chapter 16 notes
Electric Charge and Field • Chapter 16 (Giancoli) • All sections except 16.10 (Gauss’s law)
Comparison between the Electric and the Gravitational forces Both have long range,
F∝
1 r2
The electric charge of an object plays the same role in determining the electric force as does the object’s mass in the gravitational force Both are central forces (forces act along the line joining the objects)
• But!
Strength
•
The electric forces exerted by one part of an atom on another part are much stronger than their gravitational attraction.
Mass is always a positive quantity → the gravitational force is always an attractive force.
•
Electric charge can be positive or negative i.e. the electric force can be attractive or repulsive
i.e. For atomic/subatomic particles, the gravitational force is usually neglected. (We shall see an example later on)
A very useful rule for determining the direction of the force: “Like charges repel, unlike charges attract”
The law of conservation of charge
The Electronic Charge
Total amount of charge in an isolated system remains constant.
The magnitude of the smallest (elementary) free charge is the charge on the electron or the proton.
“One cannot create or destroy a charge”
This charge is denoted by e.
Example: The decay of the neutron. An isolated neutron decays in about 15 minutes.
By convention: Charge on proton = + e Charge on electron = – e
n0 → p+ + e− + v0 (neutron)
(proton)
(electron) (neutrino)
(Total charge is zero before and after the decay)
© Z. Altounian
In the SI system, the unit of charge is the coulomb [C]. e = 1.6 × 10-19 C
1
Putting these together,
Coulomb’s Law: Provides a description for the electric force. If two charges, Q1 and Q2 are separated by a distance r
Q1
F =
kQ1Q 2 r2
Q2
r
the magnitude of the force acting on any charge is proportional to the product of the two charges. (F∝ Q1Q2 )
This is Coulomb’s Law and gives the magnitude of the force between two charged particles. k, the proportionality constant, is given by
It is also proportional to the inverse square of the distance between the charges. 1
k= 8.99x109 Nm2/C2 = 9x109 Nm2/C2
Sometimes it is more convenient to write:
Example:
(F ∝
k
=
r2
)
1 4 π εo
Q1
2 1 = 8 .85 × 10 − 12 C Νm 4πk
=
r
2
If the two charges are in a medium other than vacuum, then k
F1
+
where ε0 is the permittivity of vacuum ε0 =
Q2
1 4π ε
What is the force acting on Q1?
magnitude:
F1 =
kQ1Q 2 r2
where ε is the permittivity of the medium One can also define the relative permittivity, κ = ε / ε0, a dimensionless quantity, such that ε = κ ε0
Example:
direction: Attractive force (unlike charges)
Summary: Q1
+
Q2
F1 F2 r
F =
What is the force acting on Q2?
Answer: Magnitude: F2 =
kQ1Q 2 r2
kQ1Q 2 r2
Note: The magnitude of a vector is always a positive quantity. Do not use the signs of the charges in this equation when calculating the magnitude of the force.
= F1
Direction: Attractive force (unlike charges)
© Z. Altounian
The magnitude of the force is given by Coulomb’s Law,
The direction of the force is determined from “Like charges repel, unlike charges attract”
2
Example : Two fixed charges, Q1= 3 µC and Q2 = - 6 µC are separated by 3 m. Find the electric force (magnitude and direction) acting on Q1. Q1 = 3 µC
To find the magnitude, use Coulomb’s Law.
Q2 = -6 µC
F1 =
kQ1Q 2
F1
+
=
r=3m
r2
(9 × 10 9 N ⋅ m 2 / C 2 )(3 × 10 −6 C )(6 × 10 −6 C ) (3 m) 2
To find the direction of the force, use the rule:
= 1.8 × 10 − 2 N
“like charges repel, unlike charges attract”
(do not include the signs of the charges when using Coulomb’s Law)
Example : Compare the electric and gravitational forces between an electron of mass 9.1 x 10-31 kg moving around a proton of mass 1.67 x 10-27 kg in the H-atom at a distance of 0.05 x 10-9 m.
What is the force acting on Q2?
Q1 = 3 µC
Q2 = -6 µC
+
F1 F2 mp = 1.67 x 10-27 kg
r=3m
me = 9.1 x 10-31 kg
+ r = 0.05 nm
The magnitude of F2 =1.8 x 10-2 N The direction of F2 is toward Q1
Example : Compare the electric and gravitational forces between an electron of mass 9.1 x 10-31 kg moving around a proton of mass 1.67 x 10-27 kg in the H-atom at a distance of 0.05 x 10-9 m.
Electric Force: Fe = =
mp = 1.67 x 10-27 kg
+
me = 9.1 x 10-31 kg
Fe Fg
Fe Fg
r = 0.05 nm
kQ1Q 2 r2
(9 × 109 Nm 2 / C2 )(1.6 × 10− 19 C)(1.6 × 10− 19 C) (0.05 × 10− 9 m) 2
= 9.22 × 10− 8 N
Gravitational Force: Fg = =
Gm1m2 r2
(6.67 × 10− 11 Nm2 / kg2 )(1.67 × 10− 27 kg)(9.1 × 10− 31 kg) (0.05 × 10− 9 m)2
= 4.05 × 10− 47 N
© Z. Altounian
3
What happens when there are more than two charges present? How is Coulomb’s Law applied?
Therefore, Fe/Fg= 2.27 x1039
Use the superposition principle “The electric force on a given charge is the vector sum of the electric forces due to each charge”
Hence, Fe >> Fg; the electrical force is much stronger than the gravitational force for atomic particles. Fg is neglected for such particles.
Conceptual Example: What is the force on Q? Q1
+
+
Q1
Q2
+
Q3
Conceptual Example: What is the force on Q?
Q3
Q
Conceptual Example: What is the force on Q? Q1
Q2
+
F2
Q
Example : Find the electric force acting on Q3
+Q3
F2 r31
Q3
K K K K F = F1 + F2 + F3
© Z. Altounian
F1
Q2
+
+
F3
+
+
+ Q Vector Addition!
F3 F1
+Q1
20 cm
r32 =20 cm
Where, Q1 = 6 µC Q2 = 4 µC Q3 = 2 µC
+Q2
r32=0.20 m r31= 2 (0.20 m)
4
F31
F32
F31 =
+Q3
=
20 cm
kQ1Q 3 ( r13 ) 2
(9 × 109 Nm 2 / C 2 )( 6 × 10 − 6 C)( 2 × 10 − 6 C) ( 0 .2 ×
2 m)2
= 1.35 N +Q1
20 cm
+Q2
F32 =
The forces acting on Q3 are: F31 is the force acting on Q3 due to Q1 F32 is the force acting on Q3 due to Q2
=
kQ2Q3 (r23)2
(9 × 109 Nm2 / C2 )(4 × 10− 6 C)(2 × 10− 6 C) (0.2 m)2
= 1.8 N
Add F31 and F32 vectorially to find F3
Direction of F3:
F3x = F31(x) + F32(x)
y
tan θ
= 1.35 cos 45o + 0 = 0.955 N
F31
F32
F3y = F31(y) + F32(y)
=
F31(y)
45° Q3 F (x) 31
= 1.35 sin 45o + 1.8 N = 2.75 N 45°
Q2
20 cm
F3 y
F3
y
F3 x F32
2 . 88
F31 θ
x
20 cm Q1
=
θ = 71D
Q3
x
20 cm
(measured from the x-axis) Q1
45° 20 cm
Q2
F32x + F32y = 2.92 N
F3 =
Electric Fields Consider the space surrounding a charge Q1. Any charge q placed in that space will experience an electric force. F1
F3
q
q
The electric field is a property of the charge Q1. The charge q, which is used in determining the electric field is called a test charge. By convention it is a positive charge. The test charge is not a real charge. We use it only to determine the direction of the electric field.
3
1
+
Q1
q 2
F2
We can calculate the electric force acting on q at any point in space. The collection of all these force vectors is the electric field created by a charge Q1.
© Z. Altounian
5
The direction of E is the direction of the force acting on the positive test charge at that point.
Definition of the Electric Field Define the magnitude of the electric field, E, at a distance r from a point charge Q, as the force on the test charge at that point divided by the value of the test charge.
E =
E =
⎛ kqQ ⎞ 1 F ⎟ = ⎜⎜ 2 ⎟ q ⎝ r ⎠ q
kQ r2
Q r
Q
+
E r
If we know the E-field at a point, then if a charge Q´ is placed at that point it will experience an electric force given by
K K F = Q ′E
Units for E: N/C
This is the magnitude of the E-field created by the charge Q.
Note that the E-field is not produced by Q'.
If Q´ is a positive charge, the direction of the force acting on it will be in the same direction as the E-field at that point.
Example : Three charges are positioned at 3 of the 4 corners of a square of side 10 cm. Each have a charge of 5 x 10-10 C. Calculate the E-field at the empty corner.
Q´ E
F
+
F = Q´E
E
If Q´ is a negative charge, the direction of the force acting on it will be in the opposite direction from the the E-field at that point. – Q´ F E
+Q1
10 cm +Q2
F = – Q´E
E
Example : Three charges are positioned at 3 of the 4 corners of a square of side 10 cm. Each have a charge of 5 x 10-10 C. Calculate the E-field at the empty corner.
+
+ +Q
+
3
Example : Three charges are positioned at 3 of the 4 corners of a square of side 10 cm. Each have a charge of 5 x 10-10 C. Calculate the E-field at the empty corner. E3 E2
+Q1
+
+Q1 10 cm
+Q2
+
+ +Q
3
For more than one charge use the superposition principle:
G G G G E = E1 + E2 + E3
© Z. Altounian
r1
+ r2
+Q2
+
45°
E1 r3
10 cm
+ +Q
3
For more than one charge use the superposition principle:
G G G G E = E1 + E2 + E3
6
Ex = E1 + E2 cos 45° = 450 N/C + (225 N/C) (0.707) = 608 N/C
Magnitudes: E1 =
E2 =
kQ1 r12
(9 × 109 N ⋅ m2 / C2 )(5 × 10−10 C)
=
kQ2
(0.1 m)2 (9 × 109 N ⋅ m2 / C2 )(5 × 10−10 C)
=
r22
(0.1 ×
2 m)
2
= 450 N / C
Ey = E3 + E2 sin 45° = 450 + 225 (0.707) = 608 N/C
= 225 N / C
y
E3
+Q1
E2 45°
+
x
E1 10 cm
E3 =
kQ3
9
2
(9 × 10 N ⋅ m / C )(5 × 10
=
r32
2
−10
C)
(0.1 m)2
= 450 N / C
+Q2
+
45°
+ +Q
3
If a charge of -5 X 10-10 C is placed at the 4th corner, what will be the electric force acting on this charge?
Therefore, the magnitude is; y
E =
E2x
+
E2y
= 861 N / C
E
E3
G G F = Q4 E
E E3
And the direction is;
tan θ =
Ey Ex
θ = 45
+Q1
45°
+
E1 10 cm
=1
D
© Z. Altounian
+Q2
+
45°
+ +Q
3
+Q1
– -Q
+ F
x
+Q2
+
45°
4
10 cm
+ +Q
3
G The magnitude of F = (5 x 10-10 C)(861 N/C) = 4.3 x 10 -7 N G G The direction of F is opposite to E because Q4 is negative
7
Comparison between the Electric and the Gravitational forces Both have long range,
F∝
1 r2
The electric charge of an object plays the same role in determining the electric force as does the object’s mass in the gravitational force Both are central forces (forces act along the line joining the objects)
• But!
Strength
•
The electric forces exerted by one part of an atom on another part are much stronger than their gravitational attraction.
Mass is always a positive quantity → the gravitational force is always an attractive force.
•
Electric charge can be positive or negative i.e. the electric force can be attractive or repulsive
i.e. For atomic/subatomic particles, the gravitational force is usually neglected. (We shall see an example later on)
A very useful rule for determining the direction of the force: “Like charges repel, unlike charges attract”
The law of conservation of charge
The Electronic Charge
Total amount of charge in an isolated system remains constant.
The magnitude of the smallest (elementary) free charge is the charge on the electron or the proton.
“One cannot create or destroy a charge”
This charge is denoted by e.
Example: The decay of the neutron. An isolated neutron decays in about 15 minutes.
By convention: Charge on proton = + e Charge on electron = – e
n0 → p+ + e− + v0 (neutron)
(proton)
(electron) (neutrino)
(Total charge is zero before and after the decay)
© Z. Altounian
In the SI system, the unit of charge is the coulomb [C]. e = 1.6 × 10-19 C
1
Putting these together,
Coulomb’s Law: Provides a description for the electric force. If two charges, Q1 and Q2 are separated by a distance r
Q1
F =
kQ1Q 2 r2
Q2
r
the magnitude of the force acting on any charge is proportional to the product of the two charges. (F∝ Q1Q2 )
This is Coulomb’s Law and gives the magnitude of the force between two charged particles. k, the proportionality constant, is given by
It is also proportional to the inverse square of the distance between the charges. 1
k= 8.99x109 Nm2/C2 = 9x109 Nm2/C2
Sometimes it is more convenient to write:
Example:
(F ∝
k
=
r2
)
1 4 π εo
Q1
2 1 = 8 .85 × 10 − 12 C Νm 4πk
=
r
2
If the two charges are in a medium other than vacuum, then k
F1
+
where ε0 is the permittivity of vacuum ε0 =
Q2
1 4π ε
What is the force acting on Q1?
magnitude:
F1 =
kQ1Q 2 r2
where ε is the permittivity of the medium One can also define the relative permittivity, κ = ε / ε0, a dimensionless quantity, such that ε = κ ε0
Example:
direction: Attractive force (unlike charges)
Summary: Q1
+
Q2
F1 F2 r
F =
What is the force acting on Q2?
Answer: Magnitude: F2 =
kQ1Q 2 r2
kQ1Q 2 r2
Note: The magnitude of a vector is always a positive quantity. Do not use the signs of the charges in this equation when calculating the magnitude of the force.
= F1
Direction: Attractive force (unlike charges)
© Z. Altounian
The magnitude of the force is given by Coulomb’s Law,
The direction of the force is determined from “Like charges repel, unlike charges attract”
2
Example : Two fixed charges, Q1= 3 µC and Q2 = - 6 µC are separated by 3 m. Find the electric force (magnitude and direction) acting on Q1. Q1 = 3 µC
To find the magnitude, use Coulomb’s Law.
Q2 = -6 µC
F1 =
kQ1Q 2
F1
+
=
r=3m
r2
(9 × 10 9 N ⋅ m 2 / C 2 )(3 × 10 −6 C )(6 × 10 −6 C ) (3 m) 2
To find the direction of the force, use the rule:
= 1.8 × 10 − 2 N
“like charges repel, unlike charges attract”
(do not include the signs of the charges when using Coulomb’s Law)
Example : Compare the electric and gravitational forces between an electron of mass 9.1 x 10-31 kg moving around a proton of mass 1.67 x 10-27 kg in the H-atom at a distance of 0.05 x 10-9 m.
What is the force acting on Q2?
Q1 = 3 µC
Q2 = -6 µC
+
F1 F2 mp = 1.67 x 10-27 kg
r=3m
me = 9.1 x 10-31 kg
+ r = 0.05 nm
The magnitude of F2 =1.8 x 10-2 N The direction of F2 is toward Q1
Example : Compare the electric and gravitational forces between an electron of mass 9.1 x 10-31 kg moving around a proton of mass 1.67 x 10-27 kg in the H-atom at a distance of 0.05 x 10-9 m.
Electric Force: Fe = =
mp = 1.67 x 10-27 kg
+
me = 9.1 x 10-31 kg
Fe Fg
Fe Fg
r = 0.05 nm
kQ1Q 2 r2
(9 × 109 Nm 2 / C2 )(1.6 × 10− 19 C)(1.6 × 10− 19 C) (0.05 × 10− 9 m) 2
= 9.22 × 10− 8 N
Gravitational Force: Fg = =
Gm1m2 r2
(6.67 × 10− 11 Nm2 / kg2 )(1.67 × 10− 27 kg)(9.1 × 10− 31 kg) (0.05 × 10− 9 m)2
= 4.05 × 10− 47 N
© Z. Altounian
3
What happens when there are more than two charges present? How is Coulomb’s Law applied?
Therefore, Fe/Fg= 2.27 x1039
Use the superposition principle “The electric force on a given charge is the vector sum of the electric forces due to each charge”
Hence, Fe >> Fg; the electrical force is much stronger than the gravitational force for atomic particles. Fg is neglected for such particles.
Conceptual Example: What is the force on Q? Q1
+
+
Q1
Q2
+
Q3
Conceptual Example: What is the force on Q?
Q3
Q
Conceptual Example: What is the force on Q? Q1
Q2
+
F2
Q
Example : Find the electric force acting on Q3
+Q3
F2 r31
Q3
K K K K F = F1 + F2 + F3
© Z. Altounian
F1
Q2
+
+
F3
+
+
+ Q Vector Addition!
F3 F1
+Q1
20 cm
r32 =20 cm
Where, Q1 = 6 µC Q2 = 4 µC Q3 = 2 µC
+Q2
r32=0.20 m r31= 2 (0.20 m)
4
F31
F32
F31 =
+Q3
=
20 cm
kQ1Q 3 ( r13 ) 2
(9 × 109 Nm 2 / C 2 )( 6 × 10 − 6 C)( 2 × 10 − 6 C) ( 0 .2 ×
2 m)2
= 1.35 N +Q1
20 cm
+Q2
F32 =
The forces acting on Q3 are: F31 is the force acting on Q3 due to Q1 F32 is the force acting on Q3 due to Q2
=
kQ2Q3 (r23)2
(9 × 109 Nm2 / C2 )(4 × 10− 6 C)(2 × 10− 6 C) (0.2 m)2
= 1.8 N
Add F31 and F32 vectorially to find F3
Direction of F3:
F3x = F31(x) + F32(x)
y
tan θ
= 1.35 cos 45o + 0 = 0.955 N
F31
F32
F3y = F31(y) + F32(y)
=
F31(y)
45° Q3 F (x) 31
= 1.35 sin 45o + 1.8 N = 2.75 N 45°
Q2
20 cm
F3 y
F3
y
F3 x F32
2 . 88
F31 θ
x
20 cm Q1
=
θ = 71D
Q3
x
20 cm
(measured from the x-axis) Q1
45° 20 cm
Q2
F32x + F32y = 2.92 N
F3 =
Electric Fields Consider the space surrounding a charge Q1. Any charge q placed in that space will experience an electric force. F1
F3
q
q
The electric field is a property of the charge Q1. The charge q, which is used in determining the electric field is called a test charge. By convention it is a positive charge. The test charge is not a real charge. We use it only to determine the direction of the electric field.
3
1
+
Q1
q 2
F2
We can calculate the electric force acting on q at any point in space. The collection of all these force vectors is the electric field created by a charge Q1.
© Z. Altounian
5
The direction of E is the direction of the force acting on the positive test charge at that point.
Definition of the Electric Field Define the magnitude of the electric field, E, at a distance r from a point charge Q, as the force on the test charge at that point divided by the value of the test charge.
E =
E =
⎛ kqQ ⎞ 1 F ⎟ = ⎜⎜ 2 ⎟ q ⎝ r ⎠ q
kQ r2
Q r
Q
+
E r
If we know the E-field at a point, then if a charge Q´ is placed at that point it will experience an electric force given by
K K F = Q ′E
Units for E: N/C
This is the magnitude of the E-field created by the charge Q.
Note that the E-field is not produced by Q'.
If Q´ is a positive charge, the direction of the force acting on it will be in the same direction as the E-field at that point.
Example : Three charges are positioned at 3 of the 4 corners of a square of side 10 cm. Each have a charge of 5 x 10-10 C. Calculate the E-field at the empty corner.
Q´ E
F
+
F = Q´E
E
If Q´ is a negative charge, the direction of the force acting on it will be in the opposite direction from the the E-field at that point. – Q´ F E
+Q1
10 cm +Q2
F = – Q´E
E
Example : Three charges are positioned at 3 of the 4 corners of a square of side 10 cm. Each have a charge of 5 x 10-10 C. Calculate the E-field at the empty corner.
+
+ +Q
+
3
Example : Three charges are positioned at 3 of the 4 corners of a square of side 10 cm. Each have a charge of 5 x 10-10 C. Calculate the E-field at the empty corner. E3 E2
+Q1
+
+Q1 10 cm
+Q2
+
+ +Q
3
For more than one charge use the superposition principle:
G G G G E = E1 + E2 + E3
© Z. Altounian
r1
+ r2
+Q2
+
45°
E1 r3
10 cm
+ +Q
3
For more than one charge use the superposition principle:
G G G G E = E1 + E2 + E3
6
Ex = E1 + E2 cos 45° = 450 N/C + (225 N/C) (0.707) = 608 N/C
Magnitudes: E1 =
E2 =
kQ1 r12
(9 × 109 N ⋅ m2 / C2 )(5 × 10−10 C)
=
kQ2
(0.1 m)2 (9 × 109 N ⋅ m2 / C2 )(5 × 10−10 C)
=
r22
(0.1 ×
2 m)
2
= 450 N / C
Ey = E3 + E2 sin 45° = 450 + 225 (0.707) = 608 N/C
= 225 N / C
y
E3
+Q1
E2 45°
+
x
E1 10 cm
E3 =
kQ3
9
2
(9 × 10 N ⋅ m / C )(5 × 10
=
r32
2
−10
C)
(0.1 m)2
= 450 N / C
+Q2
+
45°
+ +Q
3
If a charge of -5 X 10-10 C is placed at the 4th corner, what will be the electric force acting on this charge?
Therefore, the magnitude is; y
E =
E2x
+
E2y
= 861 N / C
E
E3
G G F = Q4 E
E E3
And the direction is;
tan θ =
Ey Ex
θ = 45
+Q1
45°
+
E1 10 cm
=1
D
© Z. Altounian
+Q2
+
45°
+ +Q
3
+Q1
– -Q
+ F
x
+Q2
+
45°
4
10 cm
+ +Q
3
G The magnitude of F = (5 x 10-10 C)(861 N/C) = 4.3 x 10 -7 N G G The direction of F is opposite to E because Q4 is negative
7
