lecture 04: fri 23 jan - louisiana state university
TRANSCRIPT
Physics 2102 Lecture 04: FRI 23 JAN
Electric Charge I
Physics 2113 Jonathan Dowling
Charles-Augustin de Coulomb (1736–1806)
Version: 12/26/14
Benjamin Franklin (1705–1790)
Let’s Get Started! Electric Charges…
• Two Types of Charges: Positive/Negative • Like Charges Repel • Opposite Charges Attract
Atomic Structure: • Negative Electron Cloud • Nucleus of Positive Protons, Uncharged Neutrons The Unit of Electric Charge is the “Coulomb” which is “C”. Proton Charge: e = 1.60 × 10–19 C
Benjamin Franklin (1705–1790)
Rules of Electric Attraction and Repulsion Discovered by Benjamin Franklin:
Electrical Insulators
Benjamin Franklin (1705–1790)
Rules of Electric Attraction and Repulsion Discovered by Benjamin Franklin:
Electric Conductors
Benjamin Franklin (1705–1790)
Rules of Electric Attraction and Repulsion: ICPP
C and D attract B and D attract
−( )
−( )+( )
+( )+( )
+−
+− +( )
+( )+( )
12F1q+ 21F 2q−
12F 1q+ 21F2q+
12F 1q− 21F2q−
Force Between Pairs of Point Charges: Coulomb’s Law
Coulomb’s Law — the Force Between Point Charges: • Lies Along the Line Connecting the Charges. • Is Proportional to the Product of the Magnitudes. • Is Inversely Proportional to the Distance Squared. • Note That Newton’s Third Law Says |F12| = |F21|!!
Charles-Augustin De Coulomb (1736–1806)
12F1q+ 21F 2q−
12F 1q+ 21F2q+
12F 1q− 21F2q−
Force Between Pairs of Point Charges: ICPP
!Fpe(a)
!Fpe +
!Fee(c)
!Fpp(b)
2q−12F1q+ 21F
12r
Coulomb’s Law
�
F12 =k q1 q2
r122
�
k = 14πε0
with ε0 = 8.85 ×10−12 C2
Nm2
k = 8.99 ×109 Nm2
C2 ∝ kg m3
s2 C2
The “k” is the electric constant of proportionality.
Usually, we write:
Units: F = [N] = [Newton]; r = [m] = [meter]; q = [C] = [Coulomb]
Coulomb’s Law: ICCP
(a) a > c > b (b) less
!Fnet
!Fnet
!Fnet
Coulomb’s Torsion Balance Experiment For Electric Force Identical to Cavendish’s Experiment For
Gravitational Force!
k = 8.99 ×109 Nm2
C2 ∝ kg m3
s2 C2
The experiment measures “k” the electric constant of proportionality and confirms inverse square law.
�
F12 =k q1 q2
r122
http://www.dnatube.com/video/11874/Application-Of-Coulombs-Torsion-Balance
Two Inverse Square Laws
Newton’s Law of Gravitational Force Coulomb’s Law of
Electrical Force
Area of Sphere = 4πr2
Number of Lines of Force is Constant. Hence #Force Lines Per-Unit-Area is Proportional to 1/r2
Superposition • Question: How Do We Figure Out the Force
on a Point Charge Due to Many Other Point Charges?
• Answer: Consider One Pair at a Time, Calculate the Force (a Vector!) In Each Case Using Coulomb’s Law and Finally Add All the Vectors! (“Superposition”)
• Useful To Look Out for SYMMETRY to Simplify Calculations!
Feel the Force! Example
• Three Equal Charges Form an Equilateral Triangle of Side 1.5 m as Shown
• Compute the Force on q1
• ICPP: What are the Forces on the Other Charges?
d
q1
d
d q2
q3
q1= q2= q3= 20 mC d = 1.0 cm
Solution: Set up a Coordinate System, Compute Vector Sum of F12 and F13
d
1
2
3 d
d
12F
13F
o60y
x
Fnet
θ
Feel the Force! Example q1= q2= q3= 20 mC
d = 1.0 cm
d
1
2
3 d
d
12F
13F
o60y
x
Fnet
F12 = F13 =kq1q2
r12( )2
= 8.99 ×109Nm2
C220 ×10−3C 20 ×10−3C
0.01m( )2
=3.60 ×106N
Fnet = Fnetx( )2
+ Fnety( )2
= F13x( )2
+ F13y + F12( )2
= F13 cos(30°)( )2+ F13 sin(30°)+ F12( )2
= 3.12 ×106N( )2+ 5.40 ×106N( )2
= 6.24 ×106N
By geometry: θ = 12 60° + 90° = 120°
θ
ICPP: What are the magnitudes and directions of the forces on 2 and 3?
What is the Force on Central Particle?
Charge +q Placed at Center
Another Example With Symmetry
+q r
All Forces Cancel Except From +2q!
�
! F =
k +2q +qr2
F
