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General Physics Exam - Problems and Solutions
1st Exam, 2nd Semester, Year 2013
① Describe all significant steps leading to your answers
Instruction
and write your solutions within the space provided.
② Present your answers as simple as possible.
The -components of the electric field
at can be obtained by
the following calculation.
sincos
sin cos
cos
sin
×
?
cos
cos
to the result in (a).)
the leading contribution in ? If we consider the ring as a dipole,
(a) (10 points) What is the magnitude and direction (relative to the
positive direction of the -axis) of the electric field
produced at
(c) (10 points) If is much larger than the size of the ring, what is
sinsin
sin sin
points, and the charge on each rod is uniformly distributed.
(You can get this by a separate calculation or by plugging
and the other with charge , form a circle of radius in an
(b) (5 points) Answer (a) if is at the origin (center of the ring).
Problem 1. (25 points) Two curved plastic rod, one with charge
- plane. The -axis passes through both of the connecting
Therefore, the electric field
at is given by
what would be the dipole moment of the ring?
.
(b) The electric field
at the origin due to the charge element is
given as follows:
< Fig. 1 >
(a) The electric field
at due to the charge element is given as
follows:
cos
sin
Here, the charge element is given as follows:
≤ ≤
≤ ≤
The -components of the electric field
at the origin can be obtained
by the following calculation.
cos
cos
sin
sin
sincos
sinsin
cos
Here, the charge element is given as follows:
≤ ≤
≤ ≤
cos
sin
Therefore, the electric field
at the origin is given by
.
By plugging to the result in (a), the same result can be
obtained.
Introductory Physics Office, Department of Physics, College of Science, Korea University
Last Update : 2013-10-13
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General Physics Exam - Problems and Solutions
1st Exam, 2nd Semester, Year 2013
(c) If ≫ and ≪ , then the electric field
at can be rewritten
as follows:
Problem 2. (25 points)
(a) (5 points) What is the equivalent resistance of this circuit?
⋯
Therefore, the leading contribution in the electric field due to the ring is
given by
.
Consider the dipole consisting of and located at and
, respectively. The electric field
at can be obtained by
< Fig. 2-1 >
the following calculation.
(b) (10 points) What is the equivalent resistance of this circuit?
< Fig. 2-2 >
(c) (10 points) The tip of a wire has the shape of a truncated cone,
which has the resistivity of . The length of the tip is , and its
larger and smaller diameters are and , respectively. What is the
sin
resistance of the tip?
Here,
corresponds to the dipole moment. Note that the dipole
moment has the direction from to . If ≫ and ≪ , then
the electric field at can be rewritten as follows:
⋯
< Fig. 2-3 >
(a) This circuit can be redrawn as follows:
Therefore, the leading contribution in the electric field due to the dipole is
given by
If
and
are compared, the dipole moment of the ring is given as
follows:
,
,
,
, , ,
,
,
, or
Since , is obtained.
Introductory Physics Office, Department of Physics, College of Science, Korea University
Last Update : 2013-10-13
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General Physics Exam - Problems and Solutions
1st Exam, 2nd Semester, Year 2013
(b) This circuit can be redrawn as follows:
Problem 3-A. (25 points) There is a long insulating cylinder of radius
whose symmetric axis is parallel to the -axis and goes through the
origin. The cylinder has a volume charge density of . A long
conducting cylindrical shell extending radially from
to
surrounds the insulating cylinder. The shell has no net charge.
(a) (5 points) Find the charge distribution of the shell.
(b) (10 points) Find the electric field and plot it as a function of ,
distance to the symmetric axis.
(c) (10 points) Find the electric potential and plot it as a function of r.
Set the potential at origin to be zero.
(c) The resistance of the volume element with radius and length
is given as follows:
< Fig. 3-1 >
(a) Since there is no electric field inside the conducting cylindrical shell, the
charge distributed on the inner surface of the shell has the equal amount
but opposite sign, compared to the charge distributed over the insulating
cylinder. If the surface charge density on the inner surface of the shell is
and the length of the system is , is given as follows:
,
Since the shell has no net charge, the charge distributed on the outer
surface of the shell has the equal amount but opposite sign, compared to
the charge distributed on the inner surface of the shell. If the surface
Therefore, the resistance of the tip is given as follows:
⋅ ,
charge density on the outer surface of the shell is and the length of
the system is , is given as follows:
⋅ ⋅ ,
(b) The electric field
can be obtained by the following
calculation.
(i) :
,
⋅ , ⋅
(ii) : ⋅ ,
⋅
(iii) : ⋅ ,
(iv) : ⋅
⋅ ⋅
,
Introductory Physics Office, Department of Physics, College of Science, Korea University
Last Update : 2013-10-13
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General Physics Exam - Problems and Solutions
1st Exam, 2nd Semester, Year 2013
Problem 3-B. (25 points) Two infinitely-long uniformly-charged wires are
parallel to the -axis. A wire ( ) with line charge density of
goes through and the other ( ) with goes through
. Consider a point which is from and
from .
(a) (10 points) Show that has the same electric potential if is
constant.
(b) (10 points) Show that the points having the same electric potential
form a circular cylinder extending parallel to the -axis.
(c) (5 points) Draw several equipotential lines in the - plane
qualitatively. Be sure to depict the case of (b).
(c) Since the electric potential at origin is set to , the potential
⋅
can be obtained by the following
calculation.
(i) :
ln ln
(ii) :
ln ln
(iii) :
(iv) :
line charge density is given as follows:
,
⋅ , ⋅
ln ln
ln
(a) The electric field and electric potential at due to with
< Fig. 3-2 >
,
ln ( : constant)
Similarly, the electric field and electric potential at due to
with line charge density is give as follows:
,
ln ( : constant)
Therefore, the electric potential at due to both and is
given as follows:
ln
This result shows that has the same electric potential if is constant.
(b) If is constant, the above result can be rewritten as follows:
exp ( : constant)
Since and go through and respectively, the
following result can be obtained in the case of .
,
, ,
Introductory Physics Office, Department of Physics, College of Science, Korea University
Last Update : 2013-10-13
PAGE 5/6
General Physics Exam - Problems and Solutions
1st Exam, 2nd Semester, Year 2013
,
(i) If and , then
Problem 4. (25 points) There is a long cylindrical capacitor. The radius
of the inner cylinder is and that of the outer cylinder is . The
so that plane forms
an equipotential plane.
and radii
(ii) If ≠ , then circular cylinders with centers
form equipotential planes.
(c) Equipotential lines in the - plane have properties such that the
center runs away from or and the radius increases as →.
length of the whole cylinder is .
(a) (5 points) What is the capacitance of the capacitor?
Now we insert a dielectric material (dielectric constant : ) in two
different configurations: in one configuration ( ), the inserted dielectric
has a shape of cylindrical shell of two radii ( and ) and length
and in the other configuration ( ), the dielectric has a shape of
cylindrical shell of two radii ( and ) and length .
(b) (10 points) Find the capacitance of those capacitors ( and ).
(c) (10 points) When the potential difference is for both capacitors,
how much work has to be done to take out the dielectric shells?
< Fig. 4 >
(a) Let us assume that and with charge per unit length are
uniformly distributed on the inner and outer surface of cylindrical capacitor,
respectively. The electric field
inside the cylindrical capacitor
with inner radius , outer radius , and length is given as follows:
:
⋅ ,
⋅ ,
The potential difference inside the cylindrical capacitor is given as
follows:
⋅
ln
Therefore, the capacitance of the cylindrical capacitor is given as follows:
ln
ln
If and , then .
ln
ln
Introductory Physics Office, Department of Physics, College of Science, Korea University
Last Update : 2013-10-13
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General Physics Exam - Problems and Solutions
1st Exam, 2nd Semester, Year 2013
(b) In the first configuration, a capacitor with two radii ( and )
, ,
ln
ln
and length inserted a dielectric is connected to a capacitor with
two radii ( and ) and length in series. Therefore, the
ln
calculation.
ln
ln
ln
ln
ln
capacitance of the first configuration can be obtained by the following
, ,
ln
(c) In the first configuration, the work done to take out a dielectric is
given as follows:
ln
,
ln
ln
In the second configuration, a capacitor with two radii ( and )
In the second configuration, the work done to take out a dielectric is
and length inserted a dielectric is connected a capacitor with two radii
given as follows:
( and ) and length in parallel. Therefore, the capacitance of
the second configuration can be obtained by the following calculation.
Introductory Physics Office, Department of Physics, College of Science, Korea University
ln
Last Update : 2013-10-13
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