midterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 20 2007, 11:00 pm
E & M - Basic Physical Concepts
Current and resistance
Current: I = ddtQ = n q vd A
Ohm’s law: V = I R, E = ρJ
I , R = ρℓ
E = Vℓ , J = A
A
Electric force and electric field
Electric force between 2 point charges:
|q | |q |
|F | = k 1r2 2
k = 8.987551787 × 109 N m2 /C2
ǫ0 = 4 π1 k = 8.854187817 × 10−12 C2 /N m2
qp = −qe = 1.60217733 (49) × 10−19 C
mp = 1.672623 (10) × 10−27 kg
me = 9.1093897 (54) × 10−31 kg
~
~ =F
Electric field: E
2
Power: P = I V = VR = I 2 R
Thermal coefficient of ρ: α = ρ ∆ρ
0 ∆T
Motion of free electrons in an ideal conductor:
a τ = vd → qmE τ = nJq → ρ = n qm2 τ
|Q|
~2 + · · ·
~ =E
~1 + E
Point charge: |E| = k r2 , E
Field patterns: point charge, dipole, k plates, rod,
spheres, cylinders,. . .
Charge distributions:
Linear charge density: λ = ∆Q
∆x
Surface charge density: σsurf =
∆Q
Volume charge density: ρ = ∆V
∆Qsurf
∆A
Electric flux and Gauss’ law
~ · n̂∆A
Flux: ∆Φ = E ∆A⊥ = E
Gauss law: Outgoing Flux from S, ΦS = Qenclosed
ǫ0
Steps: to obtain electric field
~ pattern and construct S
–Inspect E
H
~ · dA
~ = Qencl , solve for E
~
–Find Φs = surf ace E
ǫ
0
Spherical: Φs = 4 π r2 E
Cylindrical: Φs = 2 π r ℓ E
Pill box: Φs = E ∆A, 1 side; = 2 E ∆A, 2 sides
σ
k
~ = 0, Esurf
Conductor: E
= 0, E ⊥ = surf
in
surf
Potential
ǫ0
Potential energy: ∆U = q ∆V 1 eV ≈ 1.6 × 10−19 J
Positive charge moves from high V to low V
Point charge: V = krQ V = V1 + V2 = . . .
1 q2
Energy of a charge-pair: U = k rq12
Potential difference: |∆V | = |E ∆sk |,
R
~ · ∆~s, V − V = − B E
~ · d~s
∆V = −E
B
A
A
¯
¯
d
V
∆V
E = − dr , Ex = − ∆x ¯
= − ∂V
∂x , etc.
f ix y,z
Capacitances
Q=CV
Series: V = CQ = CQ + CQ + CQ + · · ·, Q = Qi
eq
1
2
3
Parallel: Q = Ceq V = C1 V + C2 V + · · ·, V = Vi
ǫ A
Q
Parallel plate-capacitor: C = V
= EQd = 0d
2
RQ
Q
Energy: U = 0 V dq = 12 C , u = 12 ǫ0 E 2
2
1
2
Uκ = 21κ Q
C0 , uκ = 2 ǫ0 κ Eκ
Q
Q
Spherical capacitor: V = 4 π ǫ r1 − 4 π ǫ r2
0
0
~
Potential energy: U = −~
p·E
Dielectrics: C = κC0 ,
V =IR
Direct current circuits
q
Area charge density: σA = ∆Q
∆A
1
Series: V = I Req = I R1 + I R2 + I R3 + · · ·, I = Ii
V + V + V + · · ·, V = V
Parallel: I = RV = R
i
R2
R3
eq
1
Steps: in application of Kirchhoff’s Rules
–Label currents: i1 , i2 , i3 , . . .
P
P
i
–Node equations:
i =
P in
Pout
–Loop equations: “ (±E) + (∓iR)=0”
–Natural: “+” for loop-arrow entering − terminal
“−” for loop-arrow-parallel to current flow
RC circuit: if ddty + R1C y = 0, y = y0 exp(− RtC )
Charging: E − Vc − R i = 0, 1c ddtq + R ddti = ci + R ddti = 0
Discharge: 0 = Vc − R i = qc + R ddtq , ci + R ddti = 0
Magnetic field and magnetic force
µ0 = 4 π × 10−7 T m/A
µ a2 i
µ i
0
Wire: B = 2 π0 r
Axis of loop: B =
2 (a2 +x2 )3/2
~ → q ~v × B
~
Magnetic force: F~M = i ~ℓ × B
~ × B,
~
Loop-magnet ID: ~τ = i A
µ
~ = i A n̂
2
r
Circular motion: F = mrv = q v B, T = f1 = 2 π
v
~ + q ~v × B
~
Lorentz force: F~ = q E
~
Hall effect: V = FM d , U = −~
µ·B
H
q
~ and magnetism of matter
Sources of B
µ
~
µ
v ×r̂
0 q~
~ = 0 i ∆ℓ×r̂
Biot-Savart Law: ∆B
4 π r2 , B = 4 π r2
2
µ0 i ∆y
∆B = 4 π
sin θ, sin θ = ar , ∆y = r a∆θ
r2
H
~ · d~s = µ I
B
Ampere’s law: M =
L
0
encircled
Steps: to obtain magnetic field
~ pattern and construct loop L
–Inspect B
~
–Find M and Iencl , and solve for B.
d (E A)
ΦE = ǫ
Displ. current: Id = ǫ0 d dt
0
dt
Magnetism in atom:
Orbital motion: µ = i A = 2 em L
L = m v r = n h̄,
QA
= d dt
h̄ = 2hπ = 1.06 × 10−34 J s
h̄ = 9.27 × 10−24 J/T
µB = 2em
µspin = µB
Magnetism in matter:
0
B = B0 + BM = (1 + χ) B0 = (1 + χ) µ0 B
µ0 = κm H
Ferromagnetic: χ ≫ 1
Diamagnetic: −1 ≪ χ < 0
Paramagnetic: 0 < χ ≪ 1, M = C
TB
µorbit = n µB ,
Spin: S = h̄2 ,
midterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 20 2007, 11:00 pm
2
.
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midterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 20 2007, 11:00 pm
3
Question 1, chap 25, sect 3.
part 1 of 1
10 points
Question 2, chap 25, sect 3.
part 1 of 1
10 points
A linear charge of nonuniform density
λ(x) = b x C/m, where b = 1.3 nC/m2 , is
distributed along the x-axis from 1.2 m to
4.7 m.
Determine the electric potential (relative to
zero at infinity) of the point y = 3 m on the
positive y-axis.
Correct answer: 27.3955.
Three charges are situated at three corners
of a rectangle, as shown.
The value of the Coulomb constant is
8.98755 × 109 N m2 /C2 and the acceleration
of gravity is 9.8 m/s2 .
7.4 µC
+
7.9 cm
Explanation:
2.8 cm
Let : b = 1.3 nC/m2 = 1.3 × 10−9 C/m2 ,
x1 = 1.2 m ,
x2 = 4.7 m ,
y = 3 m , and
ke = 8.98755 × 109 N m2 /C2 .
The potential is
dq
,
r
the distance from the point on the y axis to a
charge element is
p
r = x2 + y 2 ,
V = ke
Z
and the differential charge element is
dq = λ dx = b x dx .
+
1.5 µC
+
3.6 µC
How much electrical potential energy would
be expended in moving the 7.4 µC charge to
infinity?
Correct answer: 6.41954.
Explanation:
Let :
q1 = 7.4 µC = 7.4 × 10−6
q2 = 1.5 µC = 1.5 × 10−6
q3 = 3.6 µC = 3.6 × 10−6
r1,2 = 2.8 cm = 0.028 m ,
r2,3 = 7.9 cm = 0.079 m ,
9
V = ke
dq
r
x2
x dx
p
x2 + y 2
x1
x2
p
= k e b x2 + y 2 x1
q
q
2
2
2
2
x2 + y − x1 + y
= ke b
= 8.98755 × 109 N m2/C2
× 1.3 × 10−9 C/m2
q
(4.7 m)2 + (3 m)2
×
= ke b
Z
and
ke = 8.98755 × 10 N m /C2 .
Thus
Z
C,
C,
C,
q
− (1.2 m)2 + (3 m)2
= 27.3955 V .
r1,3 =
q
2
2 + r2
r1,2
2,3
U1,tot = U1,2 + U1,3
q1 q2
q1 q3
= ke
+ ke
r1,2
r1,3


q
q
2
3

= ke q1 
+q
r1,2
2
2
r +r
1,2
9
2,3
= (8.98755 × 10 N m /C2 )
1.5 × 10−6 C
−6
× (7.4 × 10 C)
0.028 m
#
3.6 × 10−6 C
+p
(0.028 m)2 + (0.079 m)2
= 6.41954 J .
2
midterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 20 2007, 11:00 pm
Question 3, chap 24, sect 5.
part 1 of 1
10 points
Consider a conducting spherical shell with
inner radius 0.7 m and outer radius 1.5 m.
There is a net charge 4 µC on the shell. At
its center, within the hollow cavity, there is a
point charge 2 µC.
0.7 m ,
Q′2 inside
2 µC
S
Q′′2 outside,
1.5 m
4 µC
0.1 m
Determine the flux through the spherical
Gaussian surface S, which has a radius of
0.1 m. The permitivity of free space is
8.8542 × 10−12 C2 /N m2.
Correct answer: 225882.
Explanation:
4
distance, what is the magnitude of the new
electric field?
Correct answer: 9064.
Explanation:
Let :
E = 1133 N/C ,
V ′ = 2 V , and
1
d′ = d .
4
The electric field between two parallel conV
ducting plates is E = − , where V is the
d
voltage between the plates, and d is the distance between the plates, so the new electric
field has a magnitude of
2V
V
V′
′
=8 −
E =− ′ =−
d
d
d
4
= 8E
= 8 (1133 N/C)
= 9064 N/C .
Let : q1 = 2 µC = 2 × 10−6 C and
ǫ0 = 8.8542 × 10−12 C2 /N m2 .
Question 5, chap 25, sect 3.
part 1 of 1
10 points
The surface S encloses only the point charge
q1 , so by Gauss’ Law
I
~ · dA
~ = qencl = q1
ΦS =
E
ǫ0
ǫ0
S
−6
2 × 10 C
=
8.8542 × 10−12 C2 /N m2
Consider a solid conducting sphere with an
inner radius R1 surrounded by a concentric
thick conducting spherical shell which has an
inner radius R2 and outer radius R3 . There is
a charge Q on the sphere and no net charge
on the shell.
For this problem, we adopt the standard
convention of setting the electric potential at
infinity to zero.
q2 = 0
= 225882 N m2 /C .
Question 4, chap 25, sect 4.
part 1 of 1
10 points
Two parallel conducting plates are connected to a constant voltage source. The
magnitude of the electric field between the
plates is 1133 N/C.
If the voltage is doubled and the distance
1
between the plates is reduced to the original
4
R1
R2
Ob
R3
b
B
bA
q1 = Q
Find the potential at O, where RO < R1 .
midterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 20 2007, 11:00 pm
1. VO =
2. VO =
3. VO =
4. VO =
5. VO =
6. VO =
7. VO =
kQ kQ
+
R2
R1
kQ kQ kQ
−
+
correct
R3
R2
R1
√
2kQ
R1
2kQ
R1
kQ kQ
+
R3
R1
kQ
R1
2kQ
R1 + R2
8. VO = ∞
9. VO = 0
Thus the potential due to the inner shell
(radius R2 ) is
V2 = k
qinner = −Q .
Since the net charge on the shell is zero, the
charge on the outer surface must be +Q for
the inner and outer charges to add up to zero:
qouter = +Q .
Now that we know the exact distribution
of charge, we can utilize the expression for
potential inside of a spherical charge distribution:
q
V =k ,
a
for a thin shell of radius a.
qinner
Q
= −k
R2
R2
and that due to the outer shell (radius R3 ) is
V3 = k
Q
qouter
=k
,
R3
R3
so the contribution from the two surfaces of
the shell is
V1 = k
Q
Q
−k
.
R3
R2
However, we are now looking at a point
inside the sphere. The charge is all on the
surface of the sphere, so similarly to the situation for the shell we have
√
2kQ
10. VO =
R1
Explanation:
We are still inside the spherical shell. The
potential due to the shell requires knowing
something about where charge exists on the
shell. To this end, consider a Gaussian spherical surface inside the shell. This surface can
have no flux through it since no electric field
can be upheld inside a conductor. By Gauss’s
Law, there can therefore be no charge enclosed. But we are already enclosing Q on
the sphere, so a charge −Q must reside on the
inner surface of the shell:
2
5
V =k
Q
R1
everywhere inside the sphere.
Thus the total potential at O is
VO = k
Q
Q
Q
−k
+k
.
R3
R2
R1
This potential is the same at O as it is everywhere inside the sphere. A conductor is an
equipotential object.
Question 6, chap 23, sect 1.
part 1 of 1
10 points
Two charges are located in the (x, y) plane
as shown in the figure below. The fields produced by these charges are observed at a point
p with coordinates (0, 0).
The value of the Coulomb constant is
8.98755 × 109 N · m2 /C2 .
p
1.7 m
1.7 m
−8.7 C
9C
2.7 m
2.2 m
midterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 20 2007, 11:00 pm
Use Coulomb’s law to find the x-component
of the electric field at p.
Correct answer: 1.47281 × 1010 .
Explanation:
Let :
(xp , yp ) = (0, 0) ,
(x1 , y1 ) = (2.2 m, −1.7 m) ,
(x2 , y2 ) = (−2.7 m, −1.7 m) ,
q1 = −8.7 C ,
q2 = 9 C , and
ke = 8.98755 × 109 N · m2 /C2 .
q
x21 + y12 ,
q
= (2.2 m)2 + (−1.7 m)2 ,
r1 =
= 2.78029 m and
q
r2 = x22 + y22 ,
q
= (−2.7 m)2 + (−1.7 m)2 ,
= 3.19061 m .
p
y2
y1
q2
q1
x2
x1
= − 8.98755 × 109 N · m2 /C2
2.2 m
−8.7 C
×
2
(2.78029 m) 2.78029 m
= 8.00413 × 109 N/C and
6
q2
cos θ2
r22
q2 x2
= −ke 2
r2 r2
Ex2 = −ke
= − 8.98755 × 109 N · m2 /C2
−2.7 m
9C
×
(3.19061 m)2 3.19061 m
= 6.72397 × 109 N/C , so
Ex = Ex1 + Ex2
= 8.00413 × 109 N/C
+ 6.72397 × 109 N/C
= 1.47281 × 1010 N/C .
Question 7, chap 22, sect 2.
part 1 of 1
10 points
Two electrostatic point charges of
+53.0 µC and +41.0 µC exert a repulsive
force on each other of 194 N.
The value of the Coulomb constant is
8.98755 × 109 N · m2 /C2 .
What is the distance between the two
charges?
Correct answer: 0.317285.
Explanation:
Consider the electric field vectors.
E2
θ1
θ2
E1
−8.7 C
9C
In the x-direction, the contributions from
the two charges are
q1
cos θ1
r12
q1 x1
= −ke 2
r1 r1
Ex1 = −ke
Let : q1
q2
Fe
ke
= 53.0 µC ,
= 41.0 µC ,
= 194 N , and
= 8.98755 × 109 N · m2 /C2 .
q1 q2
Fe = ke 2
r r
ke q1 q2
r=
Fe
q
= 8.98755 × 109 N · m2 /C2
midterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 20 2007, 11:00 pm
r
The potential for a point charge q is
(5.3 × 10−5 C) (4.1 × 10−5 C)
×
194 N
q
= 0.317285 m .
V = ke , so
r
7
Question 8, chap 25, sect 3.
part 1 of 1
10 points
qa qb qc
V = ke
+ +
ra rb rc
= (8.98755 × 109 N · C2 /m2 )
3 × 10−9 C 9 × 10−9 C
+
×
4.47214 m
3.16228 m
−9
−6 × 10 C
+
3.16228 m
Three charges are arranged in the (x, y)
plane as shown. (The scale is in meters.)
x (m) →
y (m)
5
4
3
9 nC
2
= 14.5554 V .
1
−6 nC
0
−1
−2
−3
Question 9, chap 23, sect 1.
part 1 of 1
10 points
3 nC
The diagram shows an isolated, positive
charge Q, where point B is twice as far away
from Q as point A.
−4
−5
−5
−3
−1
0
1
2
3
4
5
Find the electric potential Vo at the origin
(0 m,0 m). Let V = 0 at infinity.
Correct answer: 14.5554.
Explanation:
q
ra = x2a + ya2
q
= (−4 m)2 + (−2 m)2
= 4.47214 m ,
rb =
=
rc =
=
1.
2.
3.
q
(1 m)2 + (3 m)2
4.
q
x2c + yc2
= 3.16228 m ,
q
and
A
B
0
10 cm
20 cm
What is the ratio of the electric field
strength at point A to the electric field
strength at point B?
x2b + yb2
q
+Q
5.
EA
EB
EA
EB
EA
EB
EA
EB
EA
EB
=
=
=
=
=
2
1
4
correct
1
8
1
1
1
1
2
Explanation:
(3 m)2 + (1 m)2
= 3.16228 m .
Let : rB = 2 rA .
midterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 20 2007, 11:00 pm
The electric field strength E ∝
1
r2
EA
= A
1
EB
rB2
=
1
, so
r2
rB2
(2 r)2
= 4 .
=
rA2
r2
Explanation:
By convention, electric flux through a surface S is positive for electric field lines going
out of the surface S and negative for lines
going in.
Here the surface is a cube and no flux passes
through the vertical sides. The top receives
Φtop = −E a2
Question 10, chap 24, sect 2.
part 1 of 1
10 points
A cubic box of side a, oriented as shown,
contains an unknown charge. The vertically
directed electric field has a uniform magnitude E at the top surface and 2 E at the
bottom surface.
E
a
How much charge Q is inside the box?
2. Qencl = 3 ǫ0 E a2
1
ǫ0 E a2
2
E
=2
ǫ0 a2
3. Qencl =
4. Qencl
(inward is negative) and the bottom
Φbottom = 2 E a2 ,
so the total electric flux is
ΦE = −E a2 + 2 E a2 = E a2 .
Using Gauss’ Law, the charge inside the box
is
Qencl = ǫ0 ΦE = ǫ0 E a2 .
Question 11, chap 25, sect 3.
part 1 of 1
10 points
2E
1. Qencl = ǫ0 E a2 correct
8
Consider a dipole whose length is 13 m .
The dipole has two charges of magnitude 7 C
and masses of 2 kg . The insulating connecting
rod between the charges has negligible mass
and the radii of the masses are negligible; i.e.,
rcharge ≪ 1 m. They can be considered to be
point charges.
The direction of the electric field is shown
in the figure below and has a magnitude of
8 N/C .
2 kg
+7 C
5. Qencl = 2 ǫ0 E a2
E
ǫ0 a2
E
=3
ǫ0 a2
7. Qencl =
8. Qencl
9. insufficient information
10. Qencl = 0
13
m
8 N/C
38◦
6. Qencl = 6 ǫ0 E a2
2 kg
−7 C
Note: The dipole is initially at rest at 38◦
to the dashed line in the figure. The electric
midterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 20 2007, 11:00 pm
field is perpendicular to the dashed line in the
figure.
What is the magnitude of the angular acceleration of the dipole immediately after it is
released?
Correct answer: 3.39451.
Explanation:
Basic Concepts: The direction of the
electric dipole vector ~p is from the minus to
plus charge.
The moment of inertia of the dipole about
its center of mass is
2
2
Idipole = m rdipole
+ m rdipole
ℓdipole 2
= 2m
.
2
The torque equation for an electric dipole
in an electric field is
~ = I ~α
~τ = ~p × E
= p E sin θ n̂ ,
and
where θ is the measure of the angle between
the two vectors and n̂ is a unit vector according to the right-hand-rule.
The minimum potential energy is when the
electric dipole is aligned with the electric field.
The direction of the angular acceleration of
the dipole immediately after it is released is
clockwise (the angular direction is negative).
θE = 90◦ ,
θp = 142◦ , so
θ = θE − θp
= (90◦ ) − (142◦ )
= −52◦ .
~ = 8 N/C ,
Let : kEk
|q| = 7 C ,
m = 2 kg ,
ℓdipole = 13 m , and
θ = 52◦ ,
Solution: The electric dipole moment is
p = q ℓdipole
= (7 C) (13 m)
= 91 m C .
The moment of inertia of the dipole is
2
Idipole = 2 m rdipole
ℓdipole 2
= 2m
2
1
= (2 kg) (13 m)2
2
= 169 kg m2 .
Therefore, the angular acceleration of the
dipole is
τ
k~αk =
Idipole
p E sin θ
=
Irod
(91 m C) (8 N/C) sin(52◦ )
=
(169 kg m2 )
= 3.39451 rad/s2 .
First of four versions.
Question 12, chap 23, sect 1.
part 1 of 1
10 points
Consider three identical conducting spheres
of radius 4 cm arranged in an equilateral triangle, initially charged as shown in the figure
below.
6c
m
3C
E
9
p
9C
−3 C
midterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 20 2007, 11:00 pm
If the spheres are then all connected by a
thin conducting wire, what is the final charge
on the lower left-hand sphere (assuming no
charge accumulates on the wire)?
Correct answer: 3.
Explanation:
Let :
a = 6 cm = 0.06 m ,
Q1 = 9 C ,
Q2 = −3 C , and
Q3 = 3 C .
The field is parallel to the electron’s velocity
and acts to decelerate the electron.
The
charge
−19
on an electron is 1.60218 × 10
C and its
mass is 9.10939 × 10−31 kg .
How far does the electron travel before it is
brought to rest?
Correct answer: 2.34393.
Explanation:
Let :
a
Q3
Q1
Q2
Charge will flow in a conductor until the
conductor is at equipotential. When the
spheres are connected by a wire, the system becomes a single conductor, so the potentials on the three spheres must be equal.
Since the three spheres are identical, their potentials will be equal only if the charges are
identical on each. Therefore, at equilibrium,
Q′1 = Q′2 = Q′3 .
From conservation of charge,
Q′1 + Q′2 + Q′3 = 3 Q′1 = Q1 + Q2 + Q3 .
Therefore
Q1 + Q2 + Q3
3
9 C + (−3 C) + 3 C
=
3
= 3C .
Q′1 =
Question 13, chap 23, sect 4.
part 1 of 1
10 points
An electron moves at 2.9 × 106 m/s into a
uniform electric field of magnitude 1020 N/C.
10
v = 2.9 × 106 m/s ,
qe = 1.60218 × 10−19 C ,
m = 9.10939 × 10−31 kg ,
E = 1020 N/C .
and
1
The kinetic energy K = m v 2 is depleted
2
by the amount of work done by the electric
force F = qe E on the particle:
Z
W = F dx = F x = qe E x
since the force is constant. When the electron
comes to rest, all its kinetic energy has been
converted, so
1
m v 2 = qe E x .
2
m v2
2 qe E
(9.10939 × 10−31 kg)(2.9 × 106 m/s)2
=
2 (1.60218 × 10−19 C) (1020 N/C)
100 cm
×
1m
= 2.34393 cm .
x=
Question 14, chap 25, sect 3.
part 1 of 1
10 points
Consider two coaxial rings of 25.5 cm radius
and separated by 38.3 cm.
Calculate the electric potential at a point
on their common axis midway between the
two rings, assuming that each ring carries a
uniformly distributed charge of 4.77 µC.
Correct answer: 268866.
midterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 20 2007, 11:00 pm
11
Explanation:
Explanation:
Let : d = 38.3 cm = 0.383 m ,
q = 4.77 µC = 4.77 × 10−6 C ,
r = 25.5 cm = 0.255 m , and
ke = 8.98755 × 109 N · m2 /C2 .
With d the distance between the rings and
consider the potential due only to one ring of
radius r and charge Q. The point of measurement is a distance
s
2
d
2
L= r +
2
s
2
0.383 m
2
= (0.255 m) +
= 0.3189 m
2
from any point on the ring. If the charge in
an infinitesimal section of the ring is dq, then
the potential due to this section is
dV =
ke dq
,
L
and the potential due to the entire ring is then
Z
ke Q
.
V = dV =
L
Note: All infinitesimal sections are the same
distance from the point of measurement and
contribute equally.
Because of the two rings,
2 ke q
L
= 2 8.98755 × 109 N · m2 /C2
V =
4.77 × 10−6 C
0.3189 m
= 268866 V .
×
Question 15, chap 25, sect 1.
part 1 of 1
10 points
You have a potential difference of 11 V.
How much work is done to transfer 0.22 C
of charge through it?
Correct answer: 2.42.
Let : V = 11 V and
q = 0.22 C .
The potential difference is
W
q
W =V q
= (11 V) (0.22 C)
V =
= 2.42 J .
Question 16, chap 25, sect 2.
part 1 of 1
10 points
The magnitude of a uniform electric field
between the two plates is about 200000 N/C.
If the distance between these plates is
0.2 cm, find the potential difference between
the plates.
Correct answer: 400.
Explanation:
Let : E = 200000 N/C and
∆d = 0.2 cm = 0.002 m .
The magnitude of the potential difference is
∆V = E ∆d
= (200000 N/C) (0.002 m)
= 400 V .
Question 17, chap 25, sect 4.
part 1 of 1
10 points
If the potential in a region is given by the
function
V = 4 x − y 8 − cos(z) ,
what is the y-component of the electric field
at the point P = (xP , yP , zP )?
1. Ey = 4 xP + 8 yP7 − cos(zP )
midterm 01 – JYOTHINDRAN, VISHNU – Due: Sep 20 2007, 11:00 pm
2. Ey = 4 xP + yP8 − cos(zP )
1
3. Ey = − yP9
9
1 9
y − cos(zP )
9 P
1
5. Ey = 4 xP + yP9 − cos(zP )
9
1 9
6. Ey = yP
9
4. Ey = 4 xP −
7. Ey = 8 yP7 correct
8. Ey = 4 xP − yP8 − cos(zP )
9. Ey = 4 xP − 8 yP7 − cos(zP )
10. Ey = −8 yP7
Explanation:
The electric field is the gradient of the potential, so the y-component of the E-field evaluated at P is
~ = −∇V
E
∂V
∂V
∂V
k̂ ,
ı̂ −
̂ −
=−
∂x
∂y
∂z
∂V
Ey = −
∂y
∂ 4 x − y 8 − cos(x)
=−
∂y
= − −8 y 7
= 8 y7 ,
= 8 yP7 .
so
so at P = (xP , yP , zP )
12