CH 22: Electric Fields and Gauss’s Law

• Every charge generates an Electric Field.• Electric Field – Region of influence surrounding

any charged object.• Interaction between electric fields results in a

force between the sources of the electric fields.• Electric Field Lines are used to show direction of

the electric field. (electric field lines are used as a means of visualizing the electric field, but there are not really lines emanating from the charge)

• Positive charges are defined to have electric field lines directed away from the charge (source).

• Negative charges are defined to have electric field lines directed towards the charge (sink).

Electric field lines for a positive charge

Electric field lines for a negative charge

Electric Field Lines:

1. Can never cross

2. Have a direction denoted by an arrow

3. The number of field lines drawn is proportional to the amount of charge and hence the strength of the electric field generated.

Point Charge

Dipole – two opposite charges

3D Dipole

Courtesy of NASA

Courtesy of NASA

Repulsion AttractionCourtesy of UVI

Consider the four field patterns shown. Assuming there are no charges in the regions shown, which of the patterns represent(s) a possible electrostatic field:

1. (a)2. (b)3. (b) and (d)4. (a) and (c)5. (b) and (c)6. some other combination7. None of the above.

Determination of the magnitude and direction of an electric field

• The magnitude of the electric force is directly related to the strength of the electric field.

• The direction of the electric force is in the same direction as the electric field when acting on a positive charge.

• The direction of the force is opposite for a negative charge!

Fe = qEFe = electric force applied to charge q

q = charge within an electric field

E = electric field strength

Electric fields are vectors and must be dealt with appropriately!!

C

N

• The electric field strength decreases with distance from the charge creating the electric field in a manner similar to the electric force.

212

2112 r

qqkF e

212

2112 r

qqkEq e 2

12

11 r

qkE e

1212 EqF }

2ek qEr

Electric field due to a point charge

2ˆek qE r

r

3

ek qE rr

Vector forms of electric field

Example: A positive 30 m C point charge is in an isolated region of space.a) What is the electric field strength at a point P 5 cm from this charge?b) If a negative 20 m C charge suddenly appears at a distance of 10 cm from the original

charge, what is the electric field strength at point P halfway between the two charges?c) If a small 1 m C positive test charge is placed at point P, what is the net force on this

test charge?

21

11 r

qkE e

30 m C

E1

5 cm

Left

a)

P

C

Nx 81008.1

30 m C

E1

5 cm

P

b)

5 cm

20 m C

E2

E

22

22 r

qkE e

C

Nx 71019.7 Left

21 EEE

C

Nx 81080.1 Left

C

Nx 81080.1

30 m C

5 cm

P

c)

5 cm

20 m CE

F

EqF

1 m C

N180 LeftN180

Continuous Charge Distribution• Not all charged object have negligible dimensions.

• The electric field contributions from different parts of the charged object varies with the distance that small part is from the point of interest.

PEach section does not have to have the same amount of charge.

The contribution from each part must be summed to get the net electric field strength

r

r

qkE e ˆ

2

i

ii

ie r

r

qkE ˆ

2

rr

dqkEd e ˆ

2

rr

qkE

eˆ

2

2ˆe

dqE k r

r

Dq is a small segment of charge

DE is a small contribution to E

dq is an infinitesimal segment of charge

dE is an infinitesimal contribution to E

r

r

qkEEd

iqe

Eˆlimlim

200

Charge Density• We can substitute our small segment of charge dq with a charge distribution

relationship.

• This charge distribution relationship is used to simplify our calculations of the electric field of a continuous object.

• The charge distribution relationships are usually related to the geometry of the charged object and hence called charge density.

Types of charge densities:

Volume charge density (r) Surface charge density (s) Linear charge density (l)

Q

Vr

Q

As Q

ll

Q is the total charge contained within the object

dq = rdV dq = sdA dq = ldl

3m

Cr

2m

Cs

m

Cl