Transcript Coulomb’s Law - University of Delaware

22-4 Coulomb’s Law
F k
q1  q2
r2
• Unit
• Only apply to point charges
q1
q2
r
F k
q1  q2
• Force is a vector
• Direction is along the line joining two particles
• Attractive/repulsive for unlike/like charges
r2
Coulomb’s Law: Example
Consider the forces between the proton and the electron in hydrogen
atom in the ground state (lowest energy) where the distance
between is 0.53x10-10 m.
Gravitational force
19
19
1
.
61

10

1
.
61

10
F  9  109
10 2
0.53 10
8
 8.2  10
N 


11
Fgrav  6.67 10
 3.6 10 47 N 

1836 9.11031  9.11031
0.5310 
10 2
• Observations:
• At atomic level, one may ignore gravitational force
• Atoms are stable because of strong electrical interaction
• we typically experience gravitational force or electrical force
Back to the soda can experiment
Can you make soda can positively
charged
Superposition Principle
In a system with multiple charge particles, the net
force on one particle due to other charges (two or
more) is the vector sum of each force due to each
individual charge, and each individual contribution is
unaffected by the presence of the other charge

  

F   Fi F1  F2      FN
i
Superposition Principle: an example
q2
Assume q2 and q3 carry
same type of charges but
opposite to q1, calculate
the total force on q1
q1
Step 1: calculate
F12
Step 2: calculate
F13
q2
F12  k
q1
q1  q2
r12
F13  k
2
q1
q1  q3
r13
2
q3
q3
Step 3: calculate F
23-2 Electric Field
F21  k
q1  q2
r
2
E1  k
q1
r
2


F21  q2 E1
• Unit for E: N/C
• Again, only applies to point charges
• What is field?
– A “field” is any physical quantity that has a value at
every location in space. Its value can scalar (scalar field)
or a vector (vector field).
Scalar Field: Temperature T(x,y,z)
Vector Field: Air flow in a room V(x,y,z)
23-2 Electric Field: cont.
F21  k
q1  q2
r
2
E1  k
q1
r
2


F21  q2 E1
• E field is a vector
•Measure an E field. Using a test charge q2 and
measure the force acting on q2。
• Superposition principle is valid.
Concept Check
P
y
The E field at point P
x
+q
1.
2.
3.
4.
5.
along +x
along –x
along +y
along –y
another direction
-q
• Field line: always goes from positive to negative charges
• direction of field: tangent to field line
• strength of field: density of field line per unit area
Field of Continuous Charge Distribution
1. Evaluate the contribution from a
small charge element: dq, dE
Don’t choose a special point
Don’t perform detailed calculation
Do decompose the dE into
components: || and  to the
symmetric axis. (coordinate)
2. Exploit symmetry: if one component is zero, do not calculate it.
Concept Check: Symmetry
+
+
+
+
+
+
+
+
What is the direction
of E at point P?
+
P
1.
2.
3.
4,
5.
Up
+
+
Down
Left
Right
None of above
+
+
+
P
- -
-
3. Set up integral: need to evaluate dq and express it in terms
of a position parameter.
Quiz 2 (2/26/01, Friday)
Two charges are fixed on the
perimeter of a circular clock
face. +Q is placed at 2
o'clock; +2Q is at 12 o'clock.
1. Where should a third charge
be placed on the perimeter so
that the electric field is zero at
the center (point P)?
2. What relative charge strength
is required at that point for
the electric field to be zero?
Gauss’ Law
Quantitative relationship between electric field (direction and
magnitude) on a CLOSED SURFACE (Gaussian surface)
and total charges inside the Gaussian surface.
 
 0  E  dA  qenl
Flux of Electric Field
Measures how much field through a surface
 
   E  dA   E cos dA
• Direction of surface (dA): normal to the surface
• Flux is a function of E, A, and .
• Therefore, =0 does not necessarily mean E=0.
 
 0  E  dA  qenl
Apply Gauss’s law
 0 E cos   dA  qenl
r
Two types of Gaussian Surfaces:
L
r
dA

4

r


2
dA 
dA

dA
side

 t b
t b
side
 2r  4rL
2
Apply Gauss’s law
 0 E cos   dA  qenl
 
 0  E  dA  qenl
1. Point of interest has to be on Gaussian surface
2. Choose Gaussian surface so that E can be taken out of
integration: explore the symmetry of E
• E are the same everywhere on surface (sphere),
often =0 or 180 so that cos0=1 or cos180=-1
• E  surface (align on the surface): cos90=0
• Combination (cylinder)
Symmetry of E
Gaussian Surface
Spherical (point & spherical charges)
Cylindrical (line charge)
Planar (sheet of charge)
Sphere
Cylinder
Box or cylinder
Apply Gauss’s law
 0 EAr    qenl
 
 0  E  dA  qenl
3. Add all charges inside Gaussian Surface: algebraic sum
+qoutside
 outside
4. Calculate E.
1<0
2 >0
 
  E  dA 1   2  0
Concept Check: Gauss’ law
A
B
C
+q
+q
+q
p
p
p
+q
-q
Which situation has more electric flux through the closed surface?
Which situation has the largest E at the point P?
Gauss’ Law: Important Properties of Metals
• In a static (no charge motion) condition, there cannot be
excess charges in the interior of a metal, all excess charges
must be on the surface
• An equivalent statement: Ein=0
Quiz 3 (2/23/01)
In the figure, a small ball of mass m=1.0 mg and
charge q = 2.0 10-8 C stays still 1 cm above a
uniformly charged sheet (ignore the thickness).
Assuming the sheet extends far horizontally and into
and out of page, calculate the surface charge density
s of the sheet.
m, q
1 cm