Electric Charges
Electricity & Magnetism Lecture
Series
Objectives/Learning outcome:
1. Explain the concept of electric field around a
point charge and electric forces experienced by
charged particles moving in electric fields.
2. Write mathematical relations between electric
field, electric forces and acceleration of charged
particles moving in electric fields.
Electric and Magnetic
Fields and Forces
3. Describe and draw the trajectory of charged
particles moving in an electric field.
Applied Sciences Education Research Group (ASERG)
Faculty of Applied Sciences
Universiti Teknologi MARA
4. Determine final vertical velocity, final
acceleration and final vertical position by
applying Newton’s laws.
email: jjnita@salam.uitm.edu.my; drjjlanita@hotmail.com
http://www3.uitm.edu.my/staff/drjj/drjj1.html
Copyright DR JJ,FSG, UiTM
Electric Charges
Electric Charges
Objectives/Learning outcome:
Objectives/Learning outcome:
5. Draw and explain magnetic fields and magnetic
forces.
9. Describe the working of a velocity selector by
drawing and explaining the trajectory of charged
particles entering a magnetic/electric field.
6. Write mathematical relations for magnetic forces
acting on charged particles moving in magnetic
fields.
10. Describe the working of a mass spectrometer.
11. Explain the working mechanism of cathode ray
tubes.
7. Use Right-Hand-Rule 1 to determine direction of
forces exerted on charged particles moving in
magnetic fields.
12. Use mathematical relations to solve problems
related to velocity selectors and mass
spectrometers.
8. Explain the trajectory of charged particles
moving in magnetic fields and obtain the radius
of trajectory.
Copyright DR JJ,FSG, UiTM
3
Copyright DR JJ,FSG, UiTM
-
-
q1=2e
F
r
q1=4e
21
→
F
r
F
F
21
→
+ q2=2e
→
F
21
= k
Fset1 <
+ q2=2e
F
21
2e2e
4e2
= k
2
r
r2
Fset2
Copyright DR JJ,FSG, UiTM
q q
= k 22 1
r
8e 2
= k
r2
4e 2
= 2k 2
r
→
up
4
Electric Field
Electric Point Charges
Electric forces: attractive forces since opposite
charges
→
2
Electric Field: an area where any charged object will
experience an electric force
up
The electric field lines around a
negative point charge
→
F 12 = − F 21
E =
Is accset1 bigger or
smaller than accset2??
5
Fq
on q 0
q0
= k
The electric field lines around a
pair of point charges
qq 0
q
= k 2
q 0r2
r
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6
1
Electric Field
Capacitors
Electric Field: Place the point charge q, at the
center of a an imaginary spherical surface of radius r
(gaussian surface) .
E =
F q on
E = k
The electric field
strength is then directly
proportional to the
surface charge density,
amount of charges in 1
square meter.
q0
q0
= k
Parallel plate capacitors:Two metal (conducting)
plates, parallel to each other and separated by non
conductors such as air, paper, plastic etc. Use to store
charges and hence store electrical energy.
qq 0
q0r 2
The strength of the E field
from each plate depends on
the total charge within the
Gaussian surface
q
q
=
r2
4 πε 0 r 2
G
∫E
But the surface area of a
sphere is 4πr2
E =
q
q
σ
=
=
4π r 2ε 0
ε0
Aε0
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Let the surface charge
density or total charge per
square meter be, σ.
7
+
E =
G
v = vx
σ
V
=
ε0
d
+q
+++++++++++++
E =
σ
ε0
8
+V
+
E field
------------------
Charges moving at a constant velocity will
experience a force only when entering an E
field, hence it will accelerate.
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ε0
E fields-capacitors
G
q E
q V
G
a = 0
= 0
m
md
-V
EA =
Parallel plate capacitors: Positive charges moving
horizontally at velocity v, will accelerate vertically
down, consistent with Newton’s 2nd Law, due to the
electric force in the downward direction
G
G
G
F = q0E = m a
+
ε0
q
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E fields-capacitors
Parallel plate capacitors:Positive test charges
placed between plates will experience electric forces.
Hence, it will accelerate, consistent with Newton’s 2nd
Law.
+V
G
q
• dA =
Final position, xf
Initial position, xi
9
-V
Copyright DR JJ,FSG, UiTM
E fields-capacitors
10
E fields-capacitors
Parallel plate capacitors: Positive charges moving
horizontally at velocity v, will accelerate vertically
down, consistent with Newton’s 2nd Law, due to the
electric force in the downward direction
G
v = vx
+q
Vertical
Displacement
sy
sx
Initial position, xi
11
+V
F
------------------
Parallel plate capacitors: Positive charges moving at
velocity v, consistent with Newton’s 2nd Law, due to
the electric force to the right
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+
+++++++++++++
-V
vx
θ
Final position, xf
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v
vy
12
2
E fields-capacitors
E fields-capacitors
Parallel plate capacitors: Positive charges moving
horizontally at velocity v = v0, will accelerate vertically
down, consistent with Newton’s 2nd Law, due to the
electric force in the downward direction
Parallel plate capacitors: Positive charges moving
horizontally at velocity v = v0, will accelerate vertically
down, consistent with Newton’s 2nd Law, due to the
electric force in the downward direction
vx
Along x: No force along x. Then, vf same
as vi = v0 where v0 = (xf – xi ) / ∆t = sx / ∆t.
The time to travel across the length of the
v
Acceleration ax = 0. capacitor is ∆t = sx / v0
θ
+
θ
v
f
− vi
∆t
=
vy
∆t
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θ
v
vy
The vertical displacement is
found by using linear
kinematics with constant
acceleration in the vertical
direction
vy =
13
tan θ =
vy
vx
s y = vi t +
sy =
v
f
vy
− vi
vy
=
∆t
∆t
= a y∆t
F
qE
qV
=
=
m
m
md
qV
qVs x
∆t =
md
mdv 0
Copyright DR JJ,FSG, UiTM
14
E fields-capacitors
Parallel plate capacitors: Positive charges moving
horizontally. V = 100 V, d = 5 cm, q0 = 10e, where
the charge of an electron e = 1.6 x 10-19 C, and the
ratio e/me = 1.8 x 10-11 C/kg.; me is mass of electron.
Find E, F and a for positively charged particle.
Parallel plate capacitors: Positive charges moving
horizontally at velocity v = v0, will accelerate vertically
down, consistent with Newton’s 2nd Law, due to the
electric force in the downward direction
The angle θ
ay =
Then,
E fields-capacitors
vx
Then
But
acceleration
vy = a y∆t
Then
vy
v
Along y: E field pointing down, hence
electric force pushing the positive charge
+ down as long as it is in the E field. Then,
charge accelerates downward. vi = 0; vf = vy.
ay =
ay =
vx
vy
=
qVs x
2
mdv 0
a yt 2
Copyright DR JJ,FSG, UiTM
− 19
C ) x 2000 V/m
= 3.2x10 -15 N
2
qVs x
=
2
2
2 mdv 0
V
100 V
=
= 2000 V/m
d
0 . 05 m
F = qE = 10 (1 . 6 x 10
The
downward
force F
2
a yt 2
E =
Electric field E
The
acceleration
F
= ( 3 . 2 x 10
m
a =
15
− 15
N )x
1 . 8 x 10 − 11 C/kg
1 . 6 x 10 − 19 C
Copyright DR JJ,FSG, UiTM
16
Magnetic Field
Magnetic Monopole: Electric charge can be
separated until the smallest charge, the electron or
proton. Magnets have isolated north and south pole
and cannot be isolated into north only or south only.
S
N
Magnetic Fields and
Magnetic Forces
N
S
N
N
S
N
S
N
N
S
N
S
N S
S
Copyright DR JJ,FSG, UiTM
S
N
S
18
3
Magnetic Field
Magnetic Field
Magnetic Monopole: Electric charge can be
separated until the smallest charge, the electron or
proton. Magnets have isolated north and south pole
and cannot be isolated into north only or south only.
Earth as a Magnet: The earth magnetic north is
where the magnetic axis crosses the surface in the
northern hemisphere.
Fictitious Bar Magnet
Needle of compass is
labeled with north seeking
and south seeking poles.
Like poles repel
and unlike
(opposite) poles
attract.
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Its geographic north is where earth’s axis of rotation
crosses the surface in the northern hemisphere
19
Magnetic Field
20
Magnetic Field
Field lines: Magnetic field lines produced by a bar
magnet points from the north to the south as shown
below.
Field lines: Magnetic field lines produced by a bar
magnet and u-shaped magnet as shown below.
Field lines for permanent bar
magnet and The strength of
B is bigger near the poles as
seen by the lines which are
close together.
Compasses placed near the
bar magnet shows the field
directions.
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Copyright DR JJ,FSG, UiTM
21
Magnetic Field
For horse-shoe magnet, the
B field between poles are
equally strong
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22
Magnetic Field
Field lines: Magnetic field is a region where moving
charges may experience a magnetic force.
Field lines: Magnetic field lines produced by a bar
magnet.
Copyright DR JJ,FSG, UiTM
The charge is moving in the
same direction or opposite to
B. Hence charge don’t feel
magnetic force.
23
Copyright DR JJ,FSG, UiTM
The charge is moving
perpendicular to B. Hence
maximum force is
experienced
24
4
Magnetic Field
Magnetic Field
Field lines: Magnetic field is a region where moving
charges may experience a magnetic force.
Field lines: Magnetic field is a region where moving
charges may experience a magnetic force.
+
vper =
v(sin θ)
The charge is moving at an
angle θ with B. Only the
perpendicular component of
the velocity causes it to
experience force
B
θ
θ
vpar = v(cos θ)
G
G
G
F = q v x B = qv
Magnetic
field is
G
G G
F = q v x B = qv ⊥ B
= q ( v sin θ vB ) B
Copyright DR JJ,FSG, UiTM
B
θ
θ
vpar = v(cos θ)
B
v
= q ( v sin θ vB ) B
B to the right, v out of page, then
Force is upward. Hence the charge
will accelerate upward.
Negative charge accelerate downward.
27
Copyright DR JJ,FSG, UiTM
Magnetic Field
28
Magnetic Field
Thumb Rule/Corkscrew rule:
Thumb Rule/Corkscrew rule:
Determine the direction of
the resultant force for the
following cases:
Note the following vector
representation
Out of page
i) v and B into page
F
B
ii) v into page, B out of page
iii) v out of page, B to the right
v
v to the left, B out of page, hence the
force points upward. Then charge
accelerates upward.
Negative charge accelerates downward.
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B
G
G G
F = q v x B = qv ⊥ B
Force is out of page. Hence the
charge will accelerate out of page.
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= q ( v sin θ vB ) B
26
Direction of the resultant
force is determined by using
first right hand rule : Point all
fingers in B direction, thumb
in v direction, then palm will
face direction of force
Negative charge accelerate into page.
G
G G
F = q v x B = qv ⊥ B
The velocity is the
sum of the
perpendicular and
parallel components.
Right-Hand-Rule 1:
B
Direction of the resultant
force is determined by using
first right hand rule : Point all
fingers in B direction, thumb
in v direction, then palm will
face direction of force
B
v
θ
Magnetic Field
Field lines: Magnetic field is a region where moving
charges may experience a magnetic force.
+
B
θ
vpar = v(cos θ)
B
F
; Tesla
q ( v sin θ vB )
B =
25
vper =
v(sin θ)
⊥
= q ( v sin θ vB ) B
The velocity is the sum of the
perpendicular and parallel
components.
Magnetic Field
Direction of the resultant
force is determined by using
first right hand rule : Point all
fingers in B direction, thumb
in v direction, then palm will
face direction of force
+
When v & B are perpendicular,
vper =
then the angle θ = 90°, hence
v(sin θ)
F ≠ 0, F = Fmax.
B
v
Copyright DR JJ,FSG, UiTM
B
When v & B are parallel or
anti-parallel, then the angle θ
= 0, hence F = 0.
B
29
iv) X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
v
X X X
X X X
Into page
+
Copyright DR JJ,FSG, UiTM
30
5
Magnetic Force
Magnetic Force
Example: A proton moving in a B field
Magnetic force: Positive charges moving
horizontally from the left at velocity v, will accelerate
vertically down, consistent with Newton’s 2nd Law,
due to the magnetic force in the downward direction.
If it stays in the field, then its path will be a circle.
X
X
Straight line path
if no B field
FB
X
FB = q ( v sin θ vB ) B
X
X
FB
FB
+
G
v
+q
X
X
X
FB
FB = q ( v sin 90 D ) B
FB = qvB
X
X
X
X
Copyright DR JJ,FSG, UiTM
31
Copyright DR JJ,FSG, UiTM
Magnetic Force
Magnetic Force
Example: A proton moving at v = 5.0 x 106 m/s,
enters a magnetic field of 0.40 T at an angle of 30°.
Determine (a) the magnetic force and the
acceleration of the proton. (b) repeat for an electron.
Magnetic force: Positive charges moving
horizontally at velocity v, will accelerate vertically
down, consistent with Newton’s 2nd Law, due to the
magnetic force in the downward direction. If it stays in
the field, then its path will be a circle.
X
X
G
v
+q
X
X
Straight line path
if no B field
X
X
FB
X
X
B
X
+
X
X
For circular
motion, the
Xforce is
mv 2
= qvB
r
θ
vpar = v(cos θ)
B
v
Using Newton’s laws and
mp=1.67x10-27 kg. and
me=9.11x10-31 kg
33
For proton, charge is positive.
q=1.6x10-19 C.
FB = q ( v sin θ vB ) B
FB = q ( v sin 30 ) B
FB = 1.6x10 −13 N
F = ma
Then
a=
Hence, ap=9.6x1013 m/s
Copyright DR JJ,FSG, UiTM
F
m
ae=1.8x1017 m/s
34
Magnetic-Electric Forces
Magnetic-Electric Forces
Charge In Crossed Electric and Magnetic Fields:
Velocity Selector: Adjust E and B field to balance
the forces: FE = FB
If forces are
balanced
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B
θ
vper =
v(sin θ)
FB = qvB
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Parallel plate
capacitors: Positive
charges moving at
velocity v, entering an E
field is deflected
sideways
32
Velocity of
the particle
is
Magnetic force: Positive
charges entering the B
field experience a
deflection upwards. If it
stays in the field, then its
path will be a circle.
qE = qvB
v=
V
E
=
B dB
FB
+
FE
By adjusting the
potential V between
the plates, only the
charges with speed v
will be selected.
35
Copyright DR JJ,FSG, UiTM
36
6
Magnetic Force
Magnetic Force
Example: A proton moving enters a 2.0 Tesla B-field
and travels in a circle with r = 25 cm. Find its
momentum and the frequency and period of the
particle.
The charge moves in a circle and the force
Fcenter
responsible for the trajectory is the magnetic
2
force
mv
= q ( v sin 90 ) B Momentum mv
Then
r
Frequency f, is the number
of oscillations or revolution
in 1 second. Period, T, is
time for 1 complete
oscillation. So f = 1/T.
Example: A proton moving enters a 2.0 Tesla B-field
and travels in a circle with r = 25 cm. Find its
momentum and the frequency and period of the
particle.
Momentum
= FB
= rqB
displcemen t 2πr mv
v=
=
=
int erval
T
m
m p v = rq p B
Frequency f, is the number
of oscillations or revolution
in 1 second. Period, T, is
time for 1 complete
oscillation. So f = 1/T.
mp=1.67x10-27 kg
qp=1.6x10-19 C
rqB
= 2 π rf
m
The frequency is
v=
displcemen t 2πr mv
=
=
T
m
int erval
rq p B
m
= 2 π rf
p
qpB
f =
2π m
Period is
T =
2π m
p
p
qpB
mp=1.67x10-27 kg
Copyright DR JJ,FSG, UiTM
37
Magnetic-Electric Forces
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38
Magnetic-Electric Forces
Example: A proton is accelerated by the potential difference of
2100 V across the plates and entered a 0.01 T B-field which
points out of page. Find exit velocity through the slit and
determine the radius of the trajectory.
Example: A proton is accelerated by the potential difference of
2100 V across the plates and entered a 0.01 T B-field which
points out of page. Find exit velocity through the slit and
determine the radius of the trajectory.
Since plate separation, d, is
not known, then use
mechanical energy
conservation. Total
mechanical E is conserved.
The proton leaves the
electric field and enters the
magnetic field. Since v is
perpendicular to B, the force
FB acts on the proton and
pushes it to the right and
causes a circular path to be
traced.
(KE+PE) before = (KE+PE)after
Before: vA = 0 m/s,
KE = mvA2/2 = 0 kJ.
EPE = qVA
So, 0 + qVA = mvB2/2 + qVB
Then, 0 + q(VA - VB )= mvB2/2
After: KE =
EPE = qVB
Hence, vB2 = 2q(VA - VB )/m
6.3x105
vB =
mvB2/2
FC = FB ;
.
The radius is
m/s
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39
Magnetic-Electric Forces
Using energy conservation,
vB2 = 2q(VA - VB )/m
The radius is r = eB
mv
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mv
qB
Then, r = 6.6x10-2 m
Copyright DR JJ,FSG, UiTM
40
Magnetic-Electric Forces
Mass Spectrometer: A mass spectrometer is a device used to
determine the relative masses and the abundance of isotopes of
an element and identify unknown molecules produced in
chemical reactions
Atoms or molecules are
vaporized and ionized
(removal of electrons). The
positive charges are then
accelerated in a velocity
selector before entering a B
field. Only the chosen ions
will strike the detector.
r=
mv 2
= qvB
r
m2
Mass Spectrometer: A mass spectrometer is a device used to
determine the relative masses and the abundance of isotopes of
an element and identify unknown molecules produced in
chemical reactions
The radius is
m1
m2
r
r =
m1
v =
r
mv
eB
eBr
m
Replace VA – VB by just V,
vB2 = 2qV/m
v =
FC = F B ;
mv
r
2
= evB
41
Ions of different masses can
⎛ qr 2
be detected when B is
m = ⎜⎜
2
changed & qr /2V is constant
⎝ 2V
Copyright DR JJ,FSG, UiTM
e2B 2r 2
2 qV
=
m
m2
⎞ 2
⎟B
⎟
⎠
42
7
Magnetic-Electric Forces
Mass Spectrometer: Mass spectrum of naturally
occurring neon with 3 isotopes.
⎛ qr 2
m = ⎜⎜
⎝ 2V
Copyright DR JJ,FSG, UiTM
⎞ 2
⎟B
⎟
⎠
43
8