Field Lines Electric Flux Recall That We Defined the Electric Field to Be the Force Per Unit Charge at a Particular Point

Home , Electric flux, Field line

Gauss’ Law Field Lines Electric Flux Recall that we defined the electric field to be the force per unit charge at a particular point:

P For a source point charge Q and a test charge q at point P q

Q at P

If Q is positive, then the field is directed radially away from the charge.

Note: direction of arrows Note: spacing of lines Note: straight lines If Q is negative, then the field is directed radially towards the charge.

Negative Q implies anti-parallel to

Note: direction of arrows Note: spacing of lines Note: straight lines + +

Field lines were introduced by Michael Faraday to help visualize the direction and magnitude of the electric field. The direction of the field at any point is given by the direction of the field line, while the magnitude of the field is given qualitatively by the density of field lines. In the above diagrams, the simplest examples are given where the field is spherically symmetric. The direction of the field is apparent in the figures. At a point charge, field lines converge so that their density is large - the density scales in proportion to the inverse of the distance squared, as does the field. As is apparent in the diagrams, field lines start on positive charges and end on negative charges. This is all convention, but it nonetheless useful to remember. - +

This figure portrays several useful concepts. For example, near the point charges (that is, at a distance that is small compared to their separation), the field becomes spherically symmetric. This makes sense - near a charge, the field from that one charge certainly should dominate the net electric field since it is so large. Along a line (more accurately, a plane) bisecting the line joining the charges, we see that the field is directed along the -x direction as shown. + +

In this case, we see the zero-field region precisely between the two charges, and we also see a fairly rapid convergence on a spherically symmetric distribution of field lines. Example (Question) The figure shows the electric field lines for a system of two point charges.

(a) What are the relative magnitudes of the charges? (b) What are the signs of the charges? (c) In what regions of space is the electric field strong? weak? Example (Solution) The figure shows the electric field lines for a system of two point charges.

(a) What are the relative magnitudes of the charges? (b) What are the signs of the charges? (c) In what regions of space is the electric field strong? weak?

a) There are 32 lines coming from the charge on the left, while there are 8 converging on that on the right. Thus, the one on the left is 4 times larger than the one on the right. Example (Solution) The figure shows the electric field lines for a system of two point charges.

(a) What are the relative magnitudes of the charges? (b) What are the signs of the charges? (c) In what regions of space is the electric field strong? weak?

• The one on the left is positive; field lines leave it. The one on the right is negative; field lines end on it. Example (Solution) The figure shows the electric field lines for a system of two point charges.

(a) What are the relative magnitudes of the charges? (b) What are the signs of the charges? (c) In what regions of space is the electric field strong? weak?

• The field is strong near both charges. It is strongest on a line connecting the charges. Few field lines are drawn there, but this is for clarity. The field is weakest to the right of the right hand charge.

END How would you measure 'the density of electric field lines' in a vicinity of space? First think only of a discrete set of electric field lines. One o b v i o u s a n s w e r t o t h e question is that you would count the number of lines passing through an imaginary geometrical (not real!) surface. The first two are obvious, and the Uniform electric field - the following diagram lines have constant density will indicate the and are all parallel last:

The maximum number o f f i e l d l i n e s i s intercepted when the Surface is inserted in three surface is perpendicular different orientations and to the field, while no field lines will pass intersects the electric field through it when the lines. surface is parallel to the field. N u m b e r o f N u m b e r o f Number of field f i e l d l i n e s f i e l d l i n e s l i n e s p a s s i n g passing through passing through through surface is surface is large surface is zero between zero and the maximum. Depends on the orientation angle θ.

Electric flux a function of surface orientation, … relative to the electric field In general, the number of field lines passing through an area A is directly proportional to cos(θ), where θ is the angle between the electric field vector and the normal to the surface vector.

Use this symbol to represent electric flux through a surface Increasing surface area

Increasing electric field

The electric flux through the blue surface is the same in these two figures. Electric flux is proportional to the product of electric field and surface area A. The number of field lines passing through a geometrical surface of given area depends on three things: the field strength, the area, and the orientation of the surface. [E]-Electric field; Newton/Coulomb {N/C} [A]-Surface area; meter2 {m2} [θ]-Angle; degrees or radians {o, rad }

2 2 [ΦE]-Electric flux; Newton meter /Coulomb {Nm /C} Example (Question) A uniform electric field of 2.1 kN/C passes through a rectangular area 22 cm by 28 cm. The field makes an angle of 30o with the normal to the area. Determine the electric flux through the rectangle. Area A Example (Solution) A uniform electric field of 2.1 kN/C passes through a rectangular area 22 cm by 28 cm. The field makes an angle of 30o with the normal to the area. Determine the electric flux through the rectangle.

30o 2.1kN/C

0.22 m X0.28 m = 0.062 m2

22 cm 28 cm

END

It is useful also to represent the area A by a vector. The length of the vector is given by the area (a scalar quantity), while the orientation is perpendicular to the area. With this definition, the flux can be defined as:

recall A slightly different notation for electric flux:

Unit vector perpendicular to surface of area A WHAT IF THE SURFACE IS CURVED AND/OR THE FIELD VARIES WITH POSITION?

Element of surface area dA

Surface Consider the surface as made up of small elements over which the electric field is uniform. Flux for Non-Uniform Fields / Flux for Non-Uniform Surface

You might have noticed that all these equations really only work for uniform electric fields.

We can use them here provided we make them pertain to differential (small) area elements, and over a differential area the field is uniform.

We then need to integrate (sum) to get the total flux through an extended surface in a non-uniform field. Flux for Non-Uniform Fields / Flux for Non-Uniform Surface

The differential electric flux passing through a differential area is given by:

Element of surface area dA

As before, dA is a vector oriented perpendicular to the area, and the area itself is differential (i.e., it's infinitesimally small and it's shape doesn't usually matter). The total electric flux can be evaluated by integrating this differential flux over the surface. Element of surface area dA Flux for Non-Uniform Fields / Flux for Non-Uniform Surface

Flux for one segment

Surface

Total flux through surface q

THE INTEGRAL IS TAKEN OVER THE ENTIRE SURFACE.

Divide surface into small elements dA THE BASIC DEFINITION OF ELECTRIC FLUX Example (Question) A positive test charge of magnitude 3.0µC is surrounded by a sphere with radius 0.20 m centered on the charge. Find the electric flux through the sphere due to this charge.

+ At any point on the sphere the magnitude of the electric field is:

+ Point charge electric field at distance r from charge

E = 6.75 X 105 N/C From symmetry the field is perpendicular to the spherical surface at every point. +

Same direction on the surface The electric field has the same magnitude over the spherical surface. It is thus constant and may be taken outside the integral +

The integral that remains is just the surface area of the sphere. The electric flux through the spherical surface is:

E = 6.75 X 105 N/C

END Example (Question) A positive test charge of magnitude 3.0µC is surrounded by a sphere with radius 0.20 m centered on the charge. Find the electric flux through the sphere due to this charge.

Consider the same example again, but this time we save the calculations for the + end. At any point on the sphere the magnitude of the electric field is: + Point charge electric field at distance r from charge The electric field has the same magnitude over the spherical surface. It is thus constant and may be taken outside the integral +

The integral that remains is just the surface area of the sphere. The electric flux through the spherical surface is:

q ΦE = εo

€ +

END GAUSS’ LAW

Gauss' law is a generalization of the results discussed above for the single charge and spherical surface.

It relates the electric flux passing through any surface enclosing a charge distribution to the net charge enclosed. GAUSS’ LAW

It relates the electric flux passing through any surface enclosing a charge distribution to the net charge enclosed.

The flux through the two spheres is the same since they enclose the same charge. GAUSS’ LAW

It relates the electric flux passing through any surface enclosing a charge distribution to the net charge enclosed.

The flux through the two surfaces is the same since they enclose the same charge. GAUSS’ LAW

Each field line that enters the surface also exits the surface.

The flux through the surface is zero since no charge is + enclosed. GAUSS’ LAW

Flux through spherical surface is zero.

YOU CAN REASON

- + Each field line that exits the surface eventually re-enters the surface.

The net charge enclosed is zero. IN GAUSS’ LAW, THE CLOSED SURFACE IS REFERRED TO AS A + GAUSSIAN SURFACE USING GAUSS’ LAW Gauss’ law is true for any surface enclosing any charge distribution. When the charge distribution has sufficient symmetry we can chose a surface --- Gaussian Surface --- over which the evaluation of the flux integral becomes simple. Gauss’ law allows us to calculate the field far more easily than we could using Coulomb’s law and superposition. Using Gauss’ Law Step 1: Study the symmetry to see if you can construct a Gaussian surface on which the field magnitude and its direction relative to the surface are constant.

Through step 1 the dot product in the integral can be replaced by:

Then Ecos(θ) is constant and can be removed from inside the integral: Using Gauss’ Law Step 2: Evaluate the flux. This should be easy because of your choice of the Gaussian surface. Using Gauss’ Law Step 3: Evaluate the enclosed charge. This is not the same as the total charge if the Gaussian surface lies within the charge distribution.

The Gaussian surface is an imaginary surface that you place in the charge distribution. Choosing the Gauusian surface shape and location depends on the symmetry of the charge distribution AND on what exactly you want to calculate. Using Gauss’ Law Step 4: Equate the flux

and solve for E. The direction of the electric field vector should be evident for the symmetry and polarity of the charges.

In step 1 you reasoned the direction of E. Now go back and use that information. ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION. Example (Question) In this example we will follow the steps in applying Gauss’ law in order to obtain an Charged sphere ρ V expression for the electric field at a point external to the uniform spherical charge distribution.

R Finding the electric field using Coulomb’s law and superposition is lengthy and involves messy math. Using Gauss’ law makes the task easy, but, we must assume the field is radial. Example (Solution) ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION. Step 1: Study the symmetry to see if you can construct a Gaussian surface on which the field magnitude and its direction relative to the surface are constant. Imaginary Gaussian surface (sphere of radius r)

Charged sphere ρV

Assume a radial electric field normal to r P Example (Solution) ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION.

Step 2: Evaluate the flux. This should be easy because of your choice of the Gaussian surface. The integral is just the area of the Gaussian surface.

P dA Example (Solution) ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION.

Step 3: Evaluate the enclosed charge. This is not the same as the total charge if the Gaussian surface lies within the charge distribution. Imaginary Gaussian surface

Charged sphere ρV

All of the spherical charge distribution is R contained inside the spherical Gaussian surface. r

P Total charge enclosed Example (Solution) ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION.

Step 4: Equate the flux

and solve for E. The direction of the R electric field vector should be evident for the symmetry and polarity of the charges. r P Example (Solution) ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION.

Combining results and solving for the electric field

Direction of E from symmetry of charge distribution Example (Solution) ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION.

Combining results and solving for the electric field

Integral over charge distribution R r

Integral over Gaussian surface

SOLVING FOR THE ELECTRIC FIELD E Example (Solution) ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION.

Writing electric field expression in terms of total charge q inside Gaussian surface

Integral over charge distribution R r

P Integral over Gaussian surface

SOLVING FOR THE ELECTRIC FIELD E

END ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION. Example (Question) In this example we will follow the steps in applying Gauss’ law in order to obtain an Charged sphere ρ V expression for the electric field at a point internal to the uniform spherical charge distribution.

Charged sphere ρV

Assume a radial r electric field P normal to Example (Solution) ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION.

Step 2: Evaluate the flux. This should be easy because of your choice of the Gaussian surface. The integral is just the area of the Gaussian surface.

dA Example (Solution) ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION.

Step 3: Evaluate the enclosed charge. This is not the same as the total charge if the Gaussian surface lies within the charge distribution. Imaginary Gaussian surface

Charged sphere ρV Only the charge inside the Gaussian surface is R enclosed by the Gaussian surface. All charge external to the Gaussian surface is not included. r P Total charge enclosed

Volume enclosed by Gaussian surface Example (Solution) ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION.

Step 3: Evaluate the enclosed charge. This is not the same as the total charge if the Gaussian surface lies within the charge distribution.

Only the charge inside the Gaussian surface is enclosed by the Gaussian surface. All charge external to the Gaussian surface is not included. R

r External charge does not contribute to flux P External charge

Gaussian surface

E field lines intersect surface twice contribution to flux cancel Example (Solution) ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION.

Step 4: Equate the flux

and solve for E. The direction of the r electric field vector should be evident for the symmetry and polarity of the charges. R P

V = volume enclosed by Gaussian surface Example (Solution) ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION.

Combining results and solving for the electric field

Direction of E from symmetry of charge distribution Example (Solution) ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION.

Combining results and solving for the electric field

I n t e g r a l o v e r c h a r g e r distribution contained inside Gaussian surface R P

Integral over Gaussian surface

SOLVING FOR THE ELECTRIC FIELD E END Electric field expression at a point outside and inside the spherical charge distribution

R r r R P P

OUTSIDE INSIDE Direction of E is radial ELECTRIC FIELD FROM A INFINITE FLAT CHARGE DISTRIBUTION.

Example (Question) In this example we will follow the steps in applying Gauss’ law in order to obtain an

Charged surface ρS expression for the electric field at a point above a uniformly charged infinite flat z surface. y

x Finding the electric field using Coulomb’s law and superposition is lengthy and involves messy math. Using Gauss’ law makes the task easy, but, we must assume the field is uniform and normal to the flat surface. ELECTRIC FIELD FROM A INFINITE FLAT CHARGE DISTRIBUTION.

Example (Question) In this example we will follow the steps in applying Gauss’ law in order to obtain an

Charged surface ρS expression for the electric field at a point above a uniformly charged infinite flat surface.

View of the flat surface with electric field + + + + + + lines drawn. Field is uniform + + + + + + and normal to the flat surface. + + + + + + Example (Solution) ELECTRIC FIELD FROM A INFINITE FLAT CHARGE DISTRIBUTION. Step 1: Study the symmetry to see if you can construct a Gaussian surface on which the field magnitude and its direction relative to the surface are constant. One possibility is to use a cylindrical surface + + + + + + with the top and bottom surfaces parallel to the A1 A + + +A +2 + + 3 charged plane. The E field will be normal to + + + + + + the top and bottom surfaces.

The E field is parallel to the side of the cylinder and as such no field lines (flux) pass through it.

This type of surface is often called a Gaussian pill box. Example (Solution) ELECTRIC FIELD FROM A INFINITE FLAT CHARGE DISTRIBUTION. Step 2: Evaluate the flux. This should be easy because of your choice of the Gaussian surface.

+ + + + + + Surface S contains

A1 A three parts (Top + + +A +2 + + 3 A3, Bottom A1 and + + + + + + Side A2)

Top Bottom Side Example (Solution) ELECTRIC FIELD FROM A INFINITE FLAT CHARGE DISTRIBUTION. Step 2: Evaluate the flux. This should be easy because of your choice of the Gaussian surface. Top flux

+ + + + + +

A1 A + + +A +2 + + 3

+ + + + + +

Uniform over A3 parallel Outward pointing Example (Solution) ELECTRIC FIELD FROM A INFINITE FLAT CHARGE DISTRIBUTION. Step 2: Evaluate the flux. This should be easy because of your choice of the Gaussian surface. Bottom flux

+ + + + + +

A1 A + + +A +2 + + 3

+ + + + + +

Uniform over A1 parallel Outward pointing Example (Solution) ELECTRIC FIELD FROM A INFINITE FLAT CHARGE DISTRIBUTION. Step 2: Evaluate the flux. This should be easy because of your choice of the Gaussian surface. Side flux

+ + + + + +

A1 A + + +A +2 + + 3

+ + + + + +

Contained in side surface perpendicular Outward pointing Example (Solution) ELECTRIC FIELD FROM A INFINITE FLAT CHARGE DISTRIBUTION. Step 2: Evaluate the flux. This should be easy because of your choice of the Gaussian surface.

Top Bottom Side

+ + + + + + A2 Total flux through Gaussian A 1 + + + +A +3 + surface is equal to flux through top an bottom surfaces. + + + + + + Example (Solution) ELECTRIC FIELD FROM A INFINITE FLAT CHARGE DISTRIBUTION. Step 3: Evaluate the enclosed charge. This is not the same as the total charge if the Gaussian surface lies within the charge distribution.

T h e t o t a l c h a r g e + + + + + + enclosed consists of only A the charge on the disk 1 A A3 + + + +2 + + contained within the Gaussian Surface. + + + + + +

Charged surface ρS Disk of charged contained inside Gaussian surface Area A=A1=A3 Example (Solution) ELECTRIC FIELD FROM A INFINITE FLAT CHARGE DISTRIBUTION. Step 4: Equate the flux

and solve for E. The direction of the electric field vector should be evident for the symmetry and polarity of the charges.

+ + + + + +

A1 A + + +A +2 + + 3

+ + + + + + Example (Solution) ELECTRIC FIELD FROM A INFINITE FLAT CHARGE DISTRIBUTION. Solve for electric field

+ + + + + +

A1 A + + A+ 2+ + +3

Direction of E determined + + + + + + from symmetry END Example (Question) ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE. In one of the examples of lecture 2 we calculated the magnitude and direction of the electric field from a long straight line of charge using Coulomb’s law and superposition. Now we will show how Gauss’ law can be used to obtain the same result, in a few simple steps.

Uniform charge density Example (Solution) ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE. Step 1: Study the symmetry to see if you can construct a Gaussian surface on which the field magnitude and its direction relative to the surface are constant.

Imaginary Gaussian surface

Chose a cylinder of length and radius R

E normal and constant to surface A2 E parallel and to surfaces A1 and A3 Example (Solution) ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE. Step 2: Evaluate the flux. This should be easy because of your choice of the Gaussian surface.

Surface S contains three parts (Front

A1, Back A3 and Side A2)

Front Back Side Example (Solution) ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE. Step 2: Evaluate the flux. This should be easy because of your choice of the Gaussian surface. Flux front

Contained in front surface perpendicular Outward pointing Example (Solution) ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE. Step 2: Evaluate the flux. This should be easy because of your choice of the Gaussian surface. Flux back

Contained in back surface perpendicular Outward pointing Example (Solution) ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE. Step 2: Evaluate the flux. This should be easy because of your choice of the Gaussian surface. Flux side

Normal to side surface (constant on surface) parallel Outward pointing Example (Solution) ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE. Step 3: Evaluate the enclosed charge. This is not the same as the total charge if the Gaussian surface lies within the charge distribution.

The total enclosed by the Gaussian surface is only the segment of the line within the bounds of the front and back surfaces.

Uniform charge density Example (Solution) ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE. Step 4: Equate the flux

and solve for E. The direction of the electric field vector should be evident for the symmetry and polarity of the charges.

Uniform charge density Example (Solution) ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE. Step 4: Equate the flux and solve for E. Example (Solution) ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE. Step 4: Equate the flux and solve for E. The direction of the electric field vector should be evident for the symmetry and polarity of the charges.

ρ  E =  A εo 2

Surface area of a cylinder

€ Note R radial distance from charged line END