Lecture 3
Lectures Schedule Week (Week 10) Tuesday: Core Lecture 4 Wednesday: Physical Science Lecture 1 (Gauss's Law) Thursday: Core Lecture 5 Friday: Biological Electromagnetism Lecture 1 (Nerves, ECG) Problem Sheet 2 will be ready tomorrow which you will need for next week
The slides for these lectures are on my web page:
http://www.sct.gu.edu.au/~sctsang/
Last Lecture The Electric Field • The interaction between charged bodies (eg: between two electrons) is carried by an entity termed the electric field. • The electric field is defined by the following:
E=
F q
Unit = NC-1 = Vm-1 (working unit)
The electric field at a point in space is the electrostatic force per unit charge felt by a positive test charge q place there. • |E| is called the electric field strength
1
• Electric field of a point charge Applying Coulombs law for a the force for a point charge q on a test charge Q yields: r points in the 1 E=
F 4 = q
Qq rˆ 2 1 0 r = q 4
direction of the force Q would put on q
0
Q Y
r
Q rˆ r2
q
F = Eq X
• E is a vector field hence the total field due to n point charges is simply found by summing the forces and dividing by the test point charge q: 1 1 n E = ( F1 + F 2 + ... + F n ) = ∑ F i = E 1 + E 2 + ... + E n q q i =1
Electric Field Lines One positive and one negative point charge
High Field Region
Weak Field Region
Field lines begin on positive charge and terminate on the negative charge
2
Example 1: Determine E at the midpoint joining two point charges q1=1.0x10-6 C and q2=-3.0x10-6 C, separated by 1.0cm. d
E=
d
1 4
0
Q rˆ r2
()
q1 ˆ 1 q2 ˆ E1 = E2 = 2 i 2 i d d 4 4 0 0 x iˆ is the unit vector in the x-direction 1
q1
+ive
q2
+ive -ive E1 Test charge q at x=0
E2
Direction of E from each charge on the test charge?
E TOT =
E TOT = E 1 + E 2 E TOT =
(q1 + q2 ) ˆ
1 4
0
d2
i
1 (1.0 + 3.0) × 10−6 ˆ i × 2 4 × 8.854 × 10 −12 (5.0 × 10 −3) E TOT = 1.44 × 10 9 NC −1 iˆ
Lecture 3 Electrostatic Potential • Potential energy (PE) is the ability to do work, Electrostatic Potential is not PE but is proportional to it. • Since the electric field exerts a force on a charge, then a free charge will move and hence alter its PE. • The electrostatic force pushes a free charge from a region of higher potential to lower potential. Lower Potential
q E
Electrostatic potential (V) is the quantity, the gradient of which is equal to the electric field and has the unit of VOLT (V), 1V=1NmC-1: E =−
dV dx
one dimension
|E| = 10/1 =10 V/m
Note that the electrostatic potential is a scalar!
3
Relationship between Electrostatic Potential and Work (Energy) E=−
dV dx
Recall the definition of the electric field and equating: Energy lost by E field
F dV ∆V E= =− =− q dx ∆x ⇒ q∆V = − F∆x
Work=Force X Distance: ∆W = F∆x ⇒ ∆W = −q∆V Hence the energy gained by the charge due the electric field is given by ∆W = q∆V
q E This is the work done (energy gained) on a charge q moving through a potential difference ∆V. If the charge had to move against the potential it would lose this energy ∆W Potential Difference V is called voltage in electric circuits
• Electrostatic PE is analogous to the PE of a mass raised some height in a gravitational field. The greater the height raised the more the PE. • A charge has a higher PE the more it is moved against the force of an electric field. Example + Lower Potential Lower PE
Higher Potential Higher PE
+
• KE of the charge is converted to PE as it moves to the region of higher potential
4
Potential of a point charge Recall the Electric field due to a point charge Q is given by E=
1 4
dV 1 ⇒− = dr 4 ⇒ dV = −
0
Q r' 2
0
Q r' 2
1 4
0
Q dr r' 2
The potential at a distance r from the charge compared to infinity yields the potential: V (∞ )
∫
V (r )
∞
dV = − ∫
V (r ) =
r
1 4
0
Q 4
0
Q dr' r' 2
∞
1 × ( −1) Q Q 1 1 ⇒V (3 ∞ ) − V (r ) = − = − r 1 2 4 r' 4 0 0 ∞ r =0
The potential of a point charge Q
r
Exercise: Confirm that it is possible to get the electric field by taking the derivative of this function.
Example: Graphical representation of the potential felt by a charge due to a dipole field, the positive charge has +1C and negative -1C and 1m apart. Increase in potential to go up hill
Example
+ + +
+
Lower Potential Smaller PE
5
Example: Determine the potential due to the arrangement of point charges shown below where d=1.3m: q1=+12nC, q3=-24nC
r1
q2=+31nC q4=+17nC
The potential at P is just given by the algebraic sum of the potentials from each charge and remember the potential is a scalar quantity. Recall that the potential due to a point charge Q is V = V1 + V2 + V3 + V4 =
4
V (r ) =
Q 4
0
r
1 q1 q2 q3 q4 + + + r2 r3 r4 0 r1
By symmetry we see that each of the charges is an equal distance from P which is r=
d 2
⇒V = ⇒V =
2 4 4×
0
d
{q1 + q2 + q3 + q4}
2 × {(12) + (31) + (−24) + (17)} × 10 −9 = 350V × 8.85 × 10−12 × 1.3
Electric Current • Electric current is the net movement of charge and is analogous to the flow of water through a river • Imagine a point P in an object where charges are moving past with a velocity v. If a small amount dq passes in a time dt then the current is : v - - --t - - -- dq - - ⇒ dq = Idt ⇒ ∫ dq = q = I ∫ dt = It I = dt 0 P dq The unit of current is the Ampere, 1A = 1 Coulomb per second C/s
Conventional and electron currents • Currents are the result of an electric field exerting a force on charges that are free to move. • Current can be due to the flow of positive or negative charge • In a conductor the flow of charge is normally due to electron movement with q=-e, as such the current flows in the opposite direction to the charge movement. This is called Conventional Current. • Unless otherwise specified, current (I) used in the course will be conventional current. • Current is a scalar even though we attribute a direction to it (like time).
6
Example: A loop of copper No net flow of charge]I=0
e-
eCurrent in opposite direction To e- movement
ee-
e-
• Currents are conserved (this is a result of charge conservation)
i0 = i1 + i2 Example: What is the current and the direction in the lower right hand wire
8A Conservation of Current iin = iout Current in = 11A Current out = 3A
To equate these the current must be out =11-3=8A
7
Example: During a lightning strike ~ 104 A flow during the return stroke which takes around 100µs. How many e- pass between the ground and the charge? + + + ++ + + + ++ + ++ ++ + ++
- -- -- - Cloud - ----- --- - -- -- I
Current Direction?
++++++++++ Earth q = It −6
q = 10 × 100 × 10 = 1C 4
Recall that one e- has charge 1.602x10-19 C -
Number of e =
1C 18 ~ 6.2 × 10 electrons 1.602 × 10-19 C
Example: Television. The image is formed by 1 (B&W) or 3 (colour) beams of energetic electrons striking the screen. The electrons are accelerated in an electric field (the electron gun) E ~ 106 V/m. Say the beam current is 1.0x10-4 A (0.1 mA). How many e- are emitted per second by the electron gun?
q = it q = 1 × 10−4 ×1 = 1 × 10−4 C
Recall that one e- has charge 1.602x10-19 C -
Number of e =
1× 10-4 C 14 ~ 6.2 × 10 electrons 1.602 × 10-19 C
8
Current and Voltage in Conductors + + +
conductor
I
-E
How is E generated in a conductor?
• Current is the movement of charge • Charge movement is due to the electric field exerting a force on the charge: F=qE • Recall from Lecture 1 that charges move easily in conductors ⇒ Establish a potential difference between two points in the conductor
The potential difference between two points is called the voltage The electric field within a conductor is E=
V d
V is the voltage that exist between two points a distance d apart.
Note: This equation is not true in general since E may be non-uniform.
• The charges moving through a conductor with a potential difference V will have work done on them and their energy will change by W where ∆W = q∆V
• If q increases its potential, we must do work on it and hence it loses energy • If q lowers its potential, the field does work on it and the charge gains energy
9
Example.
The electron gun. An electric field is established between two parallel metal plates which have a small hole through which electrons enter and leave. The electrons are supplied by a filament, which is a wire heated to more than 2000 Kelvin so that it literally boils off electrons from its surface. Each electron gains energy eV as it is accelerated between the metal plates of the gun, where V is the potential difference (voltage) between the plates. Let d=1.0cm and V= 3000 V - typical of a CRO; TV guns run at 15-20kV. What is the E field and how fast will the electrons travel?
E=
V 3000 = = 3.0 × 105 V / m d 0.01
The electrons will gain kinetic energy: ∆W = eV = −1.602 × 10−19 × −3000 = 4.8 × 10−16 J
The velocity of the electrons can be found since K=
1 2 2K mv ⇒ v = 2 m
⇒v=
2K = m
m = 9.11×10 −31 kg
2 × 4.8 × 10 −16 = 3.3 × 10 7 ms-1 9.11 × 10 −31
10
The slides for these lectures are on my web page:
http://www.sct.gu.edu.au/~sctsang/
Last Lecture The Electric Field • The interaction between charged bodies (eg: between two electrons) is carried by an entity termed the electric field. • The electric field is defined by the following:
E=
F q
Unit = NC-1 = Vm-1 (working unit)
The electric field at a point in space is the electrostatic force per unit charge felt by a positive test charge q place there. • |E| is called the electric field strength
1
• Electric field of a point charge Applying Coulombs law for a the force for a point charge q on a test charge Q yields: r points in the 1 E=
F 4 = q
Qq rˆ 2 1 0 r = q 4
direction of the force Q would put on q
0
Q Y
r
Q rˆ r2
q
F = Eq X
• E is a vector field hence the total field due to n point charges is simply found by summing the forces and dividing by the test point charge q: 1 1 n E = ( F1 + F 2 + ... + F n ) = ∑ F i = E 1 + E 2 + ... + E n q q i =1
Electric Field Lines One positive and one negative point charge
High Field Region
Weak Field Region
Field lines begin on positive charge and terminate on the negative charge
2
Example 1: Determine E at the midpoint joining two point charges q1=1.0x10-6 C and q2=-3.0x10-6 C, separated by 1.0cm. d
E=
d
1 4
0
Q rˆ r2
()
q1 ˆ 1 q2 ˆ E1 = E2 = 2 i 2 i d d 4 4 0 0 x iˆ is the unit vector in the x-direction 1
q1
+ive
q2
+ive -ive E1 Test charge q at x=0
E2
Direction of E from each charge on the test charge?
E TOT =
E TOT = E 1 + E 2 E TOT =
(q1 + q2 ) ˆ
1 4
0
d2
i
1 (1.0 + 3.0) × 10−6 ˆ i × 2 4 × 8.854 × 10 −12 (5.0 × 10 −3) E TOT = 1.44 × 10 9 NC −1 iˆ
Lecture 3 Electrostatic Potential • Potential energy (PE) is the ability to do work, Electrostatic Potential is not PE but is proportional to it. • Since the electric field exerts a force on a charge, then a free charge will move and hence alter its PE. • The electrostatic force pushes a free charge from a region of higher potential to lower potential. Lower Potential
q E
Electrostatic potential (V) is the quantity, the gradient of which is equal to the electric field and has the unit of VOLT (V), 1V=1NmC-1: E =−
dV dx
one dimension
|E| = 10/1 =10 V/m
Note that the electrostatic potential is a scalar!
3
Relationship between Electrostatic Potential and Work (Energy) E=−
dV dx
Recall the definition of the electric field and equating: Energy lost by E field
F dV ∆V E= =− =− q dx ∆x ⇒ q∆V = − F∆x
Work=Force X Distance: ∆W = F∆x ⇒ ∆W = −q∆V Hence the energy gained by the charge due the electric field is given by ∆W = q∆V
q E This is the work done (energy gained) on a charge q moving through a potential difference ∆V. If the charge had to move against the potential it would lose this energy ∆W Potential Difference V is called voltage in electric circuits
• Electrostatic PE is analogous to the PE of a mass raised some height in a gravitational field. The greater the height raised the more the PE. • A charge has a higher PE the more it is moved against the force of an electric field. Example + Lower Potential Lower PE
Higher Potential Higher PE
+
• KE of the charge is converted to PE as it moves to the region of higher potential
4
Potential of a point charge Recall the Electric field due to a point charge Q is given by E=
1 4
dV 1 ⇒− = dr 4 ⇒ dV = −
0
Q r' 2
0
Q r' 2
1 4
0
Q dr r' 2
The potential at a distance r from the charge compared to infinity yields the potential: V (∞ )
∫
V (r )
∞
dV = − ∫
V (r ) =
r
1 4
0
Q 4
0
Q dr' r' 2
∞
1 × ( −1) Q Q 1 1 ⇒V (3 ∞ ) − V (r ) = − = − r 1 2 4 r' 4 0 0 ∞ r =0
The potential of a point charge Q
r
Exercise: Confirm that it is possible to get the electric field by taking the derivative of this function.
Example: Graphical representation of the potential felt by a charge due to a dipole field, the positive charge has +1C and negative -1C and 1m apart. Increase in potential to go up hill
Example
+ + +
+
Lower Potential Smaller PE
5
Example: Determine the potential due to the arrangement of point charges shown below where d=1.3m: q1=+12nC, q3=-24nC
r1
q2=+31nC q4=+17nC
The potential at P is just given by the algebraic sum of the potentials from each charge and remember the potential is a scalar quantity. Recall that the potential due to a point charge Q is V = V1 + V2 + V3 + V4 =
4
V (r ) =
Q 4
0
r
1 q1 q2 q3 q4 + + + r2 r3 r4 0 r1
By symmetry we see that each of the charges is an equal distance from P which is r=
d 2
⇒V = ⇒V =
2 4 4×
0
d
{q1 + q2 + q3 + q4}
2 × {(12) + (31) + (−24) + (17)} × 10 −9 = 350V × 8.85 × 10−12 × 1.3
Electric Current • Electric current is the net movement of charge and is analogous to the flow of water through a river • Imagine a point P in an object where charges are moving past with a velocity v. If a small amount dq passes in a time dt then the current is : v - - --t - - -- dq - - ⇒ dq = Idt ⇒ ∫ dq = q = I ∫ dt = It I = dt 0 P dq The unit of current is the Ampere, 1A = 1 Coulomb per second C/s
Conventional and electron currents • Currents are the result of an electric field exerting a force on charges that are free to move. • Current can be due to the flow of positive or negative charge • In a conductor the flow of charge is normally due to electron movement with q=-e, as such the current flows in the opposite direction to the charge movement. This is called Conventional Current. • Unless otherwise specified, current (I) used in the course will be conventional current. • Current is a scalar even though we attribute a direction to it (like time).
6
Example: A loop of copper No net flow of charge]I=0
e-
eCurrent in opposite direction To e- movement
ee-
e-
• Currents are conserved (this is a result of charge conservation)
i0 = i1 + i2 Example: What is the current and the direction in the lower right hand wire
8A Conservation of Current iin = iout Current in = 11A Current out = 3A
To equate these the current must be out =11-3=8A
7
Example: During a lightning strike ~ 104 A flow during the return stroke which takes around 100µs. How many e- pass between the ground and the charge? + + + ++ + + + ++ + ++ ++ + ++
- -- -- - Cloud - ----- --- - -- -- I
Current Direction?
++++++++++ Earth q = It −6
q = 10 × 100 × 10 = 1C 4
Recall that one e- has charge 1.602x10-19 C -
Number of e =
1C 18 ~ 6.2 × 10 electrons 1.602 × 10-19 C
Example: Television. The image is formed by 1 (B&W) or 3 (colour) beams of energetic electrons striking the screen. The electrons are accelerated in an electric field (the electron gun) E ~ 106 V/m. Say the beam current is 1.0x10-4 A (0.1 mA). How many e- are emitted per second by the electron gun?
q = it q = 1 × 10−4 ×1 = 1 × 10−4 C
Recall that one e- has charge 1.602x10-19 C -
Number of e =
1× 10-4 C 14 ~ 6.2 × 10 electrons 1.602 × 10-19 C
8
Current and Voltage in Conductors + + +
conductor
I
-E
How is E generated in a conductor?
• Current is the movement of charge • Charge movement is due to the electric field exerting a force on the charge: F=qE • Recall from Lecture 1 that charges move easily in conductors ⇒ Establish a potential difference between two points in the conductor
The potential difference between two points is called the voltage The electric field within a conductor is E=
V d
V is the voltage that exist between two points a distance d apart.
Note: This equation is not true in general since E may be non-uniform.
• The charges moving through a conductor with a potential difference V will have work done on them and their energy will change by W where ∆W = q∆V
• If q increases its potential, we must do work on it and hence it loses energy • If q lowers its potential, the field does work on it and the charge gains energy
9
Example.
The electron gun. An electric field is established between two parallel metal plates which have a small hole through which electrons enter and leave. The electrons are supplied by a filament, which is a wire heated to more than 2000 Kelvin so that it literally boils off electrons from its surface. Each electron gains energy eV as it is accelerated between the metal plates of the gun, where V is the potential difference (voltage) between the plates. Let d=1.0cm and V= 3000 V - typical of a CRO; TV guns run at 15-20kV. What is the E field and how fast will the electrons travel?
E=
V 3000 = = 3.0 × 105 V / m d 0.01
The electrons will gain kinetic energy: ∆W = eV = −1.602 × 10−19 × −3000 = 4.8 × 10−16 J
The velocity of the electrons can be found since K=
1 2 2K mv ⇒ v = 2 m
⇒v=
2K = m
m = 9.11×10 −31 kg
2 × 4.8 × 10 −16 = 3.3 × 10 7 ms-1 9.11 × 10 −31
10
