Download Chapter 16 – Electric Forces and Fields

College Physics 150 Chapter 16 – Electric Forces and Fields
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Electric Charge
Conductors & Insulators
Coulomb’s Law
Electric Field
Motion of a Point Charge in a Uniform E-­‐‑field
Conductors in Electrostatic Equilibrium
Gauss’s Law
What is Electric Charge?
Charge is a fundamental property of maKer just like mass. It cannot be broken down any further. Experiments show that likes charges will repel each other and unlike charges will aKract each other and that the force decreases with increasing distance between charges.
There are two kinds of electric charge: positive and negative.
A body is electrically neutral if the sum of all the charges in a body is zero.
Charge is a conserved quantity.
Electric Charge
The elementary unit of charge is e = 1.602×10-­‐‑19 C.
The charge on the electron is -­‐‑1e.
The charge on the proton is +1e.
The charge on the neutron is 0e.
This body is electrically neutral.
+
-­‐‑ -­‐‑
+ +
+
-­‐‑
+ -­‐‑
-­‐‑
An object can become polarized if the charges within it can be separated. By holding a charged rod near the body, it can be polarized.
-­‐‑
-­‐‑
+ + + + +
-­‐‑
-­‐‑
-­‐‑
+
+
+
+
+
Example (text problem 16.4): A metallic sphere has a charge of +4.0 nC. A negatively charged rod has a charge of -­‐‑6.0 nC. When the rod touches the sphere, 8.2×109 electrons are transferred. What are the charges of the sphere and the rod now?
Each electron has a charge 1.602×10-­‐‑19 C so the total charge transferred is -­‐‑1.3 nC.
The rod is left with -­‐‑6.0 nC + 1.3 nC = -­‐‑4.7 nC of charge and the sphere now has +4.0 nC -­‐‑ 1.3 nC = +2.7 nC of charge.
An insulator is made of material that does not allow electric charge to move through it easily.
A conductor is made of material that allows electric charge to move through it easily.
Coulomb’s Law
The magnitude of the force between two point charges is:
k q1 q2
F=
r2
where q1 and q2 are the charges, r is the separation between the two charges and k = 8.99×109 Nm2/C2.
1
where k =
and ε 0 = 8.85 ×10 −12 C 2 /Nm 2
4πε 0
and ε0 is called the permiKivity of free space.
The electric force is directed between the centers of the two point charges.
q1
F21
F12
q2
AKractive force between q1 and q2.
r
Repulsive force between q1 F21
and q2.
q1
q2
F12
r
The electric force is an example of a long-­‐‑range or field force, just like the force of gravity.
Example: What is the net force on the charge q1 due to the other two charges? q1 = +1.2 µC, q2 = -­‐‑0.60 µC, and q3 = +0.20 µC.
F21
θ
F31
The net force on q1 is Fnet = F21 + F31
The magnitudes of the forces are: 9
2
2
−6
−6
k q1 q2 ( 9 ×10 Nm /C ) (1.2 ×10 C)(0.60 ×10 C)
F21 =
=
2
r21
(1.2 m)2 + (0.5 m)2
= 3.8 ×10 −3 N
9
2
2
−6
−6
k q1 q3 ( 9 ×10 Nm /C ) (1.2 ×10 C)(0.20 ×10 C)
F31 =
=
2
r31
(1.2 m)2
= 1.5 ×10 −3 N
Example: What is the net force on the charge q1 due to the other two charges? q1 = +1.2 µC, q2 = -­‐‑0.60 µC, and q3 = +0.20 µC.
F21
θ
F31
The components of the net force are:
Fnet,x = F31,x + F21,x = −F31 + F21 cosθ = 2.0 ×10 −3 N
Fnet,y = F31,y + F21,y = 0 + F21 sin θ = 1.4 ×10 −3 N
1.2 m
cosθ =
= 0.92
1.3 m
Where from the figure
0.5 m
sin θ =
= 0.38
1.3 m
Example: What is the net force on the charge q1 due to the other two charges? q1 = +1.2 µC, q2 = -­‐‑0.60 µC, and q3 = +0.20 µC.
F21
θ
F31
The magnitude of the net force is:
2
2
Fnet = Fnet,x
+ Fnet,y
= 2.4 ×10 −3 N
Fnet,y
tan ϕ =
= 0.70
Fnet,x
The direction of the net force is:
ϕ = 35°
Example (text problem 16.11): What is the ratio of the electric force and gravitational force between a proton and an electron separated by 5.3×10-­‐‑11 m (the radius of a Hydrogen atom)?
k q1 q2
Fe =
r2
Gm1m2
Fg =
r2
The ratio is:
q1 = q2 = e
m1 = m p = 1.67 ×10 −27 kg
m2 = me = 9.11×10 −31 kg
Fe k q1 q2
ke 2
=
=
= 2.3×10 39
Fg Gm1m2 Gme m p
The Electric Field
Recall :
Fg = mg
Where g is the strength of the gravitational field.
Fe = qE
Similarly for electric forces we can define the strength of the electric field E.
For a point charge of charge Q, the magnitude of the force per unit charge at a distance r (the electric field) is: Fe k Q
E= = 2
q
r
The electric field at a point in space is found by adding all of the electric fields present.
E net = ∑ Ei
i
Be careful! The electric field is a vector!
Example: Find the electric field at the point P.
P
x
q1 = +e
x = 0 m
q2 = -­‐‑2e
x = 1 m
x = 2 m
E is a vector. What is its direction?
Place a positive test charge at the point of interest. The direction of the electric field at the location of the test charge is the same as the direction of the force on the test charge.
Example: Find the electric field at the point P.
P
x
q1 = +e
x = 0 m
q2 = -­‐‑2e
x = 1 m
x = 2 m
P
q1 = +e
Locate the positive test charge here.
q2 = -­‐‑2e
P
q1 = +e
x
x
q2 = -­‐‑2e
Direction of E due to charge 2
Direction of E due to charge 1
Example: Find the electric field at the point P.
P
x
q1 = +e
x = 0 m
q2 = -­‐‑2e
x = 1 m
The net electric field at point P is:
x = 2 m
E net = E1 + E 2
The magnitude of the electric field is:Enet
k q1
E1 = 2 =
r
k q2
E2 = 2 =
r
= E1 − E2
9
2
2
−19
9
×10
Nm
/C
(1.6
×10
C)
(
)
(2 m)2
= 3.6 ×10 −10 N/C
9
2
2
−19
9
×10
Nm
/C
(2
*1.6
×10
C)
(
)
The net E-­‐‑field is directed to the left.
(1 m)2
= 2.9 ×10 −9 N/C
Enet = E1 − E2 = −2.5 ×10
−9
N/C
Electric field lines
Electric field lines are a useful way to indicate what the magnitude and direction of an electric field is in space.
Rules:
1.  The direction of the E-field is tangent to the field lines at every point in space.
2.  The field is strong where there are many field lines and weak where there are few lines.
3.  The field lines start on + charges and end on - charges.
4.  Field lines do not cross.
Motion of a Point Charge in a Uniform E-­‐‑Field A region of space with a uniform electric field containing a particle of charge q (q > 0) and mass m. FBD for the charge q y
Apply Newton’s 2nd Law and solve for the acceleration.
Fe
x
∑F
x
= Fe = ma
Fe = qE = ma
q
a= E
m
One could now use the kinematic equations to solve for distance traveled in a time interval, the velocity at the end of a time interval, etc.
Example: What electric field strength is needed to keep an electron suspended in the air?
FBD for the electron:
y
To get an upward force on the electron, the electric field must be directed toward the Earth.
Fe
x
w
Apply Newton’s 2nd Law:
∑F
y
= Fe − w = 0
Fe = w
qE = eE = mg
mg
E=
= 5.6 ×10 −11 N/C
e
Example (text problem 16.50): A horizontal beam of electrons moving 4.0×107 m/s is deflected vertically by the vertical electric field between two oppositely charged parallel plates. The magnitude of the field is 2.00×104 N/C. (a) What is the direction of the field between the plates?
From the top plate to the boKom plate
(b) What is the charge per unit area on the plates?
This is the electric field between two E
charged plates.
Q
σ
=
=
ε0 A ε0
Note that E here is independent of the distance from the plates!
σ = Eε 0 = ( 2.00 ×10 4 N/C) (8.85 ×10 −12 C2 /Nm 2 )
= 1.77 ×10 −7 C/m 2
Example (text problem 16.50): A horizontal beam of electrons moving 4.0×107 m/s is deflected vertically by the vertical electric field between two oppositely charged parallel plates. The magnitude of the field is 2.00×104 N/C. y
Fe
(c) What is the vertical deflection d of the electrons as they leave the plates?
FBD for an electron in the beam:
Apply Newton’s 2nd Law and solve for the acceleration:
∑F
y
x
w
= Fe − w = may
Fe − w Fe
qE
ay =
= −g=
− g = (3.52 ×1015 − 9.8) m/s
m
m
m
Example (text problem 16.50): A horizontal beam of electrons moving 4.0×107 m/s is deflected vertically by the vertical electric field between two oppositely charged parallel plates. The magnitude of the field is 2.00×104 N/C. What is the vertical position of the electron after it travels a horizontal distance of 2.0 cm?
0
2
1
x = x0 + voxt + a xt
2
Time interval to travel 2.00 cm horizontally
x − x0
0.02 m
−10
t=
=
=
5.0
×10
sec
7
v0 x
4.0 ×10 m/s
1 2
y = y0 + voy t + a y t
2
0
1
y − y0 = d = ay t 2 = 4.4 ×10 −4 m
2
Deflection of an electron in the beam
Conductors in Electrostatic Equilibrium
Conductors are easily polarized. These materials have free electrons that are free to move around inside the material. Any charges that are placed on a conductor will arrange themselves in a stable distribution. This stable situation is called electrostatic equilibrium.
When a conductor is in electrostatic equilibrium, the E-­‐‑field inside it is zero.
Any net charge must reside on the surface of a conductor in electrostatic equilibrium.
Just outside the surface of a conductor in electrostatic equilibrium the electric field must be perpendicular to the surface. If this were not true, then any surface charge would have a net force acting on it, and the conductor would not be in electrostatic equilibrium.
Any excess charge on the surface of a conductor will accumulate where the surface is highly curved (i.e. a sharp point).
Gauss’s Law
Gauss’s Law relates the flux to the amount of charge enclosed! Enclose a point charge +Q with an imaginary sphere.
Here, E-­‐‑field lines exit the sphere.
Look at a small patch of the surface of the imaginary sphere.
+Q
With a positive charge inside the sphere you would see electric field lines leaving the surface.
Recall that
number of field lines
E∝
A
so that the number
of field lines ∝ EA
It is only the component of the electric field that is perpendicular to the surface that exits the surface.
Eperp
E
θ
Surface
Epar
Define a quantity called flux, which is related to the number of field lines that cross a surface:
flux = Φe = E⊥ A = ( E cosθ ) A
This angle θ is measured from the normal to the surface.
normal
E
θ
Flux > 0 when field lines exit the surface and flux < 0 when field lines enter the surface.
Example (text problem 16.62): Find the electric flux through each side of a cube of edge length a in a uniform electric field of magnitude E.
A cube has six sides: The field lines enter one face and exit through another. What is the flux through each of the other four faces?
There is zero electric flux though the other four faces. The electric field lines never enter/exit any of them. The flux through the left face is -­‐‑EA.
The flux through the right face is +EA. The net flux through the cube is zero.
Since E∝q, the flux through a surface can also be wriKen as Qinside
Φe =
ε0
This is Gauss’s Law.
The flux through a surface depends on the amount of charge inside the surface. Based on this, the cube in the previous example contained no net charge.
Φ e = E⊥ A
Qinside
This is Gauss’s Law.Φ e =
ε0
Consider a sphere around some charge
Qinside
E⊥ A =
ε0
Qinside
E⊥ 4 π r =
ε0
2
+Q
Qinside
E⊥ =
ε 0 4π r 2
kQinside
E⊥ =
r2
The flux relates electric field to the enclosed charge.
If you assume a closed sphere, then you return to Coulomb’s law.
Flux plays are bigger role with magnetism!