Download ph504-1213-ass1a

PH504/12/13a
UNIVERSITY OF KENT
SCHOOL OF PHYSICAL SCIENCES
2012/13
Assignment 1: Analysis
1. Calculate the gradient () of the following functions (assume Cartesian co-ordinates)
(a) f = 3x2+2y2+z2,
(b) f = x2y2z2,
(c) f = exyz,
 f = 6xi + 4yj + 2zk
 f = 2xyz(yzi + xzj + xyk)
 f = f (yzi + xzj + xyk)
(d) f = 4 sin(xyz) .  f = 4 (yzi + xzj + xyk) cos(xyz)
(e) ex sin(xyz) .
¶f /¶x = e x sin(xyz) + e x yz ´ cos(xyz)
¶f /¶y = xze x cos(xyz)
¶f /¶z = xye x cos(xyz)
which are obviously the three components i.e.
 f  (f / x, f / y, f / dz)
2. A vector is given by -iy+jx in Cartesian coordinates with unit vectors (i , j).
(a) Calculate the line integral of this vector anti-clockwise around the sides of the square
defined by the co-ordinates (0,0), (a,0), (a,a), (0,a).
Around the square sides are 4 line integrals:
From (0,0) to (a,0), A.dL = -y dx
Therefore line integral= 0 since y=0
From (a,0) to (a,a), A.dL = +x dy integrated between y=0 to y=a
Therefore line intergral = a2 , since x=a
From (a,a) to (0,a), A.dL = - y (-dx) integrated between x=0 to x=a
Therefore line intergral = a2 , since y=a
2
From (0,a) to (0,0), A.dL = +x -dy integrated between y=0 to y=a
Therefore line intergral = 0 , since x=0
Therefore, total line integral = 2a2
(b) Show that the full result is equal to the curl of the vector multiplied by the area of the
square (this is Stoke's theorem).
Curl (-iy+jx ) = 2k,
A constant within the square.
Therefore Integrating Curl (-iy+jx ) dx dy yields 2a2
3. (a) Suppose that a = 3i + 5j - 2k and b = 2i - 2j - 2k
where i, j and k are orthogonal unit vectors. Calculate the dot product and vector product.
a · b = (3i + 5j - 2k).(2i - 2j - 2k)
= 3i.2i + 5j.(-2)j + (-2)k.(-2)k
= 6 -10 + 4
= 0 vectors are perpendicular
Since from notes:
AB = (AyBz-AzBy)i + (AzBx-AxBz)j + (AxBy-AyBx)k
a ´ b = (-10 - 4,-4 - (-6),-6 -10)
=
-14i + 2j - 16k
What is the angle between the vectors? 90 degrees
(b) Suppose that
c = 2i - 3j + k and d = 4i + j - 3k
Calculate the angle between the vectors in radians.
c · d =8-3-3=2
c = c · c = 4 + 9 +1 = 14
d = d · d = 16 +1+ 9 = 26
c·d
2
cosq =
=
cd
14 ´ 26
need a calculator ……. 0.105 …. Theta = 1.467 radians
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Assignment 2: Charge
1. A line charge density is given by(x) = 3x3 C m-1. Calculate the total charge
contained between x = 0 and x = 1.
Integrate 3x3 dx yields 3/4 x4
Put in the limits to yield a charge 3/4C - 0 = 3/4 C
2. 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.
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.
The total charge is Q=
 
dx dy
Integrate wrt y to get
Q=  (3x2y + 4y3/3 – xy2/2 ) dx
Each term has a y in it. So y=a can be substituted here 9or later) on putting in
the limits of integration.
Integrate this answer wrt x to get
Q= x3y + 4xy3/3 – x2y2/4
Put in the limits to get the above answer.
Assignment 3: Electric field
1. State how the electric field is related to the electrostatic potential.
The electric field is the gradient of -V
turn over
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2.
In some region of space, the electrostatic potential takes the following functional form
in Cartesian coordinates x, y, and z: V(x,y,z) = 1/x2 + 1/(xy), where the potential is
measured in volts and the distances in meters. Find the electric field, E, at the point x
= 2 m, y = 2 m, z = 2m. What is the value of curl E?
The gradient can be expressed in vector form e.g.
V  (
2
1
1

,

,0) Therefore
x 3 x 2 y xy2
E  (3 / 8,1 / 8,0)
Curl E = 0
3. By applying Gauss' law to a charged conducting sphere of radius r and
charge Q, the magnitude of the electric field outside the sphere is found to be
E = + Q /(40r2). Detemine the electric potential and, hence, derive a formula
for the self-capacitance of the sphere.
integrating the electric field along a radial line:
4.
The capacitance is, taking b as infinity:
5.
 C = 40a
3.
4. Calculate the self-capacitance of the Earth in units of nF.
See Lecture Notes
For the planet Earth: about 709 µF = 700,000 nF
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4.
5. A conducting cylinder of radius Rc carries a uniform surface charge of +per unit
length. Determine the electric field strength inside and outside the cylinder.
E = 0 inside a conductor, on applying gauss’ Law
Outside Gauss's Law over a cylindrical surface yields (per unit length)
E 2r = o E = 2ro) for r > R.
(see Lecture Notes 2).
6. Air ionises for E-fields above 30 kV cm-1. Calculate the maximum charge per unit
length that can be placed on a conducting cylinder of radius 2 cm without the
surrounding air ionising.
30 kV/cm = 3 x 106 V/m
Using (2), the maximum = E 2ro = 3 x 106 x 6.28 x 0.02 x 8.85 10-12 = 3.3 10-6 C/m
Assignment 4: Capacitance
1. By consideration of the electric charge and potential, determine the total capacitance of a
system of N capacitors of capacitance C1, C2, …….CN, placed in parallel.
Capacitors in parallel add ... simply increasing Area A holding charge.
C = Q/V = Q1/V + Q2/V + Q3/V + …..
=
C1 + C2 + C3 + …..
2. Determine the total capacitance of a system of N capacitors of capacitance C1, C2, …….CN,
placed in series.
Since charge cannot be added or taken away from the conductor between series
capacitors, the net charge there remains zero.
Capacitors in series combine as reciprocals ... since they share the voltage V: Q/C =
V = V1 + V2 + V3 + …..
turn over
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1/C = V/Q = V1 /Q + V2 /Q + V3 /Q + …..
= 1/C1 + 1/C2 + ……….
3. A group of identical capacitors is connected first in series and then in parallel. The
combined capacitance in parallel is 529 times larger than for the series connection. Determine
the number of capacitors.
Given Cp = 529 Cs
where Cp =N C
and 1/ Cs = N/C, from the above results.
Substitution yields N2 = 529, or N = 23 capacitors.
4. A parallel plate capacitor has plates of area 10 cm2 separated by 5mm.
a). Calculate the capacitance and the potential difference between the plates for a
charge of 5x10-11 C.
b). Determine the energy stored by the capacitor and the magnitude of the E-field
between the plates?
c.) An electron is released from rest from one of the plates and is accelerated towards the
second plate. Find the speed of the electron when it hits the second plate.
(a) C =oA/d where A = 0.001 m2 ( 10 cm2 = 10-3 m2) and d = 0.005m.
So C = =o/5 = 1.77 x 10 -12 F
The potential difference is V = Q/C = 5 x 10-11 / 1.77 x 10-12 V = 28.3 V
(b) Energy stored is QV/2 = 7 x 10-10 J
Electric field strength is E = V/d = 5650 V/m
(c) The work done by the constant electric field is eEd.
Therefore the KE gained by the electron is mev2 /2 = eEd.
Hence v2 = 2eV/me .
This yields v = 3.15 x 106 m/s
5. A line charge having a constant charge density of +1 C m-1 extends along the y-axis
from -∞ to +∞. Calculate the E-field at a distance of 1 millimetre along the positive x-axis.
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Take = 8.85x10-12 C2 m-2 N-1 .
[20 marks]
Applying Gauss’ Theorem, or simply stating the result:
E = 2ro)
And substituting = 1 x 10-6 C/m, r = 0.001m, yields and E-field of 17 x 106 N/C
6. Consider an inner conducting metal sphere with total charge -q of radius a and a thin outer
conducting metal shell with total charge +2q and radius b (of negligible thickness). Write
down expressions for the electric field and electric potential in the three regions: (i) within the
sphere (ii) between the sphere and the shell and (iii) external to the shell. Sketch graphs of the
radial distributions of the electric field and the electric potential (please label your axes).
b
a
-q
+2q
(i) Within the sphere E=0 since it is a conductor
(ii) Between the sphere and shell, place a Gaussian surface. Charge inside is –q.
So E =
-q
/(40r2)
(iii) External to the shell, Gaussian surface with a total charge of +q inside yields E =
+q
/(40r2)
Taking V=0 at infinity, the potential external to the shell is found on intergration to be
V = + q /(40r)
In the region a < r < b, we again integrate E with r to get:
V = - q /(40r) + a constant.
To match on to the r = b equipotential thus requires us to fix the constant:
V = 2q /(40b) - q /(40r).
In the region r <a, no field means that V = constant. Matching on to the surface
potential thus yields a potential within the sphere of
V = 2q /(40b) - q /(40a).
turn over
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This can be positive or negative depending on the radii a and b.
Graphs should display the above with
E and V (y-axes) plotted against r (x-axis).
Assignment 5: Electric Dipole
6. Dipole Torque Problem. An electric dipole consists of charges +2e and -2e
separated by 0.95 nm. It is in an electric field of strength 2.9 x 106 N/C.
Calculate the magnitude of the torque on the dipole when the dipole moment is
(a) parallel to, (b) perpendicular to, and (c) antiparallel to the electric field.
The magnitude of the dipole moment of the electric dipole will be given as
p = qs where q= 2e and s = 0.95 nm. The vector direction is from the negative
to the positive.
Hence the torque on it in the electric field is
torque = E P sin(theta), where 'theta' is the angle
a) in parallel position theta = 0 , so that torque = zero
b) in perpendicular position theta= 90 degree and sin(90) = 1
,so that, torque = EP sin 90 = 2.9x10^6 x 2x0.95x10^-9x2x1.6x10^-19
= 6.8875 x 10^-22 N -m
c) in antiparallel position theta = 180 and as sin180 = zero
, the torque will also be zero.