Discussion

introduction

integral form

∯	E · dA =	Q
		ε0

differential form

∇ · E =	ρ
	ε0

examples

• spherical symmetry
  • point charge or any spherical charge distribution with total charge Q, the field outside the charge will be …

    ∯ E · dA =	Q	⇒	E(4πr2) =	Q	⇒	E =	1		Q	=	kQ
    	ε0			ε0			4πε0		r2		r2

  • spherical conductor with uniform surface charge density σ, the field outside the charge will be …

    ∯ E · dA =	Q	⇒	E(4πr2) =	(4πR2)σ	⇒	E =	σR2
    	ε0			ε0			ε0r2

    and the field inside will be zero since the Gaussian surface contains no charge …

    ∯ E · dA =	Q	=	0	⇒	E = 0
    	ε0		ε0

  • spherical insulator with uniform charge density ρ, the field outside the charge will be …

    ∯ E · dA =	Q	⇒	E(4πr2) =	(4/3πR3)ρ	⇒	E =	ρR3
    	ε0			ε0			3ε0r2

    and inside the field will be …

    					r
    ∯ E · dA =	Q	⇒	E(4πr2) =	1	⌠ ⌡	ρ(4πr2) dr =	4πρr3	⇒	E =	ρr
    	ε0			ε0			3ε0			3ε0
    					0

    Note that when r = R the field equations inside and outside match — as they should.
  • spherical insulator with nonuniform charge density ρ(r)
    Use the same method as the previous example, replace ρ with ρ(r), and see what happens.
• cylindrical symmetry
  • line with uniform charge density λ

    ∯ E · dA =	Q	⇒	E(2πrℓ) =	λℓ	⇒	E =	1		λ	=	2kλ
    	ε0			ε0			2πε0		r		r

  • cylindrical conductor with uniform surface charge density σ, the field outside the charge will be …

    ∯ E · dA =	Q	⇒	E(2πrℓ) =	σ 2πRℓ	⇒	E =	σR
    	ε0			ε0			ε0r

    and the field inside will be zero since the Gaussian surface contains no charge …

    ∯ E · dA =	Q	=	0	⇒	E = 0
    	ε0		ε0

  • cylindrical insulator with uniform charge density ρ, the field outside the charge will be …

    ∯ E · dA =	Q	⇒	E(2πrℓ) =	(πR2ℓ)ρ	⇒	E =	ρR2
    	ε0			ε0			2ε0r

    and inside the field will be …

    					r
    ∯ E · dA =	Q	⇒	E(2πrℓ) =	1	⌠ ⌡	ρ(2πrℓ) dr =	πρr2ℓ	⇒	E =	ρr
    	ε0			ε0			ε0			2ε0
    					0

    Once again, when r = R the field equations inside and outside match. Check it and see.
  • cylindrical insulator with nonuniform charge density ρ(r)
    Use the same method as the previous example, replace ρ with ρ(r), and see what happens.
• planar symmetry
  • nonconducting plane of infinitesimal thickness with uniform surface charge density σ
    Draw a box across the plane, with half of the box on one side and half on the other. (It is not necessary to divide the box exactly in half.) The "end caps" on the box will each capture the same amount of flux (EA).Thus …

    ∯ E · dA =	Q	⇒	2EA =	σA	⇒	E =	σ
    	ε0			ε0			2ε0

  • conducting plane of finite thickness with uniform surface charge density σ
    Draw a box across the surface of the conductor, with half of the box outside and half the box inside. (It is not necessary to divide the box exactly in half.) Only the "end cap" outside the conductor will capture flux. The other one is inside where the field is zero. Thus …

    ∯ E · dA =	Q	⇒	EA =	σA	⇒	E =	σ
    	ε0			ε0			ε0