Download Physics 8.02 Quiz One Equations Fall 2006

Physics 8.02
F=
Quiz One Equations
1 q1q2
4 !" o r 2
C=
!
E=
q
q !
rˆ =
r
2
4!" o r
4!" o r 3
!
r
rˆ =
points from source q to observer
r
Q
!V
Fall 2006
2
Q2
U = 1 C !V =
2
2C
C parallel = C1 + C2
Cseries =
C1C2
C1 + C2
Circumferences, Areas, Volumes:
!
1
dq
E=
rˆ
#
4!" 0 V r 2
1) The area of a circle of radius r is π r2
! ! Q
! E " dA = inside
$o
closed surface
2) The surface area of a sphere of radius r
is 4π r2
Its circumference is 2π r
"
##
!
dA points from inside to outside
Its volume is
! !
!V = V f " Vi = " $ E # dr
f
"
! !
E ! dr = 0
i
3) The area of the sides of a cylinder of
radius r and height h is 2π r h
closed path
Vpoint charge =
Its volume is π r2 h
q
4!" o r
N
Vmany point charges = $
i =1
U=
Definition of trig functions
qi
! "
4!" o r # ri
qi q j
! !
all pairs 4#$ o ri " r j
!
"1
2%
U=!
((( $# 2 ! o E '& dVvol
!V
!V
!V
Ex = , Ey = , Ez = !x
!y
!z
!V
Er = "
for spherical symmetry
!r
sin is opposite/hypotenuse;
cos is adjacent/hypotenuse;
tangent is opposite over adjacent;
Properties of 30, 45, and 60 degrees
(π/6, π/4, and π/3 radians):
sin(π/6) = cos(π/3) = 1/2,
sin(π/3) = cos(π/6) = 3 / 2 ;
sin(π/4) = cos(π/4) = 1 / 2 ;
Integrals that may be useful
b
" dr = b ! a
a
b
1
! r dr = ln(b / a )
a
b
1
(r
a
1
4 3
!r
3
2
&1 1#
dr = $ ' !
%a b"
MIT PHYSICS DEPARTMENT
Physics 8.02
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Quiz One Solutions
Fall 2006
Part I Concept Questions: (50 points)
(a) (10 points) A point charge is placed outside a hollow conducting sphere that was
initially uncharged. Which of the following statements are true or false?
1. The electric field inside the sphere is zero.
2. There is a (possibly non-constant) surface charge density on the inner surface of
the sphere.
3. There is a (possibly non-constant) surface charge density on the outer surface of
the sphere.
Answer: 1 and 3 are true, 2 is false. The point charge outside the sphere will induce
a non-uniform surface charge density on the outer surface of the sphere. The
superposition of this electric field and the electric field of the charge outside will give
a zero electric field everywhere inside the hollow spherical conductor. Denote by a
the inner radius of the hollow conducting sphere and b the outer radius. By Gauss’s
Law, the electric flux is zero through a Gaussian spherical surface that has a radius
r such that a < r < b . Therefore the total surface charge density on the inner
surface of the sphere is zero. Since the inner surface is screened from the charge
outside, the only possibility is that there is zero charge density everywhere on the
inner surface since there is no field that can separate the charges.
An energy argument also gives the same result. The lowest energy configuration is
for all the charge to reside on the outer surface and the electric field everywhere
inside the outer surface zero.
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MIT PHYSICS DEPARTMENT
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(b) (10 points) Circle your answer. If you place a negatively charged particle in an
electric field, the charge will move
1.
from higher to lower electric potential and from lower to higher potential energy.
2.
from higher to lower electric potential and from higher to lower potential energy.
3.
from lower to higher electric potential and from lower to higher potential energy.
4.
from lower to higher electric potential and from higher to lower potential energy.
Answer: 4. From lower to higher potential and higher to lower
potential energy. Objects always move to reduce their potential energy,
!U " U f # U i < 0 . Negative charges do this by moving towards a higher
potential. The electric potential difference is defined as the potential
energy difference per charge.
!V = !U / q
Since !U < 0 (always) and q < 0 (negative charge), !V " V f # Vi > 0 .
So
V f > Vi .
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MIT PHYSICS DEPARTMENT
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(c) (10 points) Circle your answer. Two opposite charges are placed on a line as shown
in the figure below.
The charge on the right is three times the magnitude of the charge on the
left. Besides infinite, where else can electric field possibly be zero?
1.
between the two charges.
2.
to the right of the charge on the right.
3.
to the left of the charge on the left.
4.
the electric field is nowhere zero.
!
Answer: 3. The electric field from the positive charge E+3q and the electric field
!
from the negative charge E! q are shown for different regions with the arrow size
indicative of the magnitude in the figure below.
The only place the fields can cancel is to the left of the negative charge.
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MIT PHYSICS DEPARTMENT
(d) (10 points) Circle your answer. Consider a parallel plate capacitor with circular
plates of area A separated by a distance d. Suppose we introduce a conducting circular
plate of area A parallel to the first two plates halfway between the plates. What happens
to the capacitance:
1) It doubles.
2) It halves.
3) It is unchanged.
Answer: 3. Unchanged. Two identical capacitors in series add inversely,
1
1 1 2
= + = .
Ceq C C C
So the equivalent capacitance has halved Ceq = C / 2 . A parallel plate capacitor with
circular plates of area A separated by a distance d with charge per area ! has
capacitance
C=
Q
"A
=
!V E d
The electric field of oppositely charged parallel plates is approximately given by
Gauss’s Law E = ! / " 0 . So the capacitance is
C0 =
# A
Q
"A
=
= 0 .
!V (" / # 0 ) d
d
If the distance between the plates is halved then the capacitance doubles
C=
2! 0 A
.
d
So the equivalent capacitance
Ceq = C / 2 =
!0 A
= C0 .
d
From an energy perspective, the added plate has negligible thickness, so the electric
field still occupies the same space. Therefore the energy stored is unchanged hence
the capacitance is unchanged by U = Q 2 / 2C .
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MIT PHYSICS DEPARTMENT
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(e) (10 points) Circle your answer.
Two charges lie on the x-axis. A representation of the equipotentials of the electric
potential of these two charges is shown above. The “streaks” in this representation are
parallel to the equipotential curves. Which one of the following statements is true.
1. The two charges have opposite signs and the charge on the left is smaller in
magnitude than the charge on the right.
2. The two charges have opposite signs and the charge on the left is larger in
magnitude than the charge on the right.
3. The two charges have the same sign and the charge on the left is smaller in
magnitude than the charge on the right.
4. The two charges have the same sign and the charge on the left is larger in
magnitude than the charge on the right.
Answer: 2. The equipotential line show a zero point to the right of the charge on the
right. Therefore the two charges must be of opposite signs The charge on the left is
further away from the zero point so the magnitude of the charge must be larger
than the charge on the right in order to cancel the potential from the charge on the
right.
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MIT PHYSICS DEPARTMENT
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Part II Analytic problem: (50 points)
An infinite slab of charge carries a charge per unit volume ρ, as shown in the sketch on the
left. The plot on the right shows the electric potential V (x) in volts due to this slab of
charge as a function of horizontal distance x from the center of the slab. The “slab” is 2
meters wide in the x-direction, and its boundaries are located at x = -1 meter and x = +1
meter, as indicated. The slab is infinite in the y direction and in the z direction (out of the
page).
(a) The potential V (x) is a linear function of x in the region x < -1 meter. What is the xcomponent Ex of the electric field in the region x < -1 meter? Be careful to indicate the
correct sign and units of your answer.
Ex = "
!V
"15 volts
="
= 15 volts / meter
!x
1 meter
(b) The potential V (x) is a linear function of x in the region x > 1 meter. What is the xcomponent Ex of the electric field in the region x > 1 meter. Be careful to indicate the
correct sign and units of your answer.
Ex = "
!V
15 volts
="
= "15 volts / meter
!x
1 meter
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MIT PHYSICS DEPARTMENT
(c) In the region –1 meter < x < 1 meter, the potential V (x) is a quadratic function of x
given by the equation
" 15
%
" 25 %
V (x) = $ V ! m -2 ' x 2 ( $
V .
# 2
&
# 2 '&
What is the x-component of the electric field Ex in this region? Be careful to indicate
the correct sign and units of your answer.
Ex = !
"V
= !15x volts / meter
"x
(d) Use Gauss’s Law and your answers above to find the charge density ρ of the slab.
You must indicate the Gaussian surface you use on the figure below. You can use the
symbol εo in your answer. Be careful to indicate the correct sign and units of your
answer.
Take a cylindrical gaussian pill box spanning the slab with one face in the empty space to the
left and one face in the empty space to the right. We have the electric fields on the faces of
this pillbox from above, and we find that
! ! Qinside
E
"
"" ! dA = # o = $15A $ 15A = A% 2 / # o
closed surface
where A is the area of the face of the pillbox. So
! = "# o 15coulombs / cubic meter
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MIT PHYSICS DEPARTMENT
(e) How much work do you do to move 1 micro-Coulomb from:
(i) x = 1 m to x = 2 m .
(ii) x = !1 m to x = 1 m
(iii) x = !2 m to x = !1 m
(iv) x = 1 m to x = infinity
(i) , (iii), and (iv) . Zero. For each of these cases, the potential is constant in each
region so the potential difference between any two points in each region is zero, so
the work done is zero.
(ii) Zero. Since the electric field changes direction in side this region, the work done
between any two points that are equidistant from the center is zero. This makes
sense because half the time you are doing positive work to move the charge, and half
the time you are doing negative work, and the net work turns out to be zero. From
the definition of work done by the field in moving between two points
x =1 m
W=
#
! !
F ! dr =
x = "1 m
x =1 m
#
! !
qE ! dr
x = "1 m
x =1 m
W =q
"
Ex dx
x = !1 m
x =1 m
x2
W = (1 ! 10 C) $ "(15 V # m )x dx = "(1 ! 10 C)(15 V # m )
2
x = "1 m
"6
-1
"6
x =1 m
-1
x = "1 m
W = !(1 " 10 !6 C)(15 V # m -1 )(1 / 2 ! 1 / 2) = 0
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