Chapter 24

Gauss's Law:

Before writing the statement for Gauss's law, the concept of electric flux must be understood.

Electric Flux:

If you hold a ring horizontally under rain, maximum number of rain drops pass through it (1).  If you hold it vertically, no rain drop passes through it (4).  If you hold it at some angle, some rain drops but not the maximum possible pass thorough it (2) and (3).  See figures shown belowTo show the orientation of the ring in space, it is better to measure the angle its normal makes with the vertical directionNormal to the ring means the line that is perpendicular to its plane.  The amount of rain passing through the ring is called the flux of rain through the ring.  In each case the flux of rain through the ring is different.

Electric field lines may also be viewed as rain lines if we are talking about a downward electric field.  The heavier the downpour, the stronger the electric field E

Electric Flux Through a Surface:

The electric flux Φe of a uniform electric field E through a loop of area A is defined as Φe = E A cosθ where θ is the angle between the field lines and the normal vector, n, to the plane of the loop. An appropriate figure for this formula is shown below:

 Note that if E is uniform that means it is not a function of space (x, y, or z), the calculation of flux ΦE = E A cosθ is easy.  If E is not uniform, such as the electric field around a point charge, then since E varies with x, y, or z, differentials and integrals must be used. The figure on the right shows the flux of a uniform field E through a ring of area A. ΦE means electric flux.

Symbol Φ is pronounced "phi".   The SI unit for electric flux is Nm2/Coul.  As can be visualized from the above picture, if the loop is turned such that the normal vector n becomes parallel to the field lines, θ becomes zero and the loop becomes vertical and the number of electric field lines passing through it horizontally becomes maximum.  If θ is zero, then cosθ = 1 that has its maximum value makes ΦE maximum , or ΦE = E A.   If the loop is turned such that θ = 90°, the loop becomes horizontal and no field lines pass through it that makes the flux Φe= 0 because cos90 = 0.  The above figure shows the electric flux of a uniform electric field.

Example 1:  A student is holding a 36cm by 15cm rectangular piece of cardboard such that its surface makes a 60.0° angle with an existing upward 2500N/C electric filed as shown.  If the intensity of the electric field remains constant over the entire surface of the cardboard, calculate the electric flux through the cardboard.

 Solution:   A =  ( 0.36m)( 0.15m) = 0.054 m2 The angle that E makes with n is shown by ? in the figure.  If n is normal to the cardboard, what angle does it make with E? You are right 30.0 degrees.  Thus θ= 30.0°, and Φ =EAcosθ =(2500N/C)(0.054m2)cos(30.0°)  Φ = 120Nm2/C.

Example 2:  A 32-nC point charge is at the center of a sphere 80.0cm in diameter.  Calculate (a) the magnitude of the electric field at any point on its surface, (b) the angle that the electric field at a point on the surface makes with normal to the surface at that point, and (c) the electric flux that passes through the entire surface of the sphere.

Solution:  Since the point charge is at the center of the sphere, it is equidistant to all point of its surface.  This makes the magnitude of the electric field the same at any of the points on its surface although the directions are different.  Therefore, the solution to Part (a) is

(a) E = kq/r2 = {9.00x109x32x10-9/(0.40m)2} N/C  =  1800N/C.

(b) The angle that the normal line makes with E is zero at any point on the sphere.  Why?  Since the point charge is at the center, each field line is along a radius. At any selected tiny area element, the normal line is also along the radius drawn to that element.  This makes E and n parallel to each other at each point on the sphere.  Thus θ=0 and  cosθ = 1 at any point on the surface of the sphere.

(c) With E being constant in magnitude, and cosθ = 1 for all points, the total electric flux becomes:

Φ = EAcosθ =  (1800 N/C) [(4)(3.14)(0.40m)2] (1)  = 3600 Nm2/C.   (Note that Asphere = 4πR2).

Example 3:  If the sphere in Example 2 is placed at the center of a cube that is 1.00m on each side, what flux does pass through each side of the cube?

Solution: The solution is left to the students because of simplicity.

The electric Flux of a Non-uniform Electric Field:

The following figure shows the field lines of a non-uniform electric field passing through a certain surface area.  Five field lines are shown along with five small area segments ΔA1, ΔA2, ...ΔA5.  At each segment ΔAi, field has a different direction determined by its Ei vectorOf course, the orientation of each area segment ΔAi is determined by its own normal vector ni.  The small flux ΔΦi through each ΔAi is

 ΔΦi=Ei ΔAi cosθi (1) Angles θi are not shown in order to keep the figure more clear.  As can be seen, in general, the angle between n1 and E1 is not equal to angles between n2 and E2,  n3 and E3,  and so on.

Recall the definition of the dot product of two vectors A and B making an angle of α with each other that is  A∙B = ABcosα  where bold letters denote vector quantities and regular letters their magnitudes.  This reads as: The dot product of  vectors A and B is equal to magnitude A times magnitude B times the cosine of the angle in between.  Equation 1 is exactly the definition of dot product.  Actually the dot product was originally defined for applications like this.  The results of dot product is always a scalar.  Equation 1 may be written as:

ΔΦi = Ei  ΔAi         (2)

In Equation 2, because of the symbol dot ( ), Ei  and ΔAi are shown in bold letters meaning that they are vectors.  The solution to (2) is (1) where magnitudes Ei  and ΔAi are multiplied by each other times the cosine of angle in between.  For simplicity, we will use (2) as the flux through a typical small area segment.

To calculate the total flux Φ through surface S, all such ΔΦi (five of which are shown) must be added.  The sum is

Φ    Σ ΔΦi   =   Σ Ei  ΔAi         (3)

Φ     E1  ΔA1   +   E2  ΔA2   +    ∙  ∙  ∙   +   Em  ΔAm          (4)

where m is the number of are segments that covers surface S.

To find an exact value for Φ, the electric flux, the number of tiny area segments must approach or the size of ΔAi must approach 0; in other words, ΔAi must be replaced with dA and consequently Σ must be replaced by .  Equation (3) becomes

Φ = E ∙ dA         (5)

In this integral, the normal vector n is embedded in dA because A is shown in bold letter emphasizing its vector nature.  Note that by defining n, the vector normal to the area segment,  we are giving each area segment a direction that determines its orientation in space.  The symbol dot ( ) includes cosine of the angle in the product.

Gauss's Law:

Gauss's law states that  the net flux through a closed surface is equal to Q/ε0 where Q is the net charge entrapped in that closed surface.  Another version of this statement is:

The flux of charge Q through a closed surface is equal to Q/ε0.  A sphere, a cube, or an inflated nylon balloon of any shape is treated as a closed surface.  If charge Q is entrapped inside any of these shapes, as long as the surface is closed, the total electric flux by the charge through that closed surface is equal to Q/ε0.   In other words, we may write from (5)

 (6)

This can be easily verified for a point charge Q placed at the center of a sphere of radius R.  The proof is as follows:

Since the point charge is at the sphere's center, all field lines are exactly radial lines.  Any selected area element dA has its normal vector n also along its radial line. This makes E and n parallel (coincide) at each point or element on the the sphere's surface and thus θ=0 and  cosθ = 1.

Applying the closed surface integral, we get:

and Gauss's law is verified.  If Q is not placed at C, then E and n will make different nonzero angles with each other at different points (or selected area elements) and the integral will not be that easy to evaluate; however, the final outcome will be the same, and the answer to that complicated integral will be Q/ε0 again.  The reason is that the number of electric field lines that Q can generate is limited to its amount, Q, and the permittivity of the medium, ε0, regardless of the shape of the enclosing surface and the location and distribution of Q in the closed surfaceWhat was your answer to Example 3?

Example 4:  A cube 8.0cm on each side has 15nC of point charge at its center.  Find (a) the net flux through it, and (b) the net flux through each side of it.

Solution:   The net flux through a closed surface is independent of the shape of that surface.  Using Gauss's law, the integral is equal to Q/ε0.  We may write:

(a)      Net flux  =  Q/ε0  =  (15nC) / [8.85x10-12 C2/(Nm2) ]  =  1700 Nm2/C.

(b)     Since the point charge is at the center of the cube,  the flux through each side is 1/6 of the total flux due to symmetry.

Flux through each side = (1700/6)Nm2/C  =  280 Nm2/C.

Gaussian Surface:

Gaussian surface is an imaginary surface that we assume around a point charge or any charge distribution.  Assuming such imaginary surface is helpful in determining the way the electric field varies around a given charge distribution.  Of course, any selected Gaussian surface must surround or entrap the charge distribution under study.  That way the total flux is known, simply Q/ε0.   If good symmetry exists in the charge distribution, and also a symmetrical Gaussian surface is selected, surface calculation becomes easy, and that facilitates the calculation of the unknown electric field.

Important Points in Gaussian Surface Selection:

1) Use symmetry.

2) Choose a Gaussian surface that is either parallel to the normal, n, or perpendicular to it.

3) If E is parallel to n over a certain area, then the magnitude of E is constant for that portion of the area, and that makes the integration easier.  If E is perpendicular to n over a certain area, then the flux is zero through that portion of the area.

 Example 5:  Determine the electric field E (a) outside and (b) inside of a spherical shell that has charge Q evenly distributed over its surface. Solution: (a) E outside of the shell: A uniformly charged shell is shown on the right with a spherical Gaussian surface selected around it such that the two spheres are concentric.  Choosing the Gaussian surface concentric with the shell helps making E parallel n over the entire surface.  That makes the E magnitude constant and out of the Gaussian integral.  We may write: ∫cs E ∙ dA  = Q/ε0   (cs means over closed surface) or, E∫dA  = Q/ε0 , or E(4πr2) = Q/ε0 , or E = Q / [4πε0r2] ,  or  E = kQ/r2    ;  { Note: k=1/(4πε0) } (b) E inside the shell: Looking at the imaginary or Gaussian surface assumed inside the shell, we see that there is no charge entrapped in it, and therefore Q = 0.  This makes the integral equal to zero. ∫cs E ∙ dA  = Q/ε0 = 0, or E∫dA  = 0, or E(4πr2) = 0.  Since r≠0 ;therefore, E = 0. The electric field inside the shell is zero.The graph of E(r) for the charged shell  is shown on the right.

Example 6:  An empty thin metal sphere with a diameter of 8.0cm is given 49nC of electric charge.  The sphere has an insulator mount and is far from other objects and charges.  Determine the electric field at points that are (a) 3.0cm, (b) 7.0cm, and (c) 70.0cm from its center.

Solution: The radius of the sphere is R = 4.0cm.

(a) Based on the results of Example 6, at r = 3.0cm < R, E = 0 for a charged spherical shell.

(b) Again, based on Example 6, at r = 7.0cm > R,  E = kQ/r2 = (9x109Nm2/C2)(49x10-9C) / (0.070m)2 = 9.0x104 N/C

(c) E = kQ/r2 = (9x109Nm2/C2)(49x10-9C) / (0.70m)2 = 9.0x102 N/C

 Example 7:  Determine the electric field E (a) outside and (b) inside of a solid sphere that has charge Q evenly distributed throughout its volume. Solution: (a) E outside the solid sphere: A uniformly charged solid sphere is shown on the right with a spherical Gaussian surface selected around it such that the two spheres are concentric.  Choosing the Gaussian surface concentric with the solid sphere helps making E parallel n over the entire surface.  The calculation is similar to that of a shell (Example 5) because the entrapped charge is the same as well as the Gaussian surface. ∫cs E ∙ dA  = Q/ε0   (cs means over closed surface) or, E∫dA  = Q/ε0 , or E(4πr2) = Q/ε0 , or E = Q / [4πε0r2] ,  or  E = kQ/r2. (b) E inside the solid sphere: The amount of entrapped charge in this case depends on the radius of the Gaussian surface.  If r = 0, the entrapped charge is also zero, and therefore the field is zero at the center.  If r = R, the entrapped charge is the entire charge Q.  For any 0 < r < R, The fraction of Q that is entrapped in the Gaussian surface is proportional to (4/3πr3)/(4/3πR3) =  r3/R3.  This means that for a Gaussian surface of radius r, the entrapped charge is [ r3/R3 ]Q.  The integral becomes: ∫cs E ∙ dA  = [ r3/R3 ]Q/ε0  or,  E(4πr2) = [ r3/R3 ]Q/ε0 , or E = [kQ/R3] r.  Since kQ/R3 is a constant, E varies linearly with r inside the solid sphere.  The graph of E(r) for the charged solid sphere is shown on the right.

Example 8:  A solid sphere of radius 3.00cm is made of plastic and has -15nC of electric charge uniformly distributed throughout its volume. Find (a) the magnitude of the electric field at 1.50cm from its center.  (b) At what radial distance from its surface the field strength magnitude is equal to the field strength you found in Part (a)?

Solution: The sphere is solid and uniformly charged; therefore, the use of formula E = [kQ/R3] r  for its internal field is appropriate.

(a) E = [(9x109)(15x10-9) / (0.03m)3] (0.015m) =  75,000 N/C.

.            (b) The field magnitude outside the sphere is E = kQ/r2.  We must have 2.25x10-3 N/C = kQ/r2  or,

75000 = (9x109)(15x10-9) / r2.  Solving for rr = .042m = 4.2cm.  The distance from the surface of the sphere is therefore

4.2cm – 3.00cm = 1.2cm.

 Example 9: Determine the magnitude of the electric field at a distance r from an infinite line of charge with a linear charge density of  λ C/m. Solution: For an infinite line of charge, an appropriate Gaussian surface is a cylinder of length L as shown.  Note that to have symmetry, the line of charge must be the axis of symmetry of the cylinder.  The Gaussian surface here has 3 segments: one cylindrical surface S1 and two circular surfaces S2 and S3.  Since the line is infinitely long, at any point around the line, the direction of the electric field becomes perpendicular to the line itself, and therefore perpendicular to S1 as well.  For S1, E and n are parallel and therefore, the part of Gaussian integral over S1 becomes: ∫S1 E ∙ dA  = ∫S1 E dA =  E ∫S1 dA = E(2πrL)     (a) For surface segments S2 and S3 , E and n are perpendicular and therefore, ∫S2 E ∙ dA + ∫S3 E ∙ dA = 0    (b) Adding (a) and (b) results in ∫cs E ∙ dA = E(2πrL) = Q/ε0     (c) Since λ is the linear charge density of the line, the amount of charge entrapped inside the cylinder is Q=λL.  Substituting for Q, Equation (c) becomes: E(2πrL) = λL/ε0     or,    E =  λ/2πε0r    or,     E = (2kλ)/r .

Example 10: The electric field at a distance of 3.0cm from a long line of charge is 240N/C.  Calculate the linear charge density of the line.

Solution: Using the above result, E = (2kλ)/r, we have  240N/C = 2(9x109)(Nm2/C2)(λ)/(0.030m)  from which λ = 4.0x10-10 C/m

 Example 11: Determine the magnitude of the electric field at a distance r from an infinite plane uniformly charged with a surface charge density of  s C/m2. Solution: A suitable Gaussian surface for this case is a cylinder as shown usually referred to as "pillbox."  The infinite plane of charge has electric field equally on both sides of it. The plane itself is the plane of symmetry.  The bottom and top of the cylinder are adjusted to be equidistant from the plane; therefore, E1 and E2 have the same magnitudes though different directions.  With this choice n1 is parallel to E1 and n2 parallel to E2.  This makes the Gaussian integral easy for the top and bottom circular surfaces.  Since E3 is perpendicular to n3; therefore, the flux through the cylinder's side surface is zero that makes it even easier.  The Gaussian integral for this cylinder becomes: ∫cs E ∙ dA = ∫S1 E1 dA + ∫S2 E2 dA + ∫S3 E3 dA = E1A + E2A + 0 =  sA/ε0      or,     2EA = sA/ε0      or,    E = s/2ε0 This shows that the electric field around an infinite plane of charge does not vary with distance from the plane.  the reason is the lack of presence of r in the derived formula. Surfaces S1 and S2 can be selected at any distance from the plane of charges and still result in the same formula :E = s/2ε0.

Example 12:  The electric field at a distance of 60.0cm from a large sheet of charge is 3750N/C.   Calculate (a) the charge density, and (b) the charge distributed over every 125m2 of this large plane, and (c) the field strength at 120.0cm from the plane.

Solution:  Using the above formula, the surface charge density may be calculated as

(a) E = s/2ε0   or,  s = 2ε0E  or,  s = 2[8.85x10-12 C2/(Nm2)] (3750N/C)  =  66.4x10-9 C/m2  or,  s = 66.4 nC/m2.

(b) Q = sA = [66.4x10-9C/m2](125m2) = 8.3μC

(c) 3750N/C.

Gauss's Law Applied to Conductors:  In general, the direction of electric field at any point on a surface charge whether flat or curved is perpendicular to the surface at that point.  If the surface is flat and large (theoretically infinite), the field lines being perpendicular to the surface become parallel to each other and make the field uniform on each side of a sheet of infinite size.  Note that by a sheet of charges, we mean electric charges that are distributed over a non-conducting (insulator) surface.  The figures emphasizing this argument are shown in Part (a) below.

If the electric charges are distributed over an infinite flat conducting surface, they all gather on one side and double the electric field strength on that side to become E=s/ε0.  The field doubles compared to a non-conducting sheet because all charges gather on one side of the conductor.  Part (b) of the above figure shows the field lines for an infinite flat conductor as well as a curved (closed surface ) conductor. The proof of the formula in (b), E=s/ε0 , is left as a problem for students.  The proof is similar to Example 11 with field lines passing through the top circular area only.

Cavity in a Conductor:

 The figure on the right shows a conductor that has a cavity inside.  If a charge like +Q exists inside the cavity, its effect appears on the outer surface of the conductor as if +Q was distributed over it. The reason is that charge +Q attracts as much as -Q on the inner surface of the cavity (conductor) and repels as much as +Q onto the outer surface of the conductor. This can be verified by applying a Gaussian surface.  Since within the material of the conductor, the electric field is zero, if we assume a Gaussian surface around the cavity as shown, the net flux through this surface must also be zero.  This requires a charge distribution of -Q on the inner wall of the cavity.  Since the conductor is neutral to begin with,  its external surface accumulates +Q.

Problems:

1) A cube 20cm on each side has a 55-nC charge at its center and -25nC of charge placed on the outer edge of one of its sides.  The cube is then placed inside a sphere of radius 20cm.  Calculate the net flux out of the sphere.

2) A -6.0nC point charge is placed at one corner of a cube 10.0cm on each side.  Calculate the net flux through each side of it.

3) A -9.0nC point charge is placed on one face of a cube 12.0cm on each side.  Calculate the net flux through the cube.

4) Draw a circle and select a central angle on it that is 25 degrees.  Let the radius be 44cm.  Calculate the arc-length opposite to this angle.

 5) Draw a sphere and show a few area elements on it as shown.  It is actually helpful to repeat this drawing on paper.  Show that in polar coordinates (spherical), the area of each element is given by  dA = R2cosφdφdθ where (θ must vary from 0 to 2π) and (φ must vary from -π/2  to +π/2) in order for dA to cover the entire surface of the sphere.  Note that this drawing is only for the purpose of calculating the area of sphere.  Here, the lowercase letter φ is just the vertical angle, and θ the horizontal angle. They should not be confused with the uppercase letter Φ that is used for electric flux and the θ used in Equation (1) before. Also, as can be seen, the red side of the area element keeps getting smaller as we travel from the equator to the poles.  If we go from the equator to the top pole, φ varies from 0 to +π/2 and therefore cosφ varies from 1 to 0. That's why the red side of the area element varies from its maximum length on the equator to its minimum length of zero at the top pole.  The factor cosφ takes care of this variation.  This is also true for the lower half of the sphere.  We know from trigonometry that cos(φ) = cos(-φ).

6) Use the area element you derived in Problem 5 to calculate the area of sphere.  Note that for area calculation, R remains constant.

7) Use problem 5 as a basis in order to develop a volume element for the purpose of calculating the volume of  sphere.  Can radius be taken a constant in this calculation?

8) Show that the electric field around an infinite flat conductor is E = s/ε0 where s is the surface charge density.