In a table-top plasma ball, the colorful lines emanating from the sphere give evidence of strong electric fields. Using Gauss’s law, we will learn that the electric field surrounding a charged sphere is identical to that of a point charge. (Getty Images)

ELECTRIC FLUX

•Electric flux

•Gauss’s law

•Application of Gauss’s Law

•Conductors in electrostatic

The number of lines penetrating per unit area (line density) is proportional to the magnitude of the electric field.

Note:

Electric field lines are perpendicular to the surface area

Consider an electric field that is uniform in both magnitude and direction as shown in Figure.

The field line penetrate a rectangular surface of area,A whose plane is orientated perpendicular to the field.

1. Electric flux.

Since the number of lines penetrating per unit area (line density) is proportional to the magnitude of the electric field,

Then the total number of lines penetrating the surface is proportional to the product EA

The product of the magnitude of electric field to the surface area, EA is called the electric flux

• Electric flux, ΦE = EA

• Unit: Nm2.C-1

• Electric flux is proportional to the number of electric field lines penetrating some surface

• If the electric field penetrating the surface area is not perpendicular to the surface area, consider the following….

Field lines representing a uniform electric field penetrating an area A’ that is at an angle to the field. Because the number of lines that go through the area A’ is the same as the number that go through A, the flux through A’ is equal to the flux through A and is given by ΦE = EA’ = EAcos .

When is the flux maximum ? When is the flux zero ?

The flux is maximum when the normal to the surface is parallel to the lines, that is θ = 0o

The flux is zero when the normal to the surface is perpendicular to the lines, that is θ = 90o

• We had assumed uniform electric field previously.

• In general situations, electric field may vary over a surface

• Hence consider a general surface divided up into large number of small elements, each of area ∆A

• Variation of electric field over one element can be neglected if the element is sufficiently small.

• Lets define vector ∆Ai whose magnitude represents the area of the i th element of the surface and direction defined to be perpendicular to the surface element.

A small element of surface area Ai. The electric field makes an angle i with the vector Ai, defined as being normal to the surface element, and the flux through the element is equal to Ei Ai cos i .

1.1 Definition of electric flux

iii cosθΔAEE ii A.E

surfaceA

Ei

E.dA.ΔΔE ii0

lim

The electric flux through the element:

The total flux is the sum of contributions of all elements. If the area of each element approach zero, the number of elements approaches infinity, hence electric flux is

1.2 Electric flux through closed surface

• A close surface is defined as one that divides space into an inside and outside region.

• The surface of a sphere, for example is a closed surface.

A closed surface in an electric field. The area vectors Ai are, by convention, normal to the surface and point outward.

The flux through an area element can be

positive (element 1) where θ < 90o ,

zero (element 2) where θ = 90o, or

negative (element 3) where 180o > θ > 90o.

• The net flux is proportional to the net number of lines leaving the surface,

• Net number = number leaving surface – number entering surface

• Using ∫ to representing integral over a closed surface, the net flux through a closed surface is

dAEnE .E.dA

Where En represents the component of the electric field normal to the surface

1.3 Example

• Consider a uniform electric field E oriented in the x direction. Find the net electric flux through the surface of a cube of edge l

A closed surface in the shape of a cube in a uniform electric field oriented parallel to the x axis. Side 4 is the bottom of the cube, and side 1 is opposite side 2.

Solution.

• The net flux is the sum of the fluxes through all faces of the cube.

• The flux through faces 3, 4 and the unnumbered ones is zero because E is perpendicular to dA on these faces.

• The net flux through faces 1 and 2 is

21

E.dAE.dAE

Solution.

• For face 1, E is constant and directed inward but dA is directed outward ( θ = 1800 ), therefore flux is

2

11

)180(cos lEEAdAEdAEE

Solution.

• For face 2, E is constant and directed outward and dA is also directed outward ( θ = 00 ), therefore flux is

2

22

)0(cos lEEAdAEdAEE

Therefore net flux over all six surfaces is

ΦE = -El2 + El2 + 0 +0 + 0 + 0 = 0

2. Gauss’s law

• Gauss’s law is the relation between the net electric flux through a closed surface (gaussian surface) and the charge enclosed by the surface.

• Fundamental importance in the study of electric fields.

Karl Friedrich Gauss German mathematician and astronomer (1777–1855)

Gauss received a doctoral degree in mathematics from the University of Helmstedt in 1799.

In addition to his work in electromagnetism, he made contributions to mathematics and science in number theory, statistics, non-Euclidean geometry, and cometary orbital mechanics.

He was a founder of the German Magnetic Union, which studies the Earth’s magnetic field on a continual basis.

A spherical gaussian surface of radius r surrounding a point charge q. When the charge is at the center of the sphere, the electric field is everywhere normal to the surface and constant in magnitude.

Consider a +ve point charge q located at the centre of a sphere or radius r as shown in Figure.

• The magnitude of the electric field is given by E = keq/ r2.

• The field lines are directed radially outward and hence perpendicular to the surface at every point on the surface.

• At each point on surface, E is parallel to the vector ∆Ai, therefore

• E.∆Ai = E∆Ai

E.dAEΦ dAEE.dA

24 rAdA

qkrr

qke

eE 4)4( 2

2

oek

41

oE

q

The net flux through the gaussian surface is given by:

Since surface is spherical, then

Hence,

Recalling that,

Then,

oE

q

Form the above equation:

1. The net flux through the spherical surface is proportional to the charge inside.

2. The flux is independent of the radius, r because the area of the spherical surface is proportional to r2, whereas the electric field is proportional to 1/ r2. Hence the product of area and electric field cancels the dependence on r.

Consider closed surfaces of various shapes surrounding a charge q.

The flux through S1 is q/ Є0

Number of lines thro’ S1 is equal thro’ S2 and S3.

The flux is proportional to number of electric field lines thro’ surface.

Construction of figure shows that the net electric flux is the same through all surfaces whether spherical or non-spherical.

Hence can conclude that the net flux thro’ any closed surface surrounding a point charge q is given by q/ Є0 and is independent of the shape of that surface.

Fig 24-8, p.744

A point charge located outside a closed surface. The number of lines entering the surface equals the number leaving the surface.

The net electric flux thro’ a closed surface that surrounds no charge is zero

• The electric field due to many charges is the vector sum of the electric fields produced by the individual charges.

AEEE.dA 21 d.....).(

The net electric flux through any closed surface depends only on the charge inside that surface.

The net flux through surface S is q1/ Є0 ,

The net flux through surface S’ is (q2 + q3)/ Є0 , and

The net flux through surface S’’ is zero.

Charge q4 does not contribute to the flux through any surface because it is outside all surfaces.

• Hence, Gauss’s Law states that the net flux thro’ any closed surface is

o

inE

q

E.dA

Where qin represents the net charge inside the surface and E represents the electric field at any point on the surface.

Note: Although the charge qin is the net charge inside the gaussian surface, E represents the total electric field, which includes contributions from charges both inside and outside the surface

3. Application of Gauss’s Law to various charge distributions

• Gauss’s law is useful in determining electric fields when the charge distribution is characterized by a high degree of symmetry.

• Hence important to choose surface over which the surface integral can be simplified

• The following conditions should be used:

• 1. The value of the electric field can be argued by symmetry to be constant over the surface

• 2.The dot product can be expressed as a simple algebraic product E.dA because E and dA are parallel

• 3. The dot product is zero because E and dA are perpendicular

• 4. The field can be argued to be zero over the surface.

3.1 The electric field due to a point charge.

• Calculate using Gauss’s law, the electric field due to an isolated point charge q

E.dAE o

in

ε

qE.dA

224 r

qk

r

qE e

o

E is parallel to dA, hence E.dA = E.dA which gives Gauss’s law as follows:

By symmetry, E is constant everywhere, hence E can be taken out of the integral:

o

2E ε

qr4EdAEΦ

Therefore:

3.2 A spherically symmetric charge distribution

• An insulating solid sphere of radius a has a uniform volume charge density ρ and carries a total positive charge Q

• (a) Calculate the magnitude of the electric field at a point outside the sphere.

• Because the charge distribution is spherically symmetric, the gaussian surface is of spherical shape of radius r.

• Hence, condition 1 and 2 are satisfied, and following the reasoning as previous example,

2r

QkE e

• An insulating solid sphere of radius a has a uniform volume charge density ρ and carries a total positive charge Q

• (b) Calculate the magnitude of the electric field at a point inside the sphere.

• The gaussian surface is a spherical having radius r < a, concentric with the insulating sphere.

• Let V’ = volume of smaller sphere• The charge qin within the gaussian surface of

volume V’ is less than Q• Hence, qin = ρV’ = ρ(4/3)πr3

• By symmetry, the electric field is constant everywhere on the gaussian surface and normal to the surface at each point

E.dAE o

in

ε

qE.dA

rr

r

r

qE

ooo

in

34

3

4

4 2

3

2

But

o

in2E ε

qr4EdAEΦ

Therefore:

3

3

4a

Q

oek

41

ra

Qk

a4

QrE

3e3

o

and

Therefore,

Note that the result differ from part (a)

The E → 0 as r → 0

If E varied as 1/ r2 inside the sphere, then problem would be created because as r → 0, E would be infinite

What would be the E value for both parts when r = a ?

ra

Qk

a4

QrE

3e3

o

Fig 24-12, p.747

3.3 The electric field due to a thin spherical shell

• A thin spherical shell of radius a has a total charge Q distributed uniformly over its surface. Find the electric field at points

• (a) outside and (b) inside the shell.

• The calculation for part (a) is identical to that for the solid sphere previously

• If the gaussian sphere is with radius r > a concentric with the spherical shell, the charge inside the surface is Q

• Therefore the field outside the shell is equivalent to that due to a point charge Q located at the center, hence

2r

QkE e

• The electric field inside the spherical shell is zero. This follows from Gauss’s law applied to a spherical surface of radius r < a concentric with the shell.

• E = 0 in the region where r < a

3.4 A cylindrical symmetric charge distribution.

• Find the electric field a distance r from a line of positive charge of infinite length and constant charge per unit length λ

(a) An infinite line of charge surrounded by a cylindrical gaussian surface concentric with the line. (b) An end view shows that the electric field at the cylindrical surface is constant in magnitude and perpendicular to the surface.

E.dAE oo

in

ε

λl

ε

qEAdAE

rk

rE e

o

22

E is directed outward, constant in magnitude and perpendicular to the cylindrical gaussian surface of radius r and length l that is co axial with line charge

The area of the curved surface is A = (2πrl), therefore:

oε

λπE

lrl 2

Therefore:

Note: The electric field due to a cylindrical symmetric charge distribution varies as 1/ r as compared to spherically symmetric charge distribution varies with 1/ r2

3.5 A plane of charge

• Find the electric field due to an infinite plane of positive charge with uniform surface charge density σ

A cylindrical gaussian surface penetrating an infinite plane of charge. The flux is EA through each end of the gaussian surface and zero through its curved surface.

By symmetry, E must be perpendicular to the plane and must have the same magnitude at all points equidistant from the plane.

A gaussian surface that reflects the symmetry is a small cylinder whose axis is perpendicular to the plane and whose ends each have an area A and are equidistant from the plane.

E is parallel to the curve surface. The flux thro’ each end of the cylinder is EA, hence the total flux thro’ the entire surface is just Φ = 2EA

Total charge inside the surface is qin = σA , therefore

o

oo

inE

E

AqEA

2

2

Note: the distance from each end of the cylinder does not appear in equation, hence can conclude that E = σ/ 2Єo at any distance from the plane.

What if two infinite planes of charge parallel to each other, one positively charged and the other negatively charged. Both planes have the same surface charge density. What does the electric field look like?

Answer:

In this situation, the electric fields due to the two planes add in the region between the planes, resulting in a uniform field of magnitude σ/ Єo.

4.0 Conductors in electrostatic equilibrium

• Electrostatic equilibrium means no net motion of charge within conductor. Hence, the following properties:– The electric field is zero everywhere inside the

conductor– If an isolated conductor carries a charge, the charge

resides on its surface– The electric field just outside a charged conductor is

perpendicular to the surface of the conductor and has a magnitude σ/ Єo

– The surface charge density on an irregular shaped conductor is greatest at locations where the radius of curvature of the surface is smallest

A conducting slab in an external electric field E. The charges induced on the two surfaces of the slab produce an electric field that opposes the external field, giving a resultant field of zero inside the slab.

4.1The electric field is zero everywhere inside the conductor

• If field not zero, free electrons experience force, hence accelerate.

• Motion of electrons mean no equilibrium.• How zero field accomplished?

– When field applied, free electrons acc. to left of figure, forming plane of negative charge

– Plane of positive charge results on the right– These planes create additional field inside conductor

that opposes field outside– The magnitude of internal field equals that of external,

hence net field of zero inside

4.2 If an isolated conductor carries a charge, the charge resides on its surface

• Previously stated that electric field zero inside the conductor when electrostatic equilibrium

• Therefore zero electric field at every point on the gaussian surface. Thus the net flux thro’ the gaussian surface is zero

• No net charge inside gaussian surface, hence any net charge must reside on its surface.

4.3 The electric field just outside a charged conductor is perpendicular to the surface of

the conductor and has a magnitude σ/ Єo

• If E is parallel to surface then electrons experience force, and move along surface, hence no equilibrium

• Thus E must be perpendicular to surface• Using gaussian surface to be like cylinder, no

flux thro’ curve part of cylinder and flat end inside the conductor (E=0)

• Hence, applying Gauss’s law

o

oo

inE

E

AqEAEdA

Electric field pattern surrounding a charged conducting plate placed near an oppositely charged conducting cylinder. Small pieces of thread suspended in oil align with the electric field lines. Note that (1) the field lines are perpendicular to both conductors and (2) there are no lines inside the cylinder (E= 0).