Previous Lectures: •Introduced to Coulomb’s law

•Learnt the superposition principle

•Showed how to calculate the electric field resulting from a series of discrete charges

•Showed how to calculate the resultant force on a charge placed in an electric field

To calculate E due to a continuous distribution of charges

Learning Objective:

Lecture 3: Electric fields

Continuous?

didn’t you say that charge is quantised?

Yes, but water consists of individual

H2O molecules, and it flows.

A Continuous Distribution of Charges

•Break the distribution into small pieces

•Treat each little piece as a point charge and add up the separate E-fields (as a vector of course!)

•This can get messy - typically results in a 3D integral, evaluated numerically by computer

What do we do if the object is not a point particle?

Charge can be distributed along a line, over a

surface, or through a volume. •Use to represent charge/length over the element dl

∫= dlq λ

∫= dAq σ

∫= dVq ρ

•Use to represent the charge/unit area over a surface area element dA

•Use to represent the charge/unit volume in a volume element dV

€

q = λL

€

q = σA

€

q = ρV

E-field on the axis of a ring of charge (Tipler)

Method:

1) Represent the charge distribution as a series of small elements ds each with a charge dQ (+ charge)

dQ

1

2) Use symmetry. Consider two ring elements which are diagonal to each other

2dE

dQ

x

y

No net E-field in the y direction

dE2

dE1

dETotal

€

dEx = dE cosθ =1

4πε0

dQ

x 2 + a2

x

x 2 + a2 ( ) 2/32204

1

ax

xdQ

+=

πε

P

For one charge element

( ) 232204

1)( /

ax

QxxE

+=

πε

204

1

x

QE

πε=Note that for x >> a

Coulomb’s Law

Total E-field at P

Charge distribution

is uniform!

The E-Field due to a large circular plane and uniform distribution of charges

1) Consider first an elemental ring

rdrdQ π 2=

( ) πε

cos4

122

0 rx

dQdEx +

=

2) Calculate the net E-field at P due to the ring element:

( ) ππε

cos2

4

122

0 rx

rdr

+=

€

E(x) =1

4πε0

Qx

x 2 + a2( )

3 / 2 =1

4πε0

Q

x 2 + a2( )

x

x 2 + a2

=1

4πε0

Q

x 2 + a2( )

cosθ

secxcos

xrx

dsecxdr

tanxr

==+

=

=

22

2

3) Use the following trignometrical identities

ε

ddEx sin2 0

=

5) Consider the limiting case R

∫∫ ==2

00

sin2

π

θθε

σddEE x

4) E-field due to the charged ring is

€

cosθ[ ]0

π2

02ε

−=02ε

=

Same result if x<<R

For an infinite plane sheet of charge the E-field produced is independent of the distance from the sheet

€

E =σ

2ε0

This result is true close to the surface of any charge distribution

E-field at a point along a line perpendicular to the mid point of line of uniform charge. Student exercise: show that

2204

1

axx

QEx

+=

πε

Note that for x >> a

204

1

x

QE

πε=Coulomb’s Law

dQ

y

No Ey

Further exercise:

Write = Q/2a

Obtain an expression for Ex as a

xEx

πε

=02

1

E field does not follow an inverse square law and is radially outward from the line

E-field inside a spherical shell of uniform charge

Spherical Shell of Charge

r1

r2

P

a1

a2

21

22

1

2

r

r

a

a=

21

22

1

2

1

2

r

r

a

a

q

q=

=

Total E-Field at P is ZERO

€

E2

E1

=

Δq2

r22

Δq1

r12

=Δq2

Δq1

r12

r22

=1

All parts of the surface can be paired off

What if

€

E =1

4πε0

Q

r1.99 ?

Summary•Illustrated how to calculate the electric field due to a continuous distribution of charge•For an infinite plane sheet of charge the E-field produced is independent of the distance from the sheet

02ε

=E

•E-field inside a spherical shell of uniform charge is zero

Classwork: Three charges, a, b, and c, each of 3 C, are located at x = 0, x = 0.5 m and x = 1.0 m, respectively. What is the force exerted by charges b and c on a?

xa b c

The principle of superposition of forces tells us that the force on charge a is the vector sum of the force caused by charge b and that caused by charge c – that is:

Faresultant = Fb + Fc

where Fb & Fc are respectively, the forces on charge a caused by charges b and c.

Since all the charges are positive, the resultant force is in the –x direction.

ir

qq

r

qqF

c

ca

b

battanaresul ⎥

⎦

⎤⎢⎣

⎡+

πε−= 22

041

Use Coulomb’s law:

Cqqq cba6103 −×===

m.rm.r cb 0150 ==

iN.F ttanaresul 410−=

Next . . . .

Electrical Potential Energy

Electric Potential

and

How to calculate the electric field strength from electric potential

Spherically Symmetric Charge Distribution

E outside a sphere or spherical shell carrying charge q is identical to that of a point charge q at the centre.