Transcript Ch1- Electrostatics L2 PP

Sinai University Faculty of Engineering Science
Department of Basic science
Electric Field
Electric Fields
In order to model the electrical influence of an isolated charge, we
resort to a rather fictional scenario. We first "fix" (think of it
as glueing) a charge (or charges) at specific coordinates in
space. This charge is called the
"configuration charge".
electrical field", Rx
E (r) = qc / 4 p e r 2 , in Rx direction
The force felt by a "test" particle in this field is
F x = qt Ex,
q Does not influence the field
It is just a probe
r
qt
Electric Field Lines
The magnitude of the force is proportional to the density of the field
lines ("flux density"), which decreases with the square of the distance.
Sample Problem
F=qE
A proton is placed in a uniform electric field E. What must be
the magnitude and direction of this field if the electrostatic
force acting on the proton is just to balance its weight?
Solution
mg
F
E=
=
e
q0
27
(1.67 x10 kg) X (9.8m / s 2 )
1.60 X 1019 C
= 1.0 x l0-17 N/C, directed up
E must point vertically upward to the float the (positively charged)
proton, because F = q0E and q0>0.
mg
Sample Problem
In an ionized helium atom (a helium atom in which one of the two
electrons has been removed), the electron and the nucleus are
separated by a distance of 26.5 pm. What is the electric field due
to the nucleus at the location of the electron?
Solution
We use Eq. 3, with q (the charge of the nucleus) equal to +2e:
1
q
q
E
k 2
2
4peo r
r
19
2
(
1
.
60
x
10
C)
9
2
2
= ( 8.99 x 10 N.m /C ) X
(26.5 x1012 m) 2
= 4.13x 10-12 N/C.
Sample Problem
The magnitude of the average electric field normally present in the
Earth's atmosphere just above the surface of the Earth is about
150 N/C, directed downward. What is the total net surface charge
carried by the Earth? Assume the Earth to be a conductor.
Solution
its average surface charge density s
E
q /4pr2
q
q 1

4peo r 2 4pr 2 e o
e0E 
1
q
4pr 2
E= 150 N/C
s =e0 E= (8.85 x 1012C2/N.m2)(-150 N/C)
=-l.33 x l0-9 C/m2.
r
q = s 4pR2 , s= (-1.33 x 10-9 C/m2)( 4p)(6.37 x 106 m)
= -6.8 x 105C = -680 kC.
Sample Problem
A charged drop of oil of radius R = 2.76 mm and density r = 920 kg/m3 is
maintained in equilibrium under the combined influence of its weight and a
down-ward uniform electric field of magnitude E = 1.65 x 106 N/C.
(a) Calculate the magnitude and sign of the charge on the drop. Express the
result in terms of the elementary charge e.
 F  mg  qE  0
-mg +q(-E) =0
mg
4 / 3pR3 rg
q

E
E
qE
4 / 3p (2.76)3 (106 m)3 (920kg / m3 )(9.8m / s 2 )
q
1.65 x106 N / C
q
 4.8 x10 19 C
n

3
19
 e  1.6 x10 N / C
(b) The drop is exposed to a radioactive source that emits
electrons. Two electrons strike the drop and are captured by it,
changing its charge by two units. If the electric field remains at its
constant value, calculate the resulting acceleration of the drop.
q' = (n + 2)(-e) = 5(- 1.6 x l0-19 C) = -8.0 x 10-19 C.
 F  mg  q E  ma
'
q' = (n + 2)(-e) = 5(- 1.6 x l0-19 C) = -8.0 x 10-19 C.
and, taking y components, we obtain -mg+ q'(-E)= ma
q'E
a  g 
m
= -9.80 m/s2 - (-8.0x l0-19 C)(1.65 X 106 N/C)
4/3p(2.76 X 10-6 m)3(920 kg/m3)
= -9.80 m/s2 + 16.3 m/s2 = +6.5 m/s2.
The Electric Dipole
electric dipole moment, p, p = qd.
Quadruples
Ez  k
P
P
,
E

k
x
2z3
2 x3
Monopole
Dipole
Ea 1/r2
Ea 1/r3
Quarupole
Ea 1/r4
Ex: Sodium chloride molecule
The separation between Na and Cl measured for NaCl is 0.236 nm (1 nm = l0-9 m),
and so the dipole moment is expected to be
p=ed=(l.60x l0-19C)(0.236x l0-9m) =3.78X l0-29C.m.
The measured value is 3.00 x 10-29 C.m, indicating that the electron is not entirely
removed from Na and attached to a Cl. To a certain extent, the electron is shared.
Sample Problem 9
A molecule of water vapor (H2O) has an electric dipole moment of
magnitude p = 6.2 x l0-30 C.m. (This large dipole moment is responsible
for many of the properties that make water such an important
substance, such as its ability to act as an almost universal solvent.).
E
Figure 1-15 Representation of dipole moment of water molecule in an electric
field, E
A Point Charge in an Electric
Field
What happens when we put a charged particle in a known electric field?
E
+
For a point charge
+
 F  ma
F = qE =ma
For a dipole, there is a torque t=pE sin q
(a) How far apart are the effective centers of positive and negative charge in a
molecule of H2O? (b) What is the maximum torque on a molecule of H2O in a
typical laboratory electric field of magnitude 1.5x104 N/C? (c) Suppose the
dipole moment of a molecule of H2O is initially pointing in a direction opposite
to the field. How much work is done by the electric field in rotating the
molecule into alignment with the field?
Solution
(a) There are 10 electrons and, correspondingly, 10 positive
charges in this molecule. We can write, for the magnitude of the dipole moment,
p= qd = (l0e)(d),
in which d is the separation we are seeking and e is the elementary charge. Thus
p
6.2 x10 30 C.m
12
d


3
.
9
x
10
 3.9 pm
19
10e (10)(1.60 x10 C )
This is about 4% of the OH bond distance in this molecule.
(b) The torque, t pE sinq is a maximum when q = 900. Substituting this value in that
equation yields
t pE sinq = (6.2x10-30C.m)(1.5x104N/C)(sin 900)
= 9.3x10-26 N.m
(c) The work done in rotating the dipole from q0 = 1800 to q = 00 is given by
W = pE(cos 00 - cos 1800 )
= 2pE= (2)(6.2x10-30C.m)(1.5x104 N/C)
-25
Assignment
1-Write short notes on the Electrophoreses
including:
a-Its structure
b- Its theory of operation
c- Its uses
2- Solve the following problems
1, 3, 6
Please, Hand it before next lecture
Electric potential
For our purposes, it will be much easier to deal with
another quantity, the scalar "electrical potential", rather
than the electric field. For a point charge, q:
V(r) = q / 4 p e r.
But since the electric field is a force per unit charge, the
electric potential must be energy per unit charge.
E=Force/charge
N/C
V=Energy/Charge
J/C (Volts)
Electric Potential
+
Mgh
W=Mgh
W=qV
+
+
Drift velocity
V1
V
E
h
Drift velocity
Mg
V2
qE
-
W=W1+W2 = qV= qV1+ qV2 V=V1+V2
-
Sample Problem
What must be the magnitude of an isolated positive
point charge for the electric potential at 15 cm from the
charge to be + 120 V?
V(r) = q / 4 p e r.
Solution
Solving for q yields
q = V 4pε0 r = (120 V)(4p)(8.9 x 10–12 C2/N.m2 )(0.15 m)
= 2.0 x 10–9 C = 2.0 nC
This charge is comparable to charges that can be produced
by friction, such as by rubbing a balloon.
Sample Problem
What is the electric potential at the surface of a gold
nucleus? The radius is 7.0 x 10-15 m, and the atomic
number Z is 79.
Solution
The nucleus, assumed spherically symmetric, behaves
electrically for external points as if it were a point charge. Thus
we can use Eq. 18, which gives, with q = + 79e,
1 q (9.0  109 N .m2 / C 2 )(79)(1.6  1019 C )
V 

4pe0 r
7.0  1015 m
 1.6  10 V .
7
This large positive potential has no effect outside a gold atom because it is
compensated by an equally large negative potential from the 79 atomic
electrons of gold.
Electric potential
Energy
Vca
qo
E
Qc
Vcb
L
Electric Field is a conservative Field
W ab = F x Dx = ( - q0E)(L) = - q0EL
Using the definition of potential energy difference, DU = -W
U b  U a  Wab
Vb  Va 

 EL.
q0
q0
Electric potential
Example
if in a vacuum two protons (qc1) lie in the xy plane at (x1, y1) = (-2,1) Angstroms
and an electron (qc2) lies at the origin ((x2, y2) = (0,0)), the energy of an
electron (q t) at (x t, y t) = (1,1) Angstroms would be
U = U
t c1
+ U
t c2
(= q
t
(V c 1 + V c 2))
= (- e) (2 e) / (4 p e 0 3 x 10 - 10) + (- e) (e) / (4 p e 0 x
2 10 - 10)
= 4.69 x 10 - 17 J,
where the distances between charges are
(-2,1)
(1,1)
r=((x t - x i) 2 + (y t - y i)2)
and "i" denotes either
configuration charges.
of
the
(0,0)
Note that chemists often compute energies in kJ/mol, which would
necessitate multiplying the above answer by a conversion factor of NA / 1000.
Charge acceleration
Van De Graaff Accelerator
+++++++++++++++++
+
+
Utc = E L=qt Vc
U= ½ mv2
Sample Problem
An alpha particle (q = + 2e) in a nuclear accelerator moves from one
terminal at a potential of Va = +6.5 x 106 V to another at a potential of Vb = 0.
(a) What is the corresponding change in the potential energy of the
system? (b) Assuming the terminals and their charges do not move and
that no external forces act on the system, what is the change in kinetic
energy of the particle?
Solution (a) we have
DU = U b – U a = q (V b – V a)
= (+2)(1.6 x 10–19 C)[ 0 – 6.5 x 106 V]
= - 2.1 x 10–12 J.
(b) If no external force acts on the system, then its mechanical energy E = U +
K must remain constant. That is, D E = DU + DK = 0, and so
D K = - DU = +2.l x 10-12 J.
The alpha particle gains a kinetic energy of 2.1 x 10-12 J, in the same way that a
particle falling in the Earth's gravitational field gains kinetic energy.
Problems
Solve the following problems
1, 2, 4, 6, 8, 13
Cyclotron is a device used to accelerate
electric charges.
Write a short account on the Cyclotron.
-Theory of operation
-Application