electric flux and potentialusing gauss’s law to solve for e • in certain situations gauss’s...

Electric Flux and Potential

I. Electric Flux

- flux defined

- Gauss’s Law

II. Electric Potential

- work and energy of charge

- potential defined

- potential of discrete charge(s)

- potential of charge distributions

- field related to potential

III. Conductors© Matthew W. Milligan

The student will be able to: HW:

1 Define and apply the concept of electric flux and solve related

problems.

1 – 5

2 State and apply Gauss’s Law and solve related problems using

Gaussian surfaces.

6 – 17

3 Calculate work and potential energy for discrete charges and solve

related problems including work to assemble or disassemble.

18 – 25

4 Define and apply the concept of electric potential and solve related

problems for a discrete set of point charges and/or a continuous

charge distribution.

26 – 32

5 Use the electric field to determine potential or potential difference

and solve related problems.

33 – 36

6 Use potential to determine electric field and solve related problems. 37 – 39

7 State the properties of conductors in electrostatic equilibrium and

solve related problems.

40 – 46

© Matthew W. Milligan

h

R

E

E

q

A point charge q is

located at the center of

a cylindrical surface.

What is the flux for the

entire surface?


h

R

r

r

x

y

E

E

dA

dA


h

R

r

r

x

y

E

E

dA

dA

dAEAdE y

dAEAdE x

2

04

11

2 r

q

r

hEy

xdxdA 2

RdydA 2

2

04

1

r

q

r

REx


h

R

r

r

x

y

E

E

dA

dA

fend =hq ×2p x

2 × 4pe0h2

4 + x2( )32

×dx0

R

ò

22

00 422 hR

qhqend

fside =Rq ×2p R

4pe0 R2 + y2( )

32

×dy

-h

2

h

2

ò

22

0 4 hR

qhside


h

R

r

r

x

y

E

E

dA

dA

22

00

22

0

4222

4

hR

qhq

hR

qh

0

q


Amazingly the sum of the

integrals simplifies to this!

Gauss determined that a similar

process integrating flux completely

around a closed surface containing

amount of charge q will always

simplify to Φ = q/ε0, regardless of

the shape of the surface or the

configuration or amount of charge!

where: qenc = total charge enclosed by surface

E = electric field

A = area vector (normal to surface)

Gauss’s Law

note: The circle on the integral sign indicates

integration entirely around a closed surface

– dA is an incremental piece of this surface.© Matthew W. Milligan

q = +2 nC

ΦE = ?


q = +2 nC

ΦE = 226 Nm2/C


q = +2 nC

ΦE = 0


ΦE = ?

−3 nC


ΦE = −339 Nm2/C

−3 nC


−3 nC −1 nC

ΦE = −339 Nm2/C


−3 nC −1 nC

ΦE = −452 Nm2/C


−1 nC +4 nC

ΦE = ?


−1 nC +4 nC

ΦE = 452 Nm2/C


−1 nC +4 nC

ΦE = −113 Nm2/C


−1 nC +4 nC

ΦE = 339 Nm2/C


−3 nC

1 nC

2 nC

ΦE = ?


−3 nC

1 nC

2 nC

ΦE = 0


Maxwell’s Equations

Gauss’s Law


Using Gauss’s Law to Solve for E

• In certain situations Gauss’s law is handy for

determining the electric field of a charge

distribution.

• This only works in highly symmetrical

situations!

• It is necessary to simplify the integral to the

form: E·A.

• In order to do this, one must imagine a

“Gaussian surface” along which the field is

uniform and the dot product may be found.© Matthew W. Milligan

electric flux and potentialusing gauss’s law to solve for e • in certain situations gauss’s...

Documents