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Unformatted text preview: Gauss’s Law
Chapter Outline ENGINEERING PHYSICS II
Electric Flux
PHY 303L
Gauss’s Law
Applications of Gauss’s Law Chapter 24: Gauss’ Law Superposition of Electric Fields
Conductors and Electric Fields Maxim Tsoi
Physics Department,
The University of Texas at Austin http://www.ph.utexas.edu/~tsoi
303L: Gauss’ Law (Ch.24) 303L: Gauss’ Law (Ch.24) Electric Flux Electric Flux we now treat electric field lines in a more quantitative way what if the surface is not perpendicular to the field ? • Consider an electric field that is uniform in • Normal to the surface of area A is at an both magnitude and direction angle θ to the uniform electric field • Number of lines per unit area ~ magnitude • A’ =A cosθ is a projection of A onto a of E plane oriented perpendicular to the field • Electric flux product of the E magnitude • The number of lines that cross A and A’ is and surface area A perpendicular to the field the same the flux through A equals the flux through A’ Φ = EA Φ = EA' = EA cos θ N m2/C SI units of electric flux Electric flux is proportional to the number of electric field lines
penetrating some surface 303L: Gauss’ Law (Ch.24) Electric flux is proportional to the number of electric field lines
penetrating some surface Example 1 303L: Gauss’ Law (Ch.24) Electric Flux Gauss’s Law what if the field is nonuniform ? relationship between flux and charge • The electric field varies over a surface • gaussian surface • Divide surface into a large number of small • Point charge q is at the center of a sphere of elements, each of area ∆A
∆Ai closed surface radius r
• Electric field on the surface of the sphere: a vector with magnitude representing the area of ∆Ai surface element and direction is E = ke q r 2 perpendicular to the surface element • The electric field Ei at the location of ∆Ai
makes an angle θI with ∆Ai
• The electric flux through ∆Ai is: • The net flux through the sphere (ke = 1/4πε0): r
r
∆Φ = E i ∆Ai cos θ i = E i ⋅ ∆Ai • The electric flux through a surface is: r
r
Φ = lim ∑ E i ⋅ ∆Ai =
∆Ai → 0 rr
∫ E ⋅ dA
surface rr
q
Φ = ∫ E ⋅ dA = E ∫ dA = E 4πr 2 = ε0 The net flux through the spherical surface is proportional to the charge inside Electric flux depends both on the field pattern and on the surface
303L: Gauss’ Law (Ch.24) Example 2 303L: Gauss’ Law (Ch.24) 1 Gauss’s Law Gauss’s Law relationship between flux and charge relationship between flux and charge • S1 is spherical surface, S2 and S3 are not
• The net flux through S1 is • For multiple charges or continuous distribution
of charge use the superposition principle: Φ = q ε0 The electric field due to many charges is the
vector sum of the electric fields produced by the
individual charges • The number of electric field lines is the same
through S1, S2, and S3 The net flux through any closed surface
surrounding a point charge q is given by Φ= q r r r ∫ E ⋅ dA = ∫ (E ε0 1 • For a charge outside the surface, any line that
enters the surface leaves it at another point • Gauss’s law r
r
+ E 2 + ... ⋅ dA ) the net flux through any closed surface is rrq
Φ = ∫ E ⋅ dA = in The net electric flux through a closed
surface that surrounds no charge is zero ε0 • It can be solved for E to determine the electric field due to a system of charges or a
continuous distribution of charge
303L: Gauss’ Law (Ch.24) 303L: Gauss’ Law (Ch.24) Electrostatic Equilibrium Gauss’s Law
application to various charge distributions properties of conductors Gauss’s law is useful in determining electric fields when the charge
distribution is characterized by a high degree of symmetry When there is no net motion of charge within a conductor, the conductor
is in electrostatic equilibrium rrq
Φ = ∫ E ⋅ dA = in 1) The electric field is zero everywhere inside the conductor ε0 • • 2) If an isolated conductor carries a charge, the charge resides on its surface Choose the gaussian surface that satisfies one or more of the following:
• The value of E can be argued by symmetry to be constant over the
surface 3) The electric field just outside a charged conductor is perpendicular to its
surface and has a magnitude σ/ε0, where σ is the surface charge density at
that point
4) On an irregularly shaped conductor, the surface charge density is greatest at • The dot product can be expressed as a simple algebraic product locations where the radius of curvature of the surface is smallest because E and dA are parallel
• The dot product is zero because E and dA are perpendicular
• The field can be argued to be zero over the surface Examples 37 303L: Gauss’ Law (Ch.24) 303L: Gauss’ Law (Ch.24) SUMMARY
Gauss’s law
• Electric flux is proportional to the number of electric field lines that
penetrate a surface
• Electric flux through a surface of area A is E is uniform and
make an angle θ with the normal to the surface: Φ = EA cos θ
• In general, the electric flux through a surface: r Φ=
• Gauss’s law r ∫ E ⋅ dA
surface the net electric flux through any closed gaussian surface is equal to the net charge inside the surface divided by
permittivity of free space ε0 rrq
Φ = ∫ E ⋅ dA = in ε0 • Properties of conductors in electrostatic equilibrium (14) 303L: Gauss’ Law (Ch.24) 2 ...
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 Spring '09
 TSOI
 Electric Fields

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