electric flux is independent of surface shapeelectric flux is independent of surface shape what...

Electric Flux is Independent of Surface Shape

What about the flux through some arbitrarysurface shape?

Approximate the surface by a patchwork ofradial and spherical pieces. Spherical piecesare centered on the charge. Radial pieceshave 0 flux through them.We can make the pieces arbitrarily small toget the best model of the surface.We can slide the pieces around and make asphere. So, the flux through any closedsurface is

∮~E · d~A =

Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 1 / 1

What about the flux through some arbitrarysurface shape?Approximate the surface by a patchwork ofradial and spherical pieces. Spherical piecesare centered on the charge. Radial pieceshave 0 flux through them.

We can make the pieces arbitrarily small toget the best model of the surface.We can slide the pieces around and make asphere. So, the flux through any closedsurface is

∮~E · d~A =

What about the flux through some arbitrarysurface shape?Approximate the surface by a patchwork ofradial and spherical pieces. Spherical piecesare centered on the charge. Radial pieceshave 0 flux through them.We can make the pieces arbitrarily small toget the best model of the surface.

We can slide the pieces around and make asphere. So, the flux through any closedsurface is

∮~E · d~A =

What about the flux through some arbitrarysurface shape?Approximate the surface by a patchwork ofradial and spherical pieces. Spherical piecesare centered on the charge. Radial pieceshave 0 flux through them.We can make the pieces arbitrarily small toget the best model of the surface.We can slide the pieces around and make asphere. So, the flux through any closedsurface is

∮~E · d~A =

Charge Outside the Surface

We have only been talking about charges enclosed by the surface.What about charges outside the surface?

Anything outside the surface will contribute nothing to the totalflux!Any line entering the surface in one place will exit it inanother....there is no net flux unless the charge is enclosed.

We have only been talking about charges enclosed by the surface.What about charges outside the surface?Anything outside the surface will contribute nothing to the totalflux!

Any line entering the surface in one place will exit it inanother....there is no net flux unless the charge is enclosed.

We have only been talking about charges enclosed by the surface.What about charges outside the surface?Anything outside the surface will contribute nothing to the totalflux!Any line entering the surface in one place will exit it inanother....there is no net flux unless the charge is enclosed.

Multiple Charges

What is the flux from a set of chargesthrough a surface?

We know that the electric field vectorsjust add, and so will the fluxes

Φe = Φ1 + Φ2 + Φ3 + . . .

Φe =(q1

ε0+ · · ·

)inside

+ (0)outside

∮~E · d~A =

where Qin = q1 + q2 + · · ·+ qi

Multiple Charges

What is the flux from a set of chargesthrough a surface?We know that the electric field vectorsjust add, and so will the fluxes

Φe = Φ1 + Φ2 + Φ3 + . . .

Φe =(q1

ε0+ · · ·

)inside

+ (0)outside

∮~E · d~A =

where Qin = q1 + q2 + · · ·+ qi

Multiple Charges

Φe = Φ1 + Φ2 + Φ3 + . . .

Φe =(q1

ε0+ · · ·

)inside

+ (0)outside

∮~E · d~A =

where Qin = q1 + q2 + · · ·+ qi

Multiple Charges

Φe = Φ1 + Φ2 + Φ3 + . . .

Φe =(q1

ε0+ · · ·

)inside

+ (0)outside

∮~E · d~A =

where Qin = q1 + q2 + · · ·+ qi

Multiple Charges

Φe = Φ1 + Φ2 + Φ3 + . . .

Φe =(q1

ε0+ · · ·

)inside

+ (0)outside

∮~E · d~A =

where Qin = q1 + q2 + · · ·+ qi

Using Gauss’ Law - Example 28.4

Example 28.4

What is the electric field inside auniformly charged sphere?

Solving this with Coulomb’s Law and superposition would not be veryfun.

However, there is a clear spherical symmetry in the problem. Let’stry it with a spherical Gaussian surface inside the original sphere.

Example 28.4

Solving this with Coulomb’s Law and superposition would not be veryfun. However, there is a clear spherical symmetry in the problem.

Let’stry it with a spherical Gaussian surface inside the original sphere.

Example 28.4

Solving this with Coulomb’s Law and superposition would not be veryfun. However, there is a clear spherical symmetry in the problem. Let’stry it with a spherical Gaussian surface inside the original sphere.

~E is perpendicular to the surface andhas the same strength everywhere onthe surface. The flux is therefore

Φe = EAsphere = 4πr2E =Qin

Qin is the charge inside the Gaussiansphere (ie. the little one). The chargeis uniform and the volume chargedensity is

ρ =QVR

43πR3

~E is perpendicular to the surface andhas the same strength everywhere onthe surface. The flux is therefore

Φe = EAsphere = 4πr2E =Qin

Qin is the charge inside the Gaussiansphere (ie. the little one). The chargeis uniform and the volume chargedensity is

ρ =QVR

43πR3

The charge enclosed by the sphere isthen

Qin = ρVr =

Q43πR3

(43πr3

Gauss’ Law becomes

4πr2Qin

14πε0

The charge enclosed by the sphere isthen

Qin = ρVr =

Q43πR3

(43πr3

Gauss’ Law becomes

4πr2Qin

14πε0

Conductors in Electrostatic Equilibrium (28.6)

We know that a conductor has anyexcess charge on its surface and ~E iszero anywhere on the inside (elsecharges would be moving).

If you assume E = 0 everywhere anddraw a Gaussian surface then you knowthat Qin = 0 inside the surface. Any netcharge must be on the outer edge...hey,we knew that already!The electric field at the surface isperpendicular to the surface and hasmagnitude (see your text):

~Esurface =η

where η is surface charge density.

We know that a conductor has anyexcess charge on its surface and ~E iszero anywhere on the inside (elsecharges would be moving).If you assume E = 0 everywhere anddraw a Gaussian surface then you knowthat Qin = 0 inside the surface. Any netcharge must be on the outer edge...hey,we knew that already!

The electric field at the surface isperpendicular to the surface and hasmagnitude (see your text):

~Esurface =η

where η is surface charge density.

We know that a conductor has anyexcess charge on its surface and ~E iszero anywhere on the inside (elsecharges would be moving).If you assume E = 0 everywhere anddraw a Gaussian surface then you knowthat Qin = 0 inside the surface. Any netcharge must be on the outer edge...hey,we knew that already!The electric field at the surface isperpendicular to the surface and hasmagnitude (see your text):

~Esurface =η

where η is surface charge density.Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 7 / 1

Conductors in Electrostatic Equilibrium

If we make a hole in the conductor we can analyze it with aGaussian surface close to the hole. It is easy to conclude thatQin = 0.

Since ~E is also zero inside the hole, this has practical applications:we can build a Faraday cage

Conductors in Electrostatic Equilibrium

If we make a hole in the conductor we can analyze it with aGaussian surface close to the hole. It is easy to conclude thatQin = 0.Since ~E is also zero inside the hole, this has practical applications:we can build a Faraday cage

The Electric Potential (Chapter 29)

We are working our way towards really practical stuff likecircuit-building.

To get there, we need to talk about energy. Chapter 9 is all aboutelectric potential.We have mostly focused on static charges so far. If we eventuallywant to understand the motion of charges we will need tounderstand electric potential.

We are working our way towards really practical stuff likecircuit-building.To get there, we need to talk about energy. Chapter 9 is all aboutelectric potential.

We have mostly focused on static charges so far. If we eventuallywant to understand the motion of charges we will need tounderstand electric potential.

We are working our way towards really practical stuff likecircuit-building.To get there, we need to talk about energy. Chapter 9 is all aboutelectric potential.We have mostly focused on static charges so far. If we eventuallywant to understand the motion of charges we will need tounderstand electric potential.

Mechanical Energy

The treatment of electric energy is developed by analogy tomechanical energy (in particular using comparisons togravitational interactions). So, let’s review.

Remember that mechanical energy is conserved by particleswhich interact via conservative forces (eg. gravity, electric).

∆Emech = ∆K + ∆U = 0

Kinetic energy of a system is the sum of the kinetic energies of allof the particles in the system.Potential energy is the interaction energy of the system.

Mechanical Energy

The treatment of electric energy is developed by analogy tomechanical energy (in particular using comparisons togravitational interactions). So, let’s review.Remember that mechanical energy is conserved by particleswhich interact via conservative forces (eg. gravity, electric).

∆Emech = ∆K + ∆U = 0

Mechanical Energy

∆Emech = ∆K + ∆U = 0

Kinetic energy of a system is the sum of the kinetic energies of allof the particles in the system.

Potential energy is the interaction energy of the system.

Mechanical Energy

∆Emech = ∆K + ∆U = 0

Mechanical Energy

We define the change in potential energy in terms of the workdone by forces:

∆U = Uf − Ui = −Winteraction forces

The component of force along the direction of motion does workon an object. Non-constant forces need the “chop-up andintegrate” trick:

W =∑

(Fs)j∆sj →

∫ sf

Fsds =

~F · d~s

Mechanical Energy

We define the change in potential energy in terms of the workdone by forces:

∆U = Uf − Ui = −Winteraction forces

The component of force along the direction of motion does workon an object. Non-constant forces need the “chop-up andintegrate” trick:

W =∑

(Fs)j∆sj →

∫ sf

Fsds =

~F · d~s

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