Physics: Electric Flux and Electric Flux Density

ELECTRIC FLUX
&
ELECTRIC FLUX DENSITY

In electromagnetism, electric flux is the rate
of flow of the electric field through a given
area.
The total number of lines of force passing
through the unit area of a surface.
It is scalar quantity.
Its unit is N.m²/C or V.m
ELECTRIC FLUX

Mathematically the electric flux is defined as:
"The dot product of electric field
intensity (E) and the vector area (∆A) is called
electric flux.“
 Where θ is angle between E and ∆A
MATHEMATICALLY
A
E
e 

 .

cos
. A
E
e 



PROPERTIES
Maximum Flux
 If the surface is placed perpendicular to the electric
field then maximum electric lines of force will pass
through the surface. Consequently maximum
electric flux will pass through the surface
 line are perpendicular than θ=0

Zero Flux
 If the surface is placed parallel to the electric field
then no electric lines of force will pass through the
surface. Consequently no electric flux will pass
through the surface.
CONT...

Consider a small positive point charge +q placed
at the Centre of a closed sphere of radius "r".
The relation is not applicable in this situation
because the direction of electric intensity varies
point to point over the surface of sphere.
In order to overcome this problem the sphere is
divided into a number of small and equal pieces
each of area ∆A.
The direction of electric field in each segment of
sphere is the same i.e. outward normal.
ELECTRIC FLUX THROUGH A SPHERE

Now we will determine the flux through each
segment.
Electric flux through the first segment:
But θ=0
CONT...
A
E
A
E
A
E
A
E












.
)
1
cos(
.
0
cos
.
cos
.
1
1
1
1 

Electric flux through the second segment:
Similarly, Electric flux through other segments:
CONT…
A
E 

 .
2
A
E
A
E
n 





.
.
.
.
.
3

Being a scalar quantity, the total flux through the
sphere will be equal to the algebraic sum of all
these flux i.e.
But
CONT…
i
n
i
i
i
n
i
i
n
i
i
A
E
A
E
i


















1
1
1
)
(
)
(
2
.
4
1
r
q
E




i
n
i
i
A
r
q






1
2
)
(
4
1



As (Area of sphere)
Then equation is
This expression shows that the total flux through
the sphere is 1/eO times the charge enclosed (q)
in the sphere.
2
1
4
)
( r
A
i
n
i
i






2
2
4
4
1
r
r
q






q


CONT…

The Electric Flux Density is called Electric
Displacement denoted by D, is a vector field that
appears in Maxwell's equations.
It is equal to the electric field strength
multiplied by the permittivity of the material
through which the electric field extends.
 It is measured in coulombs per square meter.
The Electric Flux Density (D) is related to the
Electric Field (E) by:
------Equation (1)
ELECTRIC FLUX DENSITY

In Equation [1], ε is the permittivity of the
medium (material) where we are measuring the
fields.
If you recall that the Electric Field is equal to the
force per unit charge (at a distance R from a
charge of value q1 [C])
Equation------ (2)
Then the Electric Flux Density is:
Equation------ (3)
CONT…

From Equation [3], the Electric Flux Density is
very similar to the Electric Field, but does not
depend on the material in which we are
measuring (that is, it does not depend on the
permittivity .
Note that the D field is a vector field, which
means that at every point in space it has a
magnitude and direction.
The Electric Flux Density has units of Coulombs
per meter squared [C/m²].
CONT…

Definitions
• Flux—The rate of flow through an area or volume. It can
also be viewed as the product of an area and the vector
field across the area
• Electric Flux—The rate of flow of an electric field through
an area or volume—represented by the number of E field
lines penetrating a surface

Charge and Electric Flux
Previously, we answered the question – how do we find
E-field at any point in space if we know charge distribution?
Now we will answer the opposite question – if we know E-field
distribution in space, what can we say about charge distribution?

Electric flux
Electric flux is associated with the flow of electric field through a surface
For an enclosed charge, there is a connection
between the amount of charge
and electric field flux.
2
2
1
~
~
E
r
S r
E S const


Calculating Electric Flux
dV
A
dt


Amount of fluid passing through
the rectangle of area A
cos
dV
A
dt
 

dV
A
dt

 


Calculating Electric Flux
• The flux for an electric field is
• For an arbitrary surface and nonuniform E field
• Where the area vector is a vector with magnitude of the
area A and direction normal to the plane of A
A
E






 
r
E d
r
A


Flux of a Uniform Electric Field
cos
E E A EA 
 
   A A n
 
 n

- unit vector in the direction of normal to the surface
Flux of a Non-Uniform Electric Field
E
S
E d A
 
  
E – non-uniform and
A- not flat

Few examples on calculating the electric flux
3
2 10 [ / ]
E N C
 
Find electric flux

Definitions
• Symmetry—The balanced structure of an object, the halves
of which are alike
• Closed surface—A surface that divides space into an inside
and outside region, so one can’t move from one region to
another without crossing the surface
• Gaussian surface—A hypothetical closed surface that has
the same symmetry as the problem we are working on—
note this is not a real surface it is just an mathematical one

Gauss’ Law
· Gauss’ Law depends on the enclosed charge only
1. If there is a positive net flux there is a net positive charge
enclosed
2. If there is a negative net flux there is a net negative charge
enclosed
3. If there is a zero net flux there is no net charge enclosed
• Gauss’ Law works in cases of symmetry
o
enc
q
A
d
E


 





Types of Symmetry
• Cylindrical symmetry—example a can
• Spherical symmetry—example a ball
• Rectangular symmetry—example a box—rarely used

Steps to Applying Gauss’ Law
To find the E field produced by a charge distribution at a point of
distance r from the center
1. Decide which type of symmetry best complements the
problem
2. Draw a Gaussian surface (mathematical not real)
reflecting the symmetry you chose around the charge
distribution at a distance of r from the center
3. Using Gauss’s law obtain the magnitude of E

Gauss’s Law
0
i
E
q
E d A

 
  



Applications of the Gauss’s Law
If no charge is enclosed within Gaussian surface – flux is zero!
Electric flux is proportional to the algebraic number of lines leaving
the surface, outgoing lines have positive sign, incoming - negative
Remember – electric field lines must start and must end on charges!

Examples of certain field configurations
Remember, Gauss’s law is equivalent to Coulomb’s law
However, you can employ it for certain symmetries to solve the reverse problem
– find charge configuration from known E-field distribution.
Field within the conductor – zero
(free charges screen the external field)
Any excess charge resides on the
surface
0
S
E d A
 



Field of a charged conducting sphere

Field of a thin, uniformly charged conducting wire
Field outside the wire can only point
radially outward, and, therefore, may
only depend on the distance from the wire
0
Q
E d A

 


0
2
E
r

 

- linear density of charge

Field of the uniformly charged sphere
r
E
0
3


Uniform charge within a sphere of radius r
3
' r
q Q
a
 
  
 
Q - total charge
Q
V
  - volume density of charge
Field of the infinitely large conducting plate
s- uniform surface charge density
Q
A
s 
0
2
E
s



Charged Isolated Conductors
• In a charged isolated conductor all the charge moves to the
surface
• The E field inside a conductor must be 0 otherwise a
current would be set up
• The charges do not necessarily distribute themselves
uniformly, they distribute themselves so the net force on
each other is 0.
• This means the surface charge density varies over a
nonspherical conductor

Charged Isolated Conductors cont
• On a conducting surface
• If there were a cavity in the isolated conductor, no charges
would be on the surface of the cavity, they would stay on
the surface of the conductor
o
E

s


Charge on solid conductor resides on surface.
Charge in cavity makes a equal but opposite charge reside on
inner surface of conductor.

Properties of a Conductor in Electrostatic Equilibrium
1. The E field is zero everywhere inside the conductor
2. If an isolated conductor carries a charge, the charge resides on its
surface
3. The electric field just outside a charged conductor is
perpendicular to the surface and has the magnitude given above
4. On an irregularly shaped conductor, the surface charge density is
greatest at locations where the radius of curvature of the surface
is smallest

Charges on Conductors
Field within conductor
E=0

Experimental Testing of the Gauss’s Law

1
Gauss’s Law
Basic Concepts
Electric Flux
Gauss’s Law
Applications of Gauss’s Law
Conductors in Equilibrium

2
Electric Flux
The electric flux, FE, through a surface is defined as the scalar product of E and
A, FE = EA. A is a vector perpendicular to the surface with a magnitude equal
to the surface area. This is true for a uniform electric field.
A = A cos so FE = EA = EA cos
FE = EA

Physics 24-Winter 2003-L03 3
Electric Flux Continued
What about the case when the electric field is not uniform and
the surface is not flat?
Then we divide the surface into small elements and add
the flux through each.
E i i
i
E d A E d A
F    
 

Electric Flux Continued
Finally, what about a closed
surface?
A closed surface is one that
completely encloses a volume.
This is handled as before, but we
need to resolve ambiguity about
direction of A. It is defined to
point outward so flux exiting the
enclosed volume is positive and
flux entering is negative.
F  

E E dA
r
r
Ñ

Worked Example 1
Compute the electric flux through a cylinder with an axis parallel to the electric
field direction.
E
The flux through the curved surface is zero since E is perpendicular to dA
there. For the ends, the surfaces are perpendicular to E, and E and A are
parallel. Thus the flux through the left end (into the cylinder) is –EA, while
the flux through right end (out of the cylinder) is +EA. Hence the net flux
through the cylinder is zero.
A

Gauss’s Law
Gauss’s Law relates the electric flux through a closed surface with
the charge Qin inside that surface.
0
F   

r
r
Ñ in
E
Q
E dA

This is a useful tool for simply determining the electric field, but
only for certain situations where the charge distribution is either
rather simple or possesses a high degree of symmetry.

Select a Gaussian surface with symmetry that
matches the charge distribution
Draw the Gaussian surface so that the electric field
is either constant or zero at all points on the
Gaussian surface
Use symmetry to determine the direction of E on
the Gaussian surface
Evaluate the surface integral (electric flux)
Determine the charge inside the Gaussian surface
Solve for E
Problem Solving Strategies for Gauss’s
Law

Worked Example 2
Starting with Gauss’s law, calculate the electric field
due to an isolated point charge q.
q
E
r dA
We choose a Gaussian surface that is a sphere of
radius r centered on the point charge. I have
chosen the charge to be positive so the field is
radial outward by symmetry and therefore
everywhere perpendicular to the Gaussian surface.
 
r
r
E dA EdA Gauss’s law then gives:
0 0
   
 
r
r
Ñ Ñ in
Q q
E dA E dA
 
Symmetry tells us that the field is
constant on the Gaussian surface.
 
2
2 2
0 0
1
4 so
4
    
  e
q q q
E dA E dA E r E k
r r

 
Ñ Ñ

Worked Example 3
An insulating sphere of radius a has a uniform charge density ρ and a total
positive charge Q. Calculate the electric field outside the sphere.
a
Since the charge distribution is spherically
symmetric we select a spherical Gaussian surface
of radius r > a centered on the charged sphere.
Since the charged sphere has a positive charge, the
field will be directed radially outward. On the
Gaussian sphere E is always parallel to dA, and is
constant.
Q
r
E
dA
 
2
Left side: 4
   
  
r
r
Ñ Ñ Ñ
E dA E dA E dA E r


0 0
Right side: in
Q Q
 


 
2
2 2
0 0
1
4 or
4
e
Q Q Q
E r E k
r r

 
  


Worked Example 3 cont’d
a
Q
Find the electric field at a point inside the sphere.
Now we select a spherical Gaussian surface with
radius r < a. Again the symmetry of the charge
distribution allows us to simply evaluate the left side
of Gauss’s law just as before.
r
The charge inside the Gaussian sphere is no longer Q. If we call the
Gaussian sphere volume V’then
 
2
Left side: 4
   
  
E dA E dA E dA E r

r
r
Ñ Ñ Ñ

 
3
2
0 0
4
4
3
in
Q r
E r


 
 


3
4
Right side:
3
in
Q V r
  

 
 
3
3 3
2
3
0 0
0
4 1
but so
4
3 4
3 4
3
e
r Q Q Q
E r E r k r
a a
r a
 

 
  
    


2
3
We found for ,
and for ,
e
e
Q
r a E k
r
k Q
r a E r
a
 
 
a
Q
Let’s plot this:
E
r
a

Conductors in Electrostatic
Equilibrium
The electric field is zero everywhere inside the
conductor
Any net charge resides on the conductor’s surface
The electric field just outside a charged conductor
is perpendicular to the conductor’s surface
By electrostatic equilibrium we mean a situation where
there is no net motion of charge within the conductor

Conductors in Electrostatic
Equilibrium
Why is this so?
If there was a field in the conductor the charges
would accelerate under the action of the field.
The electric field is zero everywhere inside the conductor
++++++++++++
---------------------
Ein
E E
The charges in the conductor
move creating an internal electric
field that cancels the applied field
on the inside of the conductor

Worked Example 4
Any net charge on an isolated conductor must reside on its surface and the
electric field just outside a charged conductor is perpendicular to its surface
(and has magnitude σ/ε0). Use Gauss’s law to show this.
For an arbitrarily shaped conductor we can
draw a Gaussian surface inside the conductor.
Since we have shown that the electric field
inside an isolated conductor is zero, the field
at every point on the Gaussian surface must be
zero.
From Gauss’s law we then conclude that the net
charge inside the Gaussian surface is zero. Since
the surface can be made arbitrarily close to the
surface of the conductor, any net charge must
reside on the conductor’s surface.
0
 

r
r
Ñ in
Q
E dA


We can also use Gauss’s law to determine the electric field just outside the
surface of a charged conductor. Assume the surface charge density is σ.
Since the field inside the conductor is zero
there is no flux through the face of the
cylinder inside the conductor. If E had a
component along the surface of the
conductor then the free charges would
move under the action of the field creating
surface currents. Thus E is perpendicular
to the conductor’s surface, and the flux
through the cylindrical surface must be
zero. Consequently the net flux through
the cylinder is EA and Gauss’s law gives:
0 0 0
or
in
E
Q A
EA E
 
  
F    

Worked Example 5
A conducting spherical shell of inner radius a and outer radius b with a net
charge -Q is centered on point charge +2Q. Use Gauss’s law to find the
electric field everywhere, and to determine the charge distribution on the
spherical shell.
a
b
-Q First find the field for 0 < r < a
This is the same as Ex. 2 and is the field due to a
point charge with charge +2Q.
2
2
e
Q
E k
r

Now find the field for a < r < b
The field must be zero inside a conductor in equilibrium. Thus from Gauss’s law
Qin is zero. There is a + 2Q from the point charge so we must have Qa = -2Q on
the inner surface of the spherical shell. Since the net charge on the shell is -Q we
can get the charge on the outer surface from Qnet = Qa + Qb.
Qb= Qnet - Qa = -Q - (-2Q) = + Q.
+2Q

a
b
-Q
+2Q
Find the field for r > b
From the symmetry of the problem, the field in
this region is radial and everywhere perpendicular
to the spherical Gaussian surface. Furthermore,
the field has the same value at every point on the
Gaussian surface so the solution then proceeds
exactly as in Ex. 2, but Qin=2Q-Q.
 
2
4
   
  
r
r
Ñ Ñ Ñ
E dA E dA E dA E r

Gauss’s law now gives:
 
2
2 2
0 0 0 0
2 1
4 or
4
in
e
Q Q Q Q Q Q
E r E k
r r

   

    

Summary
Two methods for calculating electric field
Coulomb’s Law
Gauss’s Law
Gauss’s Law: Easy, elegant method for symmetric
charge distributions
Coulomb’s Law: Other cases
Gauss’s Law and Coulomb’s Law are equivalent
for electric fields produced by static charges

Electric Flux Density, Gauss’s Law,
and Divergence

Electric Flux Density
• Faraday’s Experiment

Electric Flux Density, D
• Units: C/m2
• Magnitude: Number of flux lines (coulombs)
crossing a surface normal to the lines divided by
the surface area.
• Direction: Direction of flux lines (same direction
as E).
• For a point charge:
• For a general charge distribution,

Gauss’s Law
• “The electric flux passing through any
closed surface is equal to the total charge
enclosed by that surface.”

• The integration is performed over a closed
surface, i.e. gaussian surface.

• We can check Gauss’s law with a point
charge example.

Symmetrical Charge Distributions
• Gauss’s law is useful under two
conditions.
1. DS is everywhere either normal or
tangential to the closed surface, so that
DS
.dS becomes either DS dS or zero,
respectively.
2. On that portion of the closed surface for
which DS
.dS is not zero, DS = constant.

Gauss’s law simplifies the task of finding D near an
infinite line charge.

Differential Volume Element
• If we take a small enough closed surface,
then D is almost constant over the surface.

Divergence
Divergence is the outflow of flux from a small
closed surface area (per unit volume) as
volume shrinks to zero.

-Water leaving a bathtub
-Closed surface (water itself) is essentially incompressible
-Net outflow is zero
-Air leaving a punctured tire
-Divergence is positive, as closed surface (tire) exhibits net
outflow

Mathematical definition of divergence
div D
 
D x
x
D y
y

D z
z







- Cartesian
div D
 
0
v
S
D
v




d
lim

Surface integral as the volume element (v) approaches zero
D is the vector flux density

Cylindrical
Spherical
div D
  1



  D

 
1

D



Dz
z

div D
  1
r
2
 D r r
2





r

1
r sin 
 

 D  sin 
 

 



1
r sin 
 

D 



Divergence in Other Coordinate Systems

Maxwell’s First Equation
S
.
S
A



d Q
S
.
S
A



d
v
Q
v
Gauss’ Law…
…per unit volume
Volume shrinks to zero 0
v
S
.
S
A



d
v
lim
 0
v
Q
v
lim

Electric flux per unit volume is equal to the volume charge density

Maxwell’s First Equation
div D
   v
0
v
S
.
S
A



d
v
lim
 0
v
Q
v
lim

Sometimes called the point form of Gauss’ Law
Enclosed surface is reduced to a single point

 and the Divergence Theorem
  del operator

 ax
 
x
 ay
 
y

 az
 
z


What is del?

’s Relationship to Divergence
div D
  
V D

True for all coordinate systems

Other  Relationships
Gradient – results from  operating on a function
Represents direction of greatest change

Curl – cross product of  and
Relates to work in a field
If curl is zero, so is work

Examination of  and flux
Cube defined by 1 < x,y,z < 1.2
D 2 x
2
 y
 a x
 3 x
2
 y
2
 a y


Q
S
.
S
D



d
vol
.
v
 v



d 
Calculation of total flux
 total  left  right
  front
  back

x1 1
 x2 1.2

y1 1
 y2 1.2

z1 1
 z2 1.2

x1
z1
z2
z
y1
y2
y
2
 x1
2
 y




d



d
 y1
z1
z2
z
x1
x2
x
3
 x
2
 y1
2




d



d

x2
z1
z2
z
y1
y2
y
2 x2
2
 y




d



d
 y2
z1
z2
z
x1
x2
x
3 x
2
 y2
2




d



d


total x1 x2
 y1
 y2


total 0.103

Evaluation of  at center of cube
V D

div D
 
x
2 x
2
 y

 
d
d y
3 x
2
 y
2

 
d
d

div D
  4 x
 y
 6 x
2
 y


divD 4 1.1
( )
 1.1
( )
 6 1.1
( )
2
 1.1
( )



divD 12.826


Chapter 4
Energy and Potential

4.1 Energy to move a point
charge through a Field
• Force on Q due to an electric field
• Differential work done by an external
source moving Q
• Work required to move a charge a finite
distance
F E QE
dW QE
 dL


4.2 Line Integral
• Work expression without using vectors
EL is the component of E in the dL direction
• Uniform electric field density
W Q

initial
final
L
E L



d

W QE
 LBA


Example
E x y

( )
y
x
2








 Q 2
 A
.8
.6
1








 B
1
0
1








 Path: x
2
y
2
 1 z 1
Calculate the work to cary the charge from point B to point A.
W Q

B0
A0
x
E x y

( )
0




d
 Q
B1
A1
y
E x y

( )
1




d

 Q
B2
A2
z
E x y

( )
2




d


Plug path in for x and y in E(x,y)
W Q

B0
A0
x
E 0 1 x
2


 0



d
 Q
B1
A1
y
E 1 y
2
 0

 1



d

 Q
B2
A2
z
E 0 0

( )
2




d


 W 0.96



Example
• Same amount of work with a different path
• Line integrals are path independent
E x y

( )
y
x
2








 Q 2
 A
.8
.6
1








 B
1
0
1








 Path: y 3
 x 1

( ) z 1
(straight line)
Calculate the work to cary the charge from point B to point A.
W Q

B0
A0
x
E x y

( )
0




d
 Q
B1
A1
y
E x y

( )
1




d

 Q
B2
A2
z
E x y

( )
2




d


Plug path in for x and y in E(x,y)
W Q

B0
A0
x
E 0 3
 x 1

( )

[ ]
0




d
 Q
B1
A1
y
E
y

3
1
 0







1




d

 Q
B2
A2
z
E 0 0

( )
2




d


 W 0.96



4.3 Potential Difference
• Potential Difference
• Using radial distances from the point charge

4.3 Potential
• Measure potential difference between a
point and something which has zero
potential “ground”
VAB VA VB


Example – D4.4
E x y
 z

( )
6x
2
6y
4











a) Find Vmn
M
2
6
1









 N
3

3

2








 VMN
N0
M0
x
6x
2



d

N1
M1
y
6y



d

N2
M2
z
4



d

 VMN 139


b) Find Vm if V=0 at Q(4,-2,-35)
Q
4
2

35









 VM
Q0
M0
x
6x
2



d

Q1
M1
y
6y



d

Q2
M2
z
4



d

 VM 120


c) Find Vn if V=2 at P(1, 2, -4)
P
1
2
4









 VN
P0
N0
x
6x
2



d

P1
N1
y
6y



d

P2
N2
z
4



d
 2

 VN 19


4.4 Potential Field of a Point
Charge
• Let V=0 at infinity
• Equipotential surface:
– A surface composed of all points having the
same potential

Example – D4.5
Q 15 10
9


 P1
2

3
1









 0 8.85 10
12



Q is located at the origin
a) Find V1 if V=0 at (6,5,4)
P0
6
5
4








 V1
Q
40
1
P1
1
P0







 V1 20.677

b) Find V1 if V=0 at infinity
V1
Q
40
1
P1
 V1 36.047

c) Find V1 if V=5 at (2,0,4)
P5
2
0
4








 V1
Q
40
1
P1
1
P5







5

 V1 10.888


Potential field of single point
charge
Q1
A
|r - r1|
Move A
from infinity
V r
( )
Q1
4 
  0
 r r1



Potential due to two charges
Q1
A
|r - r1|
Q2
|r - r2|
Move A
from infinity
V r
( )
Q1
4 
  0
 r r1


Q2
4 
  0
 r r2




Potential due to n point charges
Continue adding charges
V r
( )
Q1
4 
  0
 r r1


Q2
4 
  0
 r r2


 ....

Qn
4 
  0
 r r n



V r
( )
1
n
m
Qm
4 
  0
 r r m





Potential as point charges become
infinite
Volume of charge
Line of charge
Surface of charge
V r
( ) v prime
 v r prime
 
4 
  0
 r r prime







d
V r
( ) L prime
 L r prime
 
4 
  0
 r r prime







d
V r
( ) S prime
 S r prime
 
4 
  0
 r r prime







d

Example
Find V on the z
axis for a uniform
line charge L in
the form of a ring
V r
( ) L prime
 L r prime
 
4 
  0
 r r prime







d

Conservative field
No work is done (energy is conserved) around a
closed path
KVL is an application of this

4.6
Potential gradient Relationship between
potential and electric field intensity
Two characteristics of relationship:
1. The magnitude of the electric field intensity is given
by the maximum value of the rate of change of potential
with distance
2. This maximum value is obtained when the direction
of E is opposite to the direction in which the potential is
increasing the most rapidly
V = -



d
E dL


Gradient
• The gradient of a scalar is a vector
• The gradient shows the maximum space rate of change
of a scalar quantity and the direction in which the
maximum occurs
• The operation on V by which -E is obtained
E = - grad V = - V

Gradients in different coordinate
systems
The following equations are found on page 104 and
inside the back cover of the text:
Cartesian
Cylindrical
Spherical
gradV
V
x
a x

V
y
a y


V
z
a z


gradV
V

a 

1

V

 a 


V
z
a z


gradV
V
r
a r

1
r
V

 a 


1
r sin 
 

V

 a 



Example 4.3
Given the potential field, V = 2x2y - 5z, and a point P(-4,
3, 6), find the following: potential V, electric field intensity
E
potential VP = 2(-4)2(3) - 5(6) = 66 V
electric field intensity - use gradient operation
E = -4xyax - 2x2ay + 5az
EP = 48ax - 32ay + 5az

Dipole
The name given to two point charges of equal magnitude
and opposite sign, separated by a distance which is small
compared to the distance to the point P, at which we want
to know the electric and potential fields

Physics: Electric Flux and Electric Flux Density

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