5) Coulomb’s Law

a) form

€

F = kq1q2 /r2

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

• 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

c) Fundamental unit of charge

e = 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 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

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

€

rE =

r F

q0

Units: N/C

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

€

rE =

kQ

r2ˆ 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

€

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

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:

€

Number of lines

area of sphere

€

=N

4πr2

€

∝1

r2

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

€

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:

€

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:

€

rE =

kq

r2ˆ 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