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Transcript
```PH504: problems solved in Workshop
Class
1. Show that
(A + B)  (A  B) = A2 – B2
and
(A + B)(A  B) = 2 BA .
Verify the above for A = 3i - 5j + 4k and B = i + j + 2k.
2. Consider a solid body in rotation given by v = r
where  is a constant vector and r is the radial vector. Calculate
v .
Vorticity is defined as the curl of the velocity vector
3. Consider a long cylinder (e.g. plastic rod) of length L and
a volume charge density ρ that is proportional to the distance
from the axis s of the cylinder / rod –
i.e. ρ(s) = ks where k = proportionality constant. Calculate the
total electric charge in the rod.
1. Calculate the ratio of the electrostatic to the gravitational
force of attraction of the proton for the electron in a
hydrogen atom.
2. A line charge density is given by(x) = 3x3 C m-1.
Calculate the total charge contained between x = 0 and x =
1.
3. Consider eight point charges Q located at the corners of a
cube of unit length. Calculate the strength of the electric
field at a specific point charge due to the other seven
charges. In so doing, determine and state the direction of
the electric field at a point charge.
1. A positive 1.0 mC charge is located at the origin and a
-0.3 mC charge is located at x = 2.0 cm. What is the force
on the negative charge?
Use:
F k
q1 q2
r2
Example 1
A positive 1.0 C charge is located at the origin and a
-0.3 C charge is located at x = 2.0 cm. What is the
force on the negative charge?
F k
q1 q2
r2
1.0 C
0.0 cm
-0.3 C
2.0 cm
6
6
N  m2 1.0 10 C 0.3 10 C
F  8.99 10
2
C2
 0.02m 
9
F  6.74 N

F  6.74Ni
2. A positive 0.1 mC charge is located at the origin, a
+0.2 mC charge is located at (0.0 cm, 1.5 cm), and a 0.2 mC charge is located at (1.0 cm, 0.0 cm). What is
the force on the negative
charge?
Example 2
A positive 0.1 C charge is located at the origin,
a +0.2 C charge is located at (0.0 cm, 1.5 cm),
and a -0.2 C charge is located at (1.0 cm, 0.0
cm). What is the force on the negative charge?
3
0.2 C
(0.0 cm, 1.5 cm)
1

0.1 C
(0 cm. 0cm)
2
-0.2 C
(1.0 cm, 0.0 cm)
q1 q2
F k
rˆ
2
r
Determine the force between the negative charge
and each positive charge.
3
0.2 C
(0.0 cm, 1.5 cm)
6
6
N  m2 0.110 C 0.2 10 C ˆ
F21  8.99 10
i
2
C2
0.01
m


9
1
2
0.1 C
(0 cm. 0cm)
-0.2 C
(1.0 cm, 0.0 cm)

F21  180
. Ni
The force between charges 1 and 2.
F21  k
3
q1 q2 ˆ
i
2
r21
0.2 C
(0.0 cm, 1.5 cm)
1

0.1 C
(0 cm. 0cm)
2
-0.2 C
(1.0 cm, 0.0 cm)
The force between charges 3 and 2.
F23
6
6
N  m2 0.2 10 C 0.2 10 C
 8.99 10
2
2
C2
 0.01m    0.015m 
9

F23  111
. N
3
0.2 C
(0.0 cm, 1.5 cm)
1

0.1 C
(0 cm. 0cm)
2
-0.2 C
(1.0 cm, 0.0 cm)
Determine the components of each force.

F23   F23 cos i  F23 sin j

F23  111
. N
0.01m
i  111
. N
0.015m
0.01m   0.015m 
0.01m 

F23  0.62Ni  0.92Nj
2
3
2
2
 0.015m 
2
j
0.2 C
(0.0 cm, 1.5 cm)
1

0.1 C
(0 cm. 0cm)
2
-0.2 C
(1.0 cm, 0.0 cm)
Determine the sum of the components of the forces.

F  180
. Ni  0.62Ni  0.92Nj
Determine the resultant force
F   2.42 Ni  0.92 Nj

F  2.59 N @ 159  from +x axis
2. Determine the final velocity and kinetic energy of
an electron released from rest in the presence of a
uniform electric field of 300 N/C in the x direction
after a period of 0.5 s.
Example - Motion of a charged
particle in an Electric Field
Determine the final velocity and kinetic energy of an
electron released from rest in the presence of a
uniform electric field of 300 N/C in the x direction
after a period of 0.5 s.

E
F  4.80 10 17 N iˆ
-e
F




F
 4.80 1017 N iˆ
ˆ
a 
F  qE  eEi
m
9.111031 kg

N

F   1.60 10 19 C  300 iˆ

m

C
a   5.3 10 13 2 iˆ

s 





Example - Motion of a charged
particle in an Electric Field
E
F
K
-e

m

a   5.3 10 13 2 iˆ
s 

  
v  vo  at
K

1
mv 2
2

1
m

9.1110 31 kg   2.6 10  7 
2
s

K  3.16 10 16 J


m 

v  0   5.3 10 13 2 iˆ  0.5 10 6 s
s 


m
v  2.6 10  7 iˆ
s

2
Some more problems:
Coulomb’s law and electric field (E-field)
1. A line charge density (x) is given by (x)=3x2 Cm-1.
Calculate the total charge contained between the points x=0 and
x=L.
1)
2.
L3 Coulombs
This result is obtained by integrating the equation for (x)
between the limits 0 and L.
A surface charge density (x,y) is given by
(x,y)=3x2+4y2-xy Cm-2. Calculate the total charge contained
within the area bounded by x=0+a, y=0+a.
2)
25a4/12 Coulombs
This result is obtained by integrating the equation twice with respect to x and y (the order doesn't matter) and using
the appropriate limits.
3. A plane circular sheet of radius b has a uniform charge
density  Cm-2. The E-field at a point P which is a
perpendicular distance a from the centre of the sheet is
Show that this tends to /20 in the limit of the sheet having
4. A line charge extends from x=0 to x=L and has a density
(x)=ax Cm-1. Calculate the E-field at a point b (>L) on the xaxis.
4)
Split the line charge up into a sequence of point charges of
width dx. The size of this point charge is dx=axdx.
Calculate the E-field produced at b by this point charge
and then integrate the result from x=0 to x=L to find the
total E-field.
Electric potential energy, electric potential and
capacitance
1. The E-field produced by a spherical conductor of
radius a and carrying a charge Q is identical to that
of a point charge Q placed at the centre of the
sphere for r>a and is zero for r<a. Given that
(1/2)0E2 is the energy density associated with an Efield, find the total energy associated with the
sphere and show that this is equal to (1/2)QV where
V is the potential of the sphere.
1) Q2/(80a)
Divide the space external to the charge into a series of thin
concentric shells of radii r and thickness dr. Energy stored
in each shell is (1/2)0E2 multiplied by volume of shell
(4r2dr). Use equation for E due to a spherical charge to
work out energy stored in each shell then integrate result
from infinity to r=a.
2. A capacitor consists of two concentric spheres of
radii a and b (b>a). When a charge of +Q is placed
on the inner sphere and a charge –Q is placed on the
outer sphere the E-field in the region between the
spheres is given by
where the direction of E is radial. Show that the
capacitance of the system is given by
2) Integrate the equation for the E-field between the limits
r=b and r=a. This gives the potential difference between
the spheres in terms of Q. Compare this result with the
definition of capacitance, Q=CV, to find C.
The electric dipole
1.
An electric dipole consists of an electron and a proton
separated by a distance of 1nm. Calculate the electric
potential at a point situated along the axis of the dipole, both
exactly and using the ideal dipole formula, at distances of 1,
10 and 100nm from the centre of the dipole. Calculate the
percentage error that results from the use of the ideal dipole
formula in each case.
e = 1.602 ´ 10-19 C
= 8.8542x10-12 C2 m-2 N-1
Ideal 1.44V, 1.44x10-2V, 1.44x10-4V
Exact 1.92V, 1.44x10-2V, 1.44x10-4V
Errors 33%, 0.25%, 2.5x10-3%
Use the equation from the lecture notes for the ideal
case. For the exact case treat the system as two point
charges and calculate the potential as the sum of the
potentials due to each charge.
Gauss's law
1.
Calculate the divergence of the following vectors
(Cartesian co-ordinates)
2i+3yj+4k, 4x2i, xyi+yzj+zxk
1.
2.
3, 8x, y+z+x
Two concentric spherical shells of radii a and b (b>a)
carry charges of +Q and -Q respectively. Find the resultant Efield in the regions r<a, b>r>a and r>b. Assume that the
charges are spread uniformly over the surfaces of the spheres.
2) For r<a and r>b E=0 for a<r<b E=Q/(40r2).
Remember it is only the charge contained
within the Gaussian surface that contributes
to the flux through the surface.
3.
Calculate the electric potentials for the charge distributions
of questions 2 and 3. Sketch your results.
3) For the spheres: r>b V=0, b>r>a
, for r<a
4.
Use Gauss’s law to find the E-field both inside and outside
of an infinite plane of thickness a and carrying a uniform
charge density . (S)
4) Outside of plane E=a/(20), within plane
E=r/(20) where r is the distance from the centre
of the plane.
When applying a Gaussian surface you need to
consider what fraction of the charge is contained
within the surface.
```
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