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5) Coulomb’s Law a) form F = kq 1 q 2 / r 2
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5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

Dec 22, 2015

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Page 1: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

5) Coulomb’s Law

a) form

F = kq1q2 /r2

Page 2: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

b) UnitsTwo possibilities:- define k and derive q (esu)- define q and derive k (SI) √

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.

F = kq1q2 /r2

9 ×109N = kq1q2 /r2

9 ×109N = k(1C)2 /(1m)2

⇒ k = 9 ×109N

Page 3: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

• For practical reasons, the coulomb is defined using current and magnetism giving

k = 8.988 x 109 Nm2/C2

• Permittivity of free space

ε0 =1

4πk= 8.84 ×10−12C2 /Nm2

Then

F =1

4πε0

q1q2

r2

Page 4: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

c) Fundamental unit of charge

e = 1.602 x 10-19 C

Page 5: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

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

Page 6: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

Example: Coulomb force vs gravity for electrons

m, e m, eFCFg

FC = ke2/r2 FN = Gm2/r2

Ratio:

FC

FN

=ke2

Gm2=

(9 ×109)(1.6 ×10−19)2N

(6.7 ×10-11)(9.1×10−31)2N

=4 ×1042

Page 7: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

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

Page 8: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

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

Page 9: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

6) Electric Field

- abstraction

- separates cause and effect in Coulomb’s law

a) Definition

rE =

r F

q0

Units: N/C

Page 10: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.
Page 11: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

b) Field due to a point charge

F

Q

q0

r

Coulomb’s law:

F = kQq0

r2

Electric Field:

E = F /q0

=kQ

r2

rE //

r F ⇒ direction is radial

Page 12: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

rE =

kQ

r2ˆ r

Page 13: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

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

Page 14: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

Example

16 µC 4 µCq1 q2

P

dD=3m

Find d to give E = 0 at P

EP = E1 − E2 = 0P

E1E2

⇒ E1 = E2

kq1

d2 =kq2

(D − d)2

q1

d2 =q2

(D − d)2

4

d2 =1

(D − d)2

4(D − d)2 = d2

2(D − d) = ±d

d = 2D or 23 D = 6m or 2m

Page 15: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

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)

Page 16: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

b) Number of lines proportional to charge

Q2Q

Page 17: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

c) Begin and end only on charges; never cross

E?

Page 18: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

d) Line density proportional to field strength

Line density at radius r:

Number of lines

area of sphere

=N

4πr2

∝1

r2

Lines of force model <==> inverse-square law

Page 19: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

8) Applications of lines-of-force model

a) dipole

Page 20: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

b) two positive charges

Page 21: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

c) Unequal charges

Page 22: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

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

E =σ

2ε 0

By comparison with the field from a point charge, we find:

E

q, A

Page 23: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

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:

E = σ /ε 0

• Field zero outside

Page 24: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

• Fringing fields near the edges

Page 25: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

f) Spherically symmetric charge distribution

+ +

+

+

++

+

+

• Symmetry ==> radial• number of lines prop. to charge

Outside the sphere:

rE =

kq

r2ˆ r

as though all charge concentrated at the centre (like gravity)

Page 26: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

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|

Page 27: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

• Closed conductor shields external fields

E E = 0

Page 28: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

• Field outside conducting shell not shielded

• Field lines perpendicular at surface

Page 29: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

• Field outside grounded shell is shielded

• Field larger for smaller radius E = kq/r2

(concentrated at sharp tips)

Page 30: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.

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

Page 31: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.
Page 32: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.
Page 33: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.
Page 34: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.
Page 35: 5) Coulomb’s Law a)form. b) Units Two possibilities: - define k and derive q (esu) - define q and derive k (SI) √ “Define” coulomb (C) as the quantity.