gauss's law

1. Learning Goals -
• How you can determine the amount of
charge within a closed surface by
examining the electric field on the surface!
• What is meant by electric flux and how
you can calculate it.
• How to use Gauss’s Law to calculate the
electric field due to a symmetric
distribution of charges.

2. Summer July 2004 2
Flux
Φ=vcosθ A
Φ=
r
v.
r
A
r
A = Aˆn
θ =0
Φ=vA
θ=45
Φ=0.707vA
Φ=(normalcomponent)xArea
=Φ

3. A charge inside a box can be probed with a
test charge qo to measure E field

4. The volume (V) flow rate
(dV/dt) of fluid through the
wire rectangle (a) is vA when
the area of the rectangle is
perpendicular to the velocity
vector v and (b) is vA cos φ
when the rectangle is tilted
at an angle φ.
We will next replace the
fluid velocity flow vector v
with the electric field
vector E to get to the
concept of electric flux ΦE.
Volume flow rate through
the wire rectangle.

5. (a) The electric flux through
the surface = EA.
(b) When the area vector
makes an angle φ with the field
than flux is = EA cos φ. The
flux is zero when φ = 90o
because the rectangle lies in a
plane parallel to the flow and
no fluid flows through the
rectangle
A flat surface in a uniform
electric field.

6. Summer July 2004 6
Φ=
r
E∑•∆
r
A
Φ=
r
E∫ •d
r
A
Approximate Flux
Exact Flux
Circle means you integrate
over a closed surface.
d
r
A=ˆndA

7. Summer July 2004 7
Find the electric flux through a cylindrical surface in a
uniform electric field E
Φ=
r
E∫ •d
r
A d
r
A=ˆndA= E∫ cosθdA
Φ= E∫ cos180dA=−EdA∫ =−EπR2
a.
b.
Φ= E∫ cos180dA= EdA∫ =EπR2
Φ= E∫ cos90dA=0
c.
Flux from a. + b. + c. = 0
What is the flux if the
cylinder were vertical ?
Suppose it were
any shape?

8. ΦE = ∫ E
.
dA
= ∫ E dA cos φ
= ∫ E dA = E ∫ dA
= E (4p R2
)
= (1/4p εo) q /R2
) (4π R2
)
= q / εo.
So the electric flux
ΦE = q / εo.
Now we can write
Gauss's Law:
ΦE = ∫ E
.
dA =
∫ |EdA| cos φ =Qencl /εoElectric FLUX through a sphere
centered on a point charge q.

9. The projection of an
element of area dA of a
sphere of radius R UP
onto a concentric sphere
of radius 2R.
The projection multiplies
each linear dimension by
2, so the area element
on the larger sphere
is 4 dA.
The same number of
lines of flux pass thru
each area element.
Flux ΦE from a point charge q.

10. Spherical Gaussian surfaces around
(a) positive and (b) negative point charge.

11. Gauss’s Law can be used to calculate the
magnitude of the E field vector:
C 2012 J. F. Becker
This result can be extended to any shape surface
with any number of point charges inside and
outside the surface as long as we evaluate the
net flux through it.

12. Use the following recipe for Gauss’s Law problems:
1. Carefully draw a figure - location of all charges,
direction of electric field vectors E
2. Draw an imaginary closed Gaussian surface so that
the value of the magnitude of the electric field is
constant on the surface and the surface contains the
point at which you want to calculate the field.

13. Use the following recipe for Gauss’s Law problems:
1. Carefully draw a figure - location of all charges,
direction of electric field vectors E
2. Draw an imaginary closed Gaussian surface so that
the value of the magnitude of the electric field is
constant on the surface and the surface contains the
point at which you want to calculate the field.
3. Write Gauss Law and perform dot product E o dA

14. Use the following recipe for Gauss’s Law problems:
1. Carefully draw a figure - location of all charges,
direction of electric field vectors E
2. Draw an imaginary closed Gaussian surface so that
the value of the magnitude of the electric field is
constant on the surface and the surface contains the
point at which you want to calculate the field.
3. Write Gauss Law and perform dot product E o dA
4. Since you drew the surface in such a way that the
magnitude of the E is constant on the surface, you
can factor the |E| out of the integral.

15. Use the following recipe for Gauss’s Law problems:
1. Carefully draw a figure - location of all charges,
direction of electric field vectors E
2. Draw an imaginary closed Gaussian surface so that
the value of the magnitude of the electric field is
constant on the surface and the surface contains the
point at which you want to calculate the field.
3. Write Gauss Law and perform dot product E . dA
4. Since you drew the surface in such a way that the
magnitude of the E is constant on the surface, you
can factor the |E| out of the integral.
5. Determine the value of Qencl from your figure and
insert it into Gauss's equation.

16. Use the following recipe for Gauss’s Law problems:
1. Carefully draw a figure - location of all charges,
and direction of electric field vectors E
2. Draw an imaginary closed Gaussian surface so that
the value of the magnitude of the electric field is
constant on the surface and the surface contains the
point at which you want to calculate the field.
3. Write Gauss Law and perform dot product E o dA
4. Since you drew the surface in such a way that the
magnitude of the E is constant on the surface, you
can factor the |E| out of the integral.
5. Determine the value of Qencl from your figure and
insert it into Gauss's equation.
6. Solve the equation for the magnitude of E.
C 2012 J. F. Becker

17. Summer July 2004 17
Applications of Gauss’s Law
• Find electric filed of an infinite long uniformly charged wire of
negligible radius.
• Find electric field of a large thin flat plane or sheet of charge
• Find electric field around two parallel flat planes
• Find E inside and outside of a long solid cylinder of charge
density ρ and radius r.
• Find E for a thin cylindrical shell of surface charge density σ
• Find E inside and outside a solid charged sphere of charge
density ρ