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Chapter 22 – Gauss Law
- Charge and Electric flux
- Electric Flux Calculations
- Gauss’s Law and applications
- Charges on Conductors
Child acquires electric charge by touching
a charged metal sphere. Electrons coat
each individual hair fiber and then repel
each other.
1. Charge and Electric Flux
- A charge distribution produces an electric field (E), and E exerts a force
on a test charge (q0). By moving q0 around a closed box that contains the
charge distribution and measuring F one can make a 3D map of E = F/q0
outside the box. From that map, we can obtain the value of q inside box.
- If we construct a boundary around a
charge, we can think of the flow coming out
from the charge like water through a screen
surrounding a sprinkler.
Electric Flux and Enclosed Charge:
-There is a connection between sign of net charge enclosed by a closed
surface and the direction of electric flux through surface (inward for -q,
outward for +q).
- There is a connection between magnitude of net enclosed charge and
strength of net “flow” of E.
- The net electric flux through the surface of a box is directly proportional to
the magnitude of the net charge enclosed by the box.
2q  2E q  E
E~ 1/r2
r1 = distance of q to
surface of box1.
r2 = 2r1= distance of q
to surface of box2.
In (c), E2 = E1/4, since r2=2r1, but A2=4A1
net flux constant.
- Electric flux = (perpendicular component of E) · (area of box face)
-The net electric flux due to a point charge inside a box is independent of
box’s size, only depends on net amount of charge enclosed.
- Charges outside the surface do not give net electric flux through surface.
2. Calculating Electric Flux
Flux Fluid Analogy:
- If we considered flux through a rectangle,
the flux will change as the rectangle
changes orientation to the flow.
vA
dt
dV
=
dV/dt = volume flow rate
v = flow speed
A
v
A
v
vA
vA
dt
dV 

⋅
=
=
=
= ⊥
⊥ ϕ
cos
(v ┴ A)
Flux of a Uniform Electric Field:
A
E
EA
A
E
E ⊥
=
=
⋅
=
Φ ϕ
cos


ΦE = E·A ΦE = E·A·cosφ ΦE = 0
Units: N m2/C
We can define a vector area: with n being a unit vector ┴ A.
n
A
A ˆ
⋅
=

Flux of a Non-uniform Electric Field:
3. Gauss’s Law
- The total electric flux through any closed surface is
proportional to the total electric charge inside the
surface.
Point Charge Inside a Spherical Surface:
- The flux is independent of the radius R of the
sphere.
2
0
4
1
R
q
E
πε
=
( )
0
2
2
0
4
4
1
ε
π
πε
q
R
R
q
A
E
E =
=
⋅
=
Φ
E // dA at each point
Point Charge Inside a Nonspherical Surface:
- Divide irregular surface into dA elements, compute electric flux for each
(E dA cosφ) and sum results by integrating.
- Each dA projects onto a spherical surface element  total electric flux
through irregular surface = flux through sphere.
0
ε
q
A
d
E
E =
⋅
=
Φ ∫


Integral through a closed
surface
Valid for + / - q
If enclosed q = 0  ΦE = 0
Point charge outside a closed surface that
encloses no charge. If an electric field line
enters the surface at one point it must leave at
another.
- Electric field lines can begin or end inside
a region of space only when there is a
charge in that region.
General form of Gauss’s law:
0
cos
ε
ϕ encl
E
Q
dA
E
dA
E
A
d
E =
=
=
⋅
=
Φ ∫
∫
∫ ⊥


( )
0
2
2
0
2
0
2
0
4
4
4
4 ε
π
πε
πε
πε
q
r
r
q
dA
r
q
dA
r
q
dA
E
E
−
=







 −
=







 −
=







 −
=
=
Φ ∫
∫
∫ ⊥
Example: Spherical Gaussian surface around –q
(negative inward flux)
4. Applications of Gauss’s Law
- When excess charge (charges other than ions/e- making up a neutral conductor)
is placed on a solid conductor and is at rest, it resides entirely on the surface, not
in the interior of the material.
- Electrostatic condition (charges at rest)  E = 0 inside material of
conductor, otherwise excess charges will move.
E = 0
q = 0
5. Charges on Conductors
- Excess charge only on surface.
- Cavity inside conductor with q = 0  E = 0
inside conductor, net charge on surface of
cavity = 0.
- Cavity inside conductor with +q E = 0
inside conductor, -q charge on surface of
cavity (drawn there by +q). Total charge
inside conductor = 0  +q on outer surface
(in addition to original qc).
Field at the surface of a conductor:
σ = q/A  q = σA
Field outside a charged conductor is
perpendicular to surface.
Ex. 22.2

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Gauss's Law Explained: Charge, Flux, and Applications

  • 1. Chapter 22 – Gauss Law - Charge and Electric flux - Electric Flux Calculations - Gauss’s Law and applications - Charges on Conductors Child acquires electric charge by touching a charged metal sphere. Electrons coat each individual hair fiber and then repel each other.
  • 2. 1. Charge and Electric Flux - A charge distribution produces an electric field (E), and E exerts a force on a test charge (q0). By moving q0 around a closed box that contains the charge distribution and measuring F one can make a 3D map of E = F/q0 outside the box. From that map, we can obtain the value of q inside box. - If we construct a boundary around a charge, we can think of the flow coming out from the charge like water through a screen surrounding a sprinkler.
  • 3.
  • 4. Electric Flux and Enclosed Charge: -There is a connection between sign of net charge enclosed by a closed surface and the direction of electric flux through surface (inward for -q, outward for +q). - There is a connection between magnitude of net enclosed charge and strength of net “flow” of E. - The net electric flux through the surface of a box is directly proportional to the magnitude of the net charge enclosed by the box.
  • 5. 2q 2E q E E~ 1/r2 r1 = distance of q to surface of box1. r2 = 2r1= distance of q to surface of box2. In (c), E2 = E1/4, since r2=2r1, but A2=4A1 net flux constant. - Electric flux = (perpendicular component of E) · (area of box face) -The net electric flux due to a point charge inside a box is independent of box’s size, only depends on net amount of charge enclosed. - Charges outside the surface do not give net electric flux through surface.
  • 6. 2. Calculating Electric Flux Flux Fluid Analogy: - If we considered flux through a rectangle, the flux will change as the rectangle changes orientation to the flow. vA dt dV = dV/dt = volume flow rate v = flow speed A v A v vA vA dt dV ⋅ = = = = ⊥ ⊥ ϕ cos (v ┴ A)
  • 7. Flux of a Uniform Electric Field: A E EA A E E ⊥ = = ⋅ = Φ ϕ cos ΦE = E·A ΦE = E·A·cosφ ΦE = 0 Units: N m2/C We can define a vector area: with n being a unit vector ┴ A. n A A ˆ ⋅ = Flux of a Non-uniform Electric Field:
  • 8. 3. Gauss’s Law - The total electric flux through any closed surface is proportional to the total electric charge inside the surface. Point Charge Inside a Spherical Surface: - The flux is independent of the radius R of the sphere. 2 0 4 1 R q E πε = ( ) 0 2 2 0 4 4 1 ε π πε q R R q A E E = = ⋅ = Φ E // dA at each point
  • 9. Point Charge Inside a Nonspherical Surface: - Divide irregular surface into dA elements, compute electric flux for each (E dA cosφ) and sum results by integrating. - Each dA projects onto a spherical surface element total electric flux through irregular surface = flux through sphere. 0 ε q A d E E = ⋅ = Φ ∫ Integral through a closed surface Valid for + / - q If enclosed q = 0 ΦE = 0
  • 10. Point charge outside a closed surface that encloses no charge. If an electric field line enters the surface at one point it must leave at another. - Electric field lines can begin or end inside a region of space only when there is a charge in that region. General form of Gauss’s law: 0 cos ε ϕ encl E Q dA E dA E A d E = = = ⋅ = Φ ∫ ∫ ∫ ⊥ ( ) 0 2 2 0 2 0 2 0 4 4 4 4 ε π πε πε πε q r r q dA r q dA r q dA E E − =         − =         − =         − = = Φ ∫ ∫ ∫ ⊥ Example: Spherical Gaussian surface around –q (negative inward flux)
  • 11. 4. Applications of Gauss’s Law - When excess charge (charges other than ions/e- making up a neutral conductor) is placed on a solid conductor and is at rest, it resides entirely on the surface, not in the interior of the material. - Electrostatic condition (charges at rest) E = 0 inside material of conductor, otherwise excess charges will move. E = 0 q = 0
  • 12. 5. Charges on Conductors - Excess charge only on surface. - Cavity inside conductor with q = 0 E = 0 inside conductor, net charge on surface of cavity = 0. - Cavity inside conductor with +q E = 0 inside conductor, -q charge on surface of cavity (drawn there by +q). Total charge inside conductor = 0 +q on outer surface (in addition to original qc).
  • 13. Field at the surface of a conductor: σ = q/A q = σA Field outside a charged conductor is perpendicular to surface.