first charge in the box be q. Then
the other four charges have magnitude 2 q, 4 q,
qtotal
8 q, and 16 q. By Gauss’ law, Φ =
. Our
ǫ0
closed box is a Gaussian surface and we know
the total flux through this closed surface, so
(1 + 2 + 4 + 8 + 16) q
31 q
=
ǫ0
ǫ0
1
q = (7.5 × 107 N · m2 /C)
31
× (8.85419 × 10−12 C2 /N · m2 )
1 × 106 µC
×
1C
Φ=
= 21.4214 µC .
6
q1
R1
R2 R3 R4 R5
What is the charge Qr3 on the inner surface of the larger spherical conducting shell?
Under static conditions, the charge on a conductor resides on the surface of the conductor.
1. Qr3 = 0
2. Qr3 = +q1 − q2
3. Qr3 = −q1 − q2 − q3
4. Qr3 = −q1 − q2 correct
5. Qr3 = +q1
6. Qr3 = −q1
7. Qr3 = −q1 + q2
8. Qr3 = +q1 + q2
9. Qr3 = +q1 + q2 + q3
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 18 2007, 11:00 pm
10. Qr3 = −q1 − q2 + q3
Explanation:
The net charge inside a Gaussian surface
located at r = R4 must be zero, since the field
in the conductor must be zero. Therefore,
the charge on the inner surface of the large
spherical conducting shell must be −(q1 + q2 ).
Question 11, chap 24, sect 3.
part 1 of 1
10 points
A uniformly charged sphere (an insulating
sphere with radius R) is shown in the figure
below.
R
7
Pick a spherical Gaussian surface with raR
dius r =
concentric with the sphere of
2
charge Q.
From Gauss’ law, Φ = E · 4 π r 2 .
3Q
Q
=
ρ =
, so the net charge
4
4 π R3
3
πR
3
R
is
inside the Gaussian surface of radius
2
4
3
πr
qen = ρ
3
3Q
4
3
=
πr
4 π R3 3
r3
=Q 3,
R
and the electric field at r is
Φc
4 π r2
Q r3 1
=
ǫ0 R3 4 π r 2
Qr
=
4 π ǫ0 R3
p
E=
R
2
If Q = 8.1 × 10−6 C is the total charge
~
inside the sphere, what is the magnitude kEk
1
R
.
of the electric field at r = ? k =
2
4 π ǫ0
~ = kQ
1. kEk
R2
~ = kQ
2. kEk
8 R2
~ = kQ
3. kEk
32 R2
~ = kQ
4. kEk
4 R2
~ = 4kQR
5. kEk
~ = 2kQR
6. kEk
~ = kQ
7. kEk
16 R2
~ = k Q correct
8. kEk
2 R2
~ = kQR
9. kEk
~ = 8kQR
10. kEk
Explanation:
and at r =
R
2
E=
kQ
Q
1
=
.
2
4 π ǫ0 2 R
2 R2
Question 12, chap 25, sect 1.
part 1 of 2
10 points
Consider the figure
+Q
+
+
+
+
+
+
+ C
+
+
+
+
#1
A −Q
− y
−
−
−
−
−
D −
−
−
−
−
B #2
x
Of the following elements, identify all that
correspond to an equipotential line or surface.
1. neither AB nor CD
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 18 2007, 11:00 pm
2. both AB and CD
8
C
3. line CD only
A
B
4. line AB only correct
−
+
Explanation:
Consider the electric field
+Q
A −Q
+
− y
+
−
+
−
+
−
+
−
+ C D −
+
−
+
−
+
−
+
−
+
−
#1
B #2
D
An equipotential line or surface (CD) is
normal to the electric field lines.
x
An equipotential line or surface (AB) is
normal to the electric field lines.
Question 14, chap 25, sect 3.
part 1 of 1
10 points
Through what potential difference would
an electron need to be accelerated for it to
achieve a speed of 3 % of the speed of light
(2.99792 × 108 m/s), starting from rest?
Correct answer: 229.95 V (tolerance ± 1 %).
Explanation:
Question 13, chap 25, sect 1.
part 2 of 2
10 points
Consider the figure
C
A
−
−q
B
+
+q
D
Of the following elements, identify all that
correspond to an equipotential line or surface.
1. both AB and CD
2. neither AB nor CD
3. line CD only correct
4. line AB only
Let : s = 3 % = 0.03 ,
c = 2.99792 × 108 m/s ,
me = 9.10939 × 10−31 kg ,
qe = 1.60218 × 10−19 C .
and
The speed of the electron is
v = 0.03 c
= 0.03 2.99792 × 108 m/s
= 8.99377 × 106 m/s ,
By conservation of energy
1
me v 2 = −(−qe ) ∆V
2
v2
∆V = me
2 qe
= 9.10939 × 10−31 kg
2
8.99377 × 106 m/s
×
2 (1.60218 × 10−19 C)
= 229.95 V .
Explanation:
Consider the electric field:
Question 15, chap 25, sect 2.
part 1 of 3
10 points
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 18 2007, 11:00 pm
Consider a solid conducting sphere with a
radius a and charge Q1 on it. There is a
conducting spherical shell concentric to the
sphere. The shell has an inner radius b (with
b > a) and outer radius c and a net charge
Q2 on the shell. Denote the charge on the
inner surface of the shell by Q′2 and that on
the outer surface of the shell by Q′′2 .
b , Q′2
Q1 , a
P
Q′′2 , c
Q2
Q1
Find the charge Q′′2 .
1. Q′′2 =
Q1 − Q2
2
2. Q′′2 = 2 (Q2 − Q1 )
3. Q′′2 = Q2 − Q1
4. Q′′2 =
Q2 − Q1
2
5. Q′′2 = 2 (Q1 + Q2 )
6. Q′′2 = Q1 − Q2
7. Q′′2 =
Q1 + Q2
2
8. Q′′2 = Q1 + Q2 correct
9. Q′′2 = 2 (Q1 − Q2 )
Explanation:
Basic Concepts: Gauss’ Law
Sketch a concentric Gaussian surface S
(dashed line) within the shell.
9
r
Since the electrostatic field in a conducting
medium is zero, according to Gauss’s Law,
ΦS =
Q1 + Q′2
=0
ǫ0
Q′2 = −Q1
But the net charge on the shell is
Q2 = Q′2 + Q′′2 ,
so the charge on the outer surface of the shell
is
Q′′2 = Q2 − Q′2
= Q2 + Q1 .
Question 16, chap 25, sect 2.
part 2 of 3
10 points
Find the
magnitudeof the electric field at
~ P k ≡ EP , where the distance
point P kE
a+b
from P to the center is r =
.
2
4 ke Q 1
1. EP =
correct
(a + b)2
2 ke (Q1 − Q2 )
2. EP =
(a + b)2
4 ke (Q1 − Q2 )
3. EP =
(a + b)2
4 ke (Q1 + Q2 )
4. EP =
(a + b)2
2 ke (Q1 + Q2 )
5. EP =
(a + b)2
2 ke Q 1
6. EP =
(a + b)2
2 ke Q 2
7. EP =
(a + b)2
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 18 2007, 11:00 pm
4 ke Q 2
8. EP =
(a + b)2
9. EP = 0
Explanation:
Choose as your Gaussian surface concentric
with the spherical surface S, which passes
through P . Here,
Z
~ ·A
~ = 4 π r 2 EP
E
Q1
ǫ0
Q1
EP =
4 π ǫ0 r 2
ke Q 1
=
r2
4 ke Q 1
=
.
(a + b)2
=
Question 17, chap 25, sect 2.
part 3 of 3
10 points
10
Using the superposition principle, adding
the 3 concentric charge distributions; i.e., Q1
at a, −Q1 at b and Q1 + Q2 at c, gives
Z
~ r · d~r , by symmetry,
VP = − E
Z
= − Er dr
Z c
Z b
Z (a+b)/2
=−
Er dr −
0 dr −
Er dr
∞
c
b
Z c
k (Q1 + Q2 )
=−
dr
r2
∞
Z b
Z (a+b)/2
k (Q1 + Q2 )
dr
−
0 dr −
r2
c
b
c
1 = −k (Q1 + Q2 )
r ∞
(a+b)/2
1 −0 − k (Q1 + Q2 )
r b
=
2 ke Q1 ke Q1 ke (Q1 + Q2 )
.
−
+
a+b
b
c
Assume: The potential at r = ∞ is zero.
Find the potential VP at point P .
Question 18, chap 25, sect 3.
part 1 of 2
10 points
2 ke Q1 ke Q1 ke (Q1 + Q2 )
1. VP =
−
+
cora+b
b
c
rect
The figure below shows two particles, each
with a charge of +Q, that are located at the
opposite corners of a square of side d.
Q
+
2 ke Q 1 2 ke Q 2
−
a+b
b
2 ke Q 1
3. VP =
a+b
2 ke Q1 ke Q1 ke (Q1 − Q2 )
4. VP =
+
−
a+b
b
c
2 ke (Q1 − Q2 )
5. VP =
a+b
2. VP =
d
+
Q
P
What is the direction of the net electric field
at point P ?
6. VP = 0
2 ke Q 1 ke Q 2
7. VP =
+
a+b
c
2 ke Q 1 ke Q 2
−
8. VP =
a+b
c
ke Q 1 ke Q 2
−
9. VP =
a+b
b
Explanation:
1.
2.
3.
correct
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 18 2007, 11:00 pm
11
The two charges Q are fixed at the vertices
of an equilateral triangle with sides of length
a as shown.
q
4.
5.
a
Explanation:
At point P , the electric fields due to the two
positive charged particles have the same magnitude. One points downward, and the other
horizontally to the left. Thus the direction of
the net electric field points left and downward,
forming a 45◦ angle with the horizontal.
Q
What is the potential energy of a particle
of charge q that is held at point P ?
1. U =
1 qQ
correct
2 π ǫ0 d
2. U = 0
√
2 qQ
2 π ǫ0 d
√
2 qQ
4. U =
4 π ǫ0 d
1 qQ
5. U =
4 π ǫ0 d
3. U =
Explanation:
The potential energy of the particle charged
+q is the sum of the potential energies due to
the two +Q charged particles.
a
Q
How much work is required to move a
charge q from the other vertex to the center of the line joining the fixed charges?
1. W =
Question 19, chap 25, sect 3.
part 2 of 2
10 points
a
2. W =
3. W =
4. W =
5. W =
6kQq
a
4kQq
a
2kQq
correct
a
kQq
a
√
2kQq
a
6. W = 0
Explanation:
Qq
Qq
Qq
+k
= +2 k
a
a
a
Qq
Qq
Qq
U2 = +k a + k a = +4 k
a
2
2
Qq
W = U2 − U1 = 2 k
.
a
U1 = +k
Question 21, chap 25, sect 4.
part 1 of 1
10 points
U = U1 + U2
1 qQ
1 qQ
=
+
4 π ǫ0 d
4 π ǫ0 d
1 qQ
.
=
2 π ǫ0 d
If the potential in a region is given by the
function
Question 20, chap 25, sect 3.
part 1 of 1
10 points
what is the y-component of the electric field
at the point P = (x′ , y ′ , z ′ )?
V = 2 x − y 2 − cos(z) ,
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 18 2007, 11:00 pm
12
Explanation:
′
1. Ey = sin(z )
Let : λ = 1.2 µC/m = 1.2 × 10−6 C/m ,
R = 4.2 m , and
ke = 8.98755 × 109 N · m2 /C2 .
2. Ey = − sin(z ′ )
3. Ey = 2
4. Ey = 2 y ′ correct
5. Ey =
y ′3
3
R
2R
6. Ey = −2 y ′
7. Ey = y ′2
8. Ey =
x′2
4
9. Ey = cos(z ′ )
10. Ey = 2 x′
Explanation:
The electric field is the gradient of the potential, so the y-component of the E-field evaluated at P is
∂V
Ey = −
∂y
∂ =−
2 x − y 2 − cos(x)
∂y
= − [−2 y]
= 2y.
Question 22, chap 25, sect 3.
part 1 of 1
10 points
p
2R
Let p be the origin. Consider the potential
due to the line of charge to the right of p.
Z
Vright = dV
Z
dq
= ke
r
Z 3R
λdx
= ke
x
R
3 R
= ke λ ln x = ke λ ln 3 .
R
By symmetry, the contribution from the line
of charge to the left of p is the same. The
contribution from the semicircle is
Z π
λRdθ
Vsemi = ke
R
0
Z π
= ke λ
dθ
0
π
= ke λ θ 0
A wire that has a uniform linear charge
density of 1.2 µC/m is bent into the shape as
shown below, with radius 4.2 m.
4.2 m
8.4 m
p
8.4 m
The value of the Coulomb constant is
8.98755 × 109 N · m2 /C2 .
Find the electrical potential at point p.
Correct answer: 57579.5 V (tolerance ± 1
%).
= ke λ π .
Thus the electric potential at p is
Vp = Vright + Vlef t + Vsemi
= 2 ke λ ln 3 + ke λ π
= ke λ (2 ln 3 + π)
= (8.98755 × 109 N · m2 /C2 )
× (1.2 × 10−6 C/m)
× (2 ln 3 + π)
= 57579.5 V .