5) coulomb’s law
DESCRIPTION
5) Coulomb’s Law. form. “Define” coulomb (C) as the quantity of charge that produces a force of 9 x 10 9 N on objects 1 m apart. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √. - PowerPoint PPT PresentationTRANSCRIPT
5) Coulomb’s Law
a) form
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F = kq1q2 /r2
b) UnitsTwo possibilities:- define k and derive q (esu)- define q and derive k (SI) √
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F = kq1q2 /r2
“Define” coulomb (C) as the quantity of charge that produces a force of 9 x 109 N on objects 1 m apart.
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F = kq1q2 /r2
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9 ×109N = kq1q2 /r2
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9 ×109N = k(1C)2 /(1m)2
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⇒ k = 9 ×109N
• For practical reasons, the coulomb is defined using current and magnetism giving
k = 8.988 x 109 Nm2/C2
• Permittivity of free space
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ε0 = 14πk
= 8.84 ×10−12C2 /Nm2
Then
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F = 14πε0
q1q2
r2
c) Fundamental unit of chargee = 1.602 x 10-19 C
Example: Force between two 1 µC charges1 mm apart
F = kq1q2/r2 = 9•109(10-6)2/(10-3)2 N = 9000 N
• ~weight of 1000-kg object (1 tonne))• same as force between two 1-C charges 1 km apart
Example: Coulomb force vs gravity for electronsm, e m, e
FCFg
FC = ke2/r2 FN = Gm2/r2
Ratio:
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FC
FN
= ke2
Gm2 = (9 ×109)(1.6 ×10−19)2N(6.7 ×10-11)(9.1×10−31)2N
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=4 ×1042
Example: Velocity of an electron in the Bohr Atom
Coulomb force: F = kq1q2/r2 (attractive)
Circular motion requires: F = mv2/r
So,
v2 = kq1q2/mr
For r = 5.29 x 10-11m, v = 2.18 x 106 m/s
d) Superposition of electric forces
Net force is the vector sum of forces from each charge
q1
q2
q3
q
F3
F2
F1
Net force on q: F = F1 + F2 + F3
F
6) Electric Field
- abstraction- separates cause and effect in Coulomb’s law
a) Definition
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r E =
r F q0
Units: N/C
b) Field due to a point charge
F
Q
q0
r
Coulomb’s law:
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F = kQq0
r2
Electric Field:
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E = F /q0
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=kQr2
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r E //
r F ⇒ direction is radial
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r E =
kQr2 ˆ r
c) Superposition of electric fields
Net field is the vector sum of fields from each charge
P
E3
E2
E1
Net field at P: E = E1 + E2 + E3
E
q1
q2
q3
Example
16 µC 4 µCq1 q2
P
dD=3m
Find d to give E = 0 at P
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EP = E1 − E2 = 0P
E1E2
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⇒ E1 = E2
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kq1
d2 =kq2
(D − d)2
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q1
d2 =q2
(D − d)2
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4d2 =
1(D − d)2
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4(D − d)2 = d2
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2(D − d) = ±d
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d = 2D or 23 D = 6m or 2m
7) Electric Field Lines (lines of force)
a) Direction of force on positive charge
radial for point chargesout for positive (begin)in for negative (end)
b) Number of lines proportional to charge
Q2Q
c) Begin and end only on charges; never cross
E?
d) Line density proportional to field strength
Line density at radius r:
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Number of linesarea of sphere
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=N
4πr2
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∝1r2
Lines of force model <==> inverse-square law
8) Applications of lines-of-force model
a) dipole
b) two positive charges
c) Unequal charges
d) Infinite plane of charge
++
+
++
+
++
+
++
+
Field is uniform and constant to ∞, in both directions
Electric field is proportional to the line density, and therefore to the charge density, =q/A
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E =σ
2ε 0
By comparison with the field from a point charge, we find:
E
q, A
e) Parallel plate capacitor (assume separation small compared to the size)
++
+
+
++
--
-
-
--
E+
E-
E=2E+
E+
E-
ER=0
E+
E-
EL=0
• Strong uniform field between:
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E = σ /ε 0
• Field zero outside
• Fringing fields near the edges
f) Spherically symmetric charge distribution
+ +
+
+
++
+
+
• Symmetry ==> radial• number of lines prop. to charge
Outside the sphere:
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r E =
kqr2 ˆ r
as though all charge concentrated at the centre (like gravity)
9) Electric Fields and Conductors
• Excess charge resides on surface at equilibrium
E1E1
E2
• Field inside is zero at eq’m; charges move until |E1| = |E2|
• Closed conductor shields external fields
E E = 0
• Field outside conducting shell not shielded
• Field lines perpendicular at surface
• Field outside grounded shell is shielded
• Field larger for smaller radius E = kq/r2
(concentrated at sharp tips)
Demonstration: Van de Graaf generator- purpose: to produce high field by concentrating charge -- used to accelerate particles for physics expts- principle: charge on conductors moves to the surface