PROBLEM SOLVING STRATEGIES
There is a 3-step process for applying Gauss’ law to derive the form of the electric field.
Step 1: Pick a good Gaussian Surface
• Symmetry condition (pulls E = constant out of the integral sign)
• Dot-product condition (determines the sign of the flux)
Step 2: Evaluate the enclosed charge qenc
Step 3: Apply Gauss’ law to derive the E-field expression
Step 1: Picking a good Gaussian surface (tricky to apply at first)
A good Gaussian surface satisfies what I personally call the Symmetry and Dot-Product
conditions.
Symmetry Condition: The symmetry of the charge distribution must also be the symmetry of
the Gaussian surface (spherically, cylindrical, cubic). The job of the symmetry condition is to
pull the E-field variable out of the flux integral sign:
∫
surface
 
E ⋅ dA= Ecos θ
∫
surface
dA
That means that the electric field will have the same value everywhere on this surface. Picture
wise, we get
Dot-Product Condition:
The Dot Product between E and dA is
 
E ⋅ dA =(E ⋅ dA)cos θ
The cosθ term determines the flux direction through the Gaussian surface. That is, we look at
the direction of the area verses the E-field. There are three possible values for cos θ:
•
•
•
•
θ=0°→ cos(0°) = 1 (E is parallel to dA)
(E ⋅ dA)cos=
θ +E ⋅ dA 
→ positive flux
θ=180°→ cos(180°) = −1 (E is antiparallel to dA)
(E ⋅ dA)cos =
θ −E ⋅ dA 
→ negative flux
θ=90°→ cos(90°) = 0 (E is perpendicular to dA)
(E ⋅ dA)cos(90°) =0 
→ zero flux
E = 0 everywhere on the Gaussian surface:
 
E ⋅ dA = 0 ⇒ E is zero everywhere
The result of step 1 is
∫
surface
 
=
⋅ dA E cos θ
E
±1 (parallel or antiparallel)

→
=
θ
dA
cos

∫ surface
0 (perpendicular)
Step 2: Evaluating the Enclosed Charge qenc
The enclosed charge will be distributed over volumes, surfaces and lines, which I call the
enclosed charge equations (qtotal = Q):
q
L enc
qenc
A enc
Q
Q
λ ≡ = enc → =
qenc
Q; σ ≡ =
→ =
qenc
Q
L total
A total
L total L enc
A total A enc
q
Venc
Q
ρ ≡ = enc →=
qenc
Q
Vtotal
Vtotal Venc
There are two cases to consider:
• Point P outside the charge distribution
If the Gaussian encloses the total charge, then the enclosed is the total charge:
qenc (point outside the surface) =
Q=
(λ
L total ,
σA total ,

line
•
surface
ρVtotal )

volume
Point P inside the charge distribution
If the Gaussian encloses only part of the total charge, then the enclosed charge is
qenc (point inside the surface) =
L enc
L total
A enc
Q;
A total
Q;
Venc
Vtotal
Q
Step 3: Applying Gauss’ Law
Combining Steps 1 & 2 into Gauss’ Law and solving for the E-field variable gives
=
Φ
∫
surface
 
E ⋅=
dA E cosθ∫
surface
=
dA
qenc
0
⇒=
E
1
∫
Gaussian
surface
qenc
dA 0
One is left with an integral over the area of the Gaussian surface:
Cylinder
 A cylinder = ∫
 A sphere = ∫
dA = 2πrL
dA = 4 πr 2
cylinder
sphere
Sphere 
=
4
πr 2L
πr 3
 Vcylinder =
 Vsphere =
3