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?
© Matthew W. Milligan
h
R
r
r
x
y
E
E
dA
dA
© Matthew W. Milligan
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
© Matthew W. Milligan
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
© Matthew W. Milligan
h
R
r
r
x
y
E
E
dA
dA
22
00
22
0
4222
4
hR
qhq
hR
qh
0
q
© Matthew W. Milligan
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 = ?
© Matthew W. Milligan
q = +2 nC
ΦE = 226 Nm2/C
© Matthew W. Milligan
q = +2 nC
ΦE = 226 Nm2/C
© Matthew W. Milligan
q = +2 nC
ΦE = 226 Nm2/C
© Matthew W. Milligan
q = +2 nC
ΦE = 226 Nm2/C
© Matthew W. Milligan
q = +2 nC
ΦE = 0
© Matthew W. Milligan
ΦE = ?
−3 nC
© Matthew W. Milligan
ΦE = −339 Nm2/C
−3 nC
© Matthew W. Milligan
−3 nC −1 nC
ΦE = −339 Nm2/C
© Matthew W. Milligan
−3 nC −1 nC
ΦE = −452 Nm2/C
© Matthew W. Milligan
−1 nC +4 nC
ΦE = ?
© Matthew W. Milligan
−1 nC +4 nC
ΦE = 452 Nm2/C
© Matthew W. Milligan
−1 nC +4 nC
ΦE = −113 Nm2/C
© Matthew W. Milligan
−1 nC +4 nC
ΦE = 339 Nm2/C
© Matthew W. Milligan
−3 nC
1 nC
2 nC
ΦE = ?
© Matthew W. Milligan
−3 nC
1 nC
2 nC
ΦE = 0
© Matthew W. Milligan
Maxwell’s Equations
Gauss’s Law
© Matthew W. Milligan
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