1. Electric Field Computation Using Gauss’s Law
Lecture 5
Electric Field Computation
Using Gauss’s Law

2. LEARNING OUTCOMES After completing this lesson you will be able to:
use Gauss’ Law to calculate the electric field of a high-symmetry charge distribution.
use Gauss’ Law to calculate the electric field due to a long line of charge, with linear charge density .
use Gauss’ Law to calculate the electric field due to an infinite sheet of charge, with surface charge density .
use Gauss’ Law to calculate the electric field due to a sphere, with volume charge density v.

3. Definition of Flux
Flux is the action or process of flowing or flowing out through a surface, frequently also with time variation.
In electronics, flux is depicted as "lines" in a plane that contains or intersects electric charge poles or magnetic poles.
Figure 1 shows the geometric orientation of the lines of flux emanating from a point charge (the source of the flux).
(b)
(a)
Figure 1. (a) Electric flux lines emanating from a point charge (b) Variation of flux density with distance.

4. Notes
The total number of flux lines depends on the strength of the source and is constant with increasing distance.
A greater density of flux lines (lines per unit area) means a stronger field.
The density of flux lines is inversely proportional to the square of the distance from the source because the surface area of a sphere increases with the square of the radius. Thus the strength of the field is inversely proportional to the square of the distance from the source.
The terms "flux", "current", "flux density", "current density", can sometimes be used interchangeably and unambiguously.

5. Electric Flux
In electromagnetism, electric flux is the measure of “flow” of the electric field through a given area.
Consider the surface shown in the figure below. Let be defined as the area vector having a magnitude of the area of the surface, A, and pointing in the normal direction, . If the surface is placed in a uniform electric field that points in the same direction as , i.e., perpendicular to the surface A, the flux through the surface is
Figure 2 Electric field lines passing through a surface of area A.
The electric flux is a scalar quantity. Its unit is N.m2.C-1.

6. On the other hand, if the electric field makes an angle θ with (Figure 3), the electric flux becomes
Note that with the definition for the normal vector , the electric flux is positive if the electric field lines are leaving the surface, and negative if entering the surface.
Figure 3 Electric field lines passing through a surface of area A whose normal makes an angle θ with the field.

7. In general, a surface S can be curved and the electric field may vary over the surface. We shall be interested in the case where the surface is closed. A closed surface is a surface which completely encloses a volume. E
Consider a closed surface S in a non-uniform electric field (see figure). Over a very small area ds on this surface the electric field E over ds can be considered to be constant. The electric flux d through the area ds is, ds E  E ds E
where the direction of ds is normal to the surface outward and E and ds makes an angle  with each other.
Surface S

8. The total flux through the closed surface S is obtained by integrating the above equation over the surface.
Electric flux, 
The circle on the integral indicates that, the integration is to be taken over the closed surface.

9. GAUSSIAN SURFACE
A Gaussian surface is a surface to which the electric field is normal and over which equal to a constant value.

10. GAUSS’S LAW
The law relates the flux through any closed surface and the net charge enclosed within the surface. The law states that the total flux of the electric field E over any closed surface is equal to 1/εoεr times the net charge enclosed by the surface.
Gaussian surface S 
This closed imaginary surface is called Gaussian surface. Gauss’s law tells us that the flux  through a closed surface S depends only on the value of the net charge inside the surface and not on the location of the charges. Charges outside the surface will not contribute to the flux. Q

11. GAUSS’S LAW – INTEGRAL FORM
This statement can be expressed mathematically by the integral equation ds Q
Gaussian surface S
It is convenient to define a vector D such that
Then we can write Gauss’s law as or
The vector D is known as the electric flux density and has units of coulomb per metre.

12. Electric Field Due to an Infinitely Long Straight Charged Wireds r l E + P
Consider a uniformly charged wire of infinite length having a constant linear charge density  (charge per unit length). Let P be a point at a distance r from the wire (see figure on the right) and E be the electric field at point P. A cylinder of length l, radius r, enclosed at each end by plane caps normal to the axis is chosen as the Gaussian surface. Consider a very small area ds on the Gaussian surface.
Figure showing an infinitely long straight charged wire.

13. By symmetry, the magnitude of the electric field will be the same at all point on the curved surface of the cylinder and directed radially outward. E and ds are along the same direction.
The electric flux through the curved surface is
Since E and ds are always at right angles to each other anywhere on the Gaussian surface,  = 0o . Therefore,
Using the fact that the surface area of the curved path is 2rl, and the fact that electric flux through the end caps are zero, we thus obtain the total flux through the Gaussian surface

14. The net charge enclosed by the Gaussian surface is Q = l, where  is the linear charge density. Therefore, by Gauss’s law,
Therefore,
The direction of electric field E is radially outward, if the charge is positive and inward, if the line charge is negative. Thus, in vector form, the electric field E can be expressed as

15. Example
Find D in the region about the line charge 8nC/m lying along the z axis in free space.
Linear charge density,  = 8 nC/m
Gaussian surface S D
z axis

16. Solution The electric Field E V/m At r = 3 m and E = 47.9 V/m.
The value of the flux density at r = 3 m is
nC/m

17. EXAMPLE : HIGH VOLTAGE POWER LINE
A long, straight power-line wire has radius 1.0 cm and carries line charge density  = 2.6 C/m.
Assuming no other charges are present,
what’s the potential difference between the wire & the ground, 22 m below?
VAB
rB = 22 m A B
rA = 1 cm

18. EXAMPLE : HIGH VOLTAGE POWER LINE
Solution
VAB
rB = 22 m A B
rA = 1 cm

19. Example
Two infinite lines of charge, each carrying a charge density  C/m are parallel to the x-axis and located at x = 1 and x = -1. Determine E at an arbitrary point in free space along the y-axis.

20. Solution
The distance between either line charge and a point y on the y-axis is
we obtain
For line 1,
For line 2,
Therefore,
Using

21. Example: Electric field under high-voltage power line
A 765-kV rms, 3-phase, 60-Hz transmission line has conductors spaced 16 m. Their height is 12 m above ground as shown in the figure. Each conductor is a bundle of smaller conductors. It has an effective diameter of 0.6 m. What is the magnitude of the rms electric field at P, which is 2 m above ground?
16 m
12 m P
Conductor 1
0.6 m
Conductor 2
Conductor 3
10 m E1 E2 E3
Ground

22. The potential difference between conductors 1 and 2 is V = 765 kV.
Solution
The potential difference between conductors 1 and 2 is V = 765 kV.
Now
15.7 m
Therefore, charge on conductor is
Conductor 1
Conductor 2

23. Therefore, the field at P from conductor 3 is
V.m-1
Field at P from conductor 2 is
V.m-1
V.m-1
Field at P from conductor 2 is
V.m-1
V.m-1

24. E (total max.)
Ans.

25. ELECTRIC FIELD DUE TO AN INFINITE CHARGED PLANE SHEET
Consider am infinite plane sheet of charge with surface charge density . Let P be a point at a distance r from the sheet (see figure below) and E be the electric field at P. Consider a Gaussian surface in the form of cylinder of cross-sectional area A and length 2r perpendicular to the sheet of charge. E ds P r A + + + + + + + r P’ ds
Infinite plane sheet E

26. By symmetry, the electric field is at right angles to the end caps and away from the plane. Its magnitude is the same at P and at the other cap at P’.
Therefore, the total flux through the closed surface is given by
(because  = 0, cos  = 1)
If  is the charge per unit area in the plane sheet, then the net positive charge Q within the Gaussian surface is, Q = A.
Using Gauss’s law,

27. Therefore,
Notes:
The electric field due to an infinitely large sheet of charge is uniform in space, and
is not affected by the distance from the plane containing the sheet of charge.

28. In vector notation, we have
Figure showing the electric field profile of an infinitely large sheet of charge.

29. Worked Example
Find the electric field at P(1,5,2) in free space due to an infinite sheet of charge with over the plane z = -1.
Solution
The plane z = -1 is parallel to x-y plane and the normal of the plane is n = k, as point P is above the plane. ^

30. Solution (continued)
At all the points above z = -1 plane the electric field E is constant along direction. Therefore,

31. Worked Example
Charge lies in y = -5 m plane in the form of an infinite square sheet with a uniform charge density of  = 29 nC/m2. Determine the electric field E at all the points.
Solution
The plane y = - 5 m is parallel to the x-z plane as shown in the following figure.

32. For y > -5, the E component will be along as normal direction to the plane y = -5 m is Therefore, and

33. For y < -5, the E component will be acting along direction, with same magnitude. Therefore,
At any point to the left or right of the plane, is constant and acts normal to the plane.

34. ELECTRIC FIELD DUE TO TWO PARALLEL CHARGED SHEETS
Consider two plane parallel infinite sheets with equal and opposite charge densities + and - as shown in the figure. The magnitude of electric field on either side of a plane sheet is P2 E1 E1 + + + + + + + + + +
+ E1 E1 P1 E2 E2 _ _ _ _ _ _ _ _ _ _
and acts perpendicular to the sheet, directed outward (if the charge is positive) or inward (if the charge is negative).
- E2 E2
Electric field between two (large) oppositely charged parallel plates.

35. ELECTRIC FIELD DUE TO TWO PARALLEL CHARGED SHEETS
(i) When the point P1 is in the sheets, the field due to two sheets will be equal in magnitude and in the same direction. The resultant field at P1 is, P2 E1 E1 + + + + +
(downwards) + + + + +
+ E1 E1
At a point P2 outside the sheets, the electric field will be equal in magnitude and opposite in direction. The resultant field at P2 is, P1 E2 E2 _ _ _ _ _ _ _ _ _ _
- E2 E2

36. POTENTIAL GRADIENT ACROSS TWO METAL PLATES
Consider a battery, whose terminal voltage is V, connected to two parallel plates, as shown on the right. Initially, positive charges will flow out from the positive terminal of the battery and accumulate at the upper plate A, and negative charges will flow from the negative terminal of the battery and accumulate at the lower plate. An electric field is developed across the two plates, from A to B. E V VA VB

37. A potential is developed across A-B and whose magnitude will increase until VAB equals V of the battery whereby charges will stop flowing. Because the medium (air) between the two plates is homogenous we have every reason to believe that the electric field will be uniform from A to B. What we want to do here is to find an expression that relates the electric field E between the metal plates to VAB and d. VA
Potential E
Electric Field V E d
Distance VB

38. Recall that E = - dV/dx. Thus, we can write
Integrating both sides of the equation, we obtain the expression
Potential VB VA
Given that E is constant we thus obtain, upon integration, the expression
Distance
x = d
Thus, x

39. Example
The plates of a cathode ray tube are parallel and are 5 mm apart. The potential difference between them is 12 kV. Find the force on an electron passing between these plates. What is the acceleration of the electron? e- E
VAB

40. Solution
Charged parallel plates produce a constant electric field.
Thus,
The direction of E is away from the + plate. That is, E always points in the direction of decreasing electrical potential. Since we know E, we can determine the force on the electron.
Felec = q E = (-1.6 x 10‑19)(2.4 x 106)
= x 10‑13 N.

41. The direction of the force is opposite to the electric field direction since q is negative. Thus, while E points from the + plate towards the ‑ plate, the electron is attracted to the + plate.
a = F/m
= (3.84 x 10‑13)/(9.11 x 10-31)
= x 1017 m/sec2

42. Electric Field Around a Spherical Shell
Consider a thin spherical shell of radius a has a charge Q+ evenly distributed over its surface. We want to find the electric field both inside and outside the shell.
The charge distribution is spherically symmetric, with a surface charge density
where A = 4πa2 is the surface area of the sphere.
Figure Electric field for uniform spherical shell of charge

43. We treat the regions r  a and r  a separately.
Case 1: r  a
We choose our Gaussian surface to be a sphere of radius r≤ a , as shown in the figure.
The charge enclosed by the Gaussian surface is qenc = 0 since all the charge is located on the surface of the shell.
Thus, from Gauss’s law,
Figure Gaussian surface for uniformly charged spherical shell for r  a.
we conclude
E = 0 for r < a

44. Case 2: r a
In this case, the Gaussian surface is a sphere of radius, as shown in the figure. Since the radius of the “Gaussian sphere” is greater than the radius of the spherical shell, all the charge is enclosed:
Since the flux through the Gaussian surface is
by applying Gauss’s law, we obtain
Figure Gaussian surface for uniformly charged spherical shell for r  a.

45. Note that the field outside the sphere is the same as if all the charges were concentrated at the center of the sphere. The qualitative behavior of E as a function of r is plotted in the figure
Figure Electric field as a function of r due to a uniformly charged spherical shell.
As in the case of a non-conducting charged plane, we again see a discontinuity in E as we cross the boundary at r = a.

46. Worked Example
The dome of a van der Graaf generator is charged. The dome has a diameter of 30 cm and its charge is 4 µC. A 5 µC point charge is placed 7 cm from the surface of the dome.
Calculate:
(a) the electric field strength at a point 7 cm from the dome
(b) the electrostatic force exerted on the 5 µC point charge.

47. Solution
(a)
N/C
(b)

48. Worked Example
If the magnitude of an electric field in air is as great as 3.0 × 106 N/C, the air becomes ionized and begins to conduct electricity. This phenomenon is called dielectric breakdown. A charge of 18 μC is to be placed on a conducting sphere. What is the minimum radius of a sphere that can hold this charge without breakdown?

49. Solution
From Gauss’s law we know that the electric field at the surface of the charged sphere is given by E = kQ/r2 where Q is the charge on the sphere and R is its radius. The minimum radius for dielectric breakdown corresponds to the maximum electric field at the surface of the sphere.
Use Gauss’s law to express the electric field at the surface of the charged sphere:
Express the relationship between Emax and Rmin for dielectric breakdown:

50. Therefore,
Substitute numerical values and evaluate Rmin:
Picture showing a van der Graaf generator discharging when the electric field at the sphere’s surface exceeds the air breakdown voltage of 3 MV/m.

51. ELECTRIC POTENTIAL DUE TO A SPHERICAL SHELL
Consider a metallic spherical shell of radius a and charge Q, as shown in the figure below. Here we want to find the electric potential everywhere around the charged sphere.
Recall that the electric field for a spherical shell of is given by
The electric potential difference between two points A and B outside the sphere may be calculated by using the following equation:

52. For r > a, we have
where we have chosen V() = 0 as our reference point. On the other hand, for r < a, the potential becomes

53. A plot of the electric potential is shown in the figure below
A plot of the electric potential is shown in the figure below. Note that the potential V is constant inside a conductor.
Figure showing how the electric potential varies as a function of r for a spherical conducting shell.

54. Worked Example
A particular van der Graaf generator has a metallic spherical dome of radius R = 2.30 m and develops a charge Q = 640 C. Considering the dome as a single isolated sphere, find
the potential at its surface,
the work needed to bring a proton from infinity to the sphere’s surface,
the potential difference between the sphere’s surface and a point 2R from its centre.

55. Solution
(a)
(b) V
(c)
Radial distance, r R 2R

56. ELECTRIC FIELD INSIDE AND OUTSIDE A COAXIAL CABLE
Consider a cylindrical capacitor formed by two concentric cylinders (see figure below), the inner conductor is typically a solid cylinder, the outer conductor is a thin cylindrical shell. rA r rB l

57. i. Region r < rA, (inside the solid cylinder)
We can identify three distinctly different regions for the application of Gauss’s Law:
i. Region r < rA, (inside the solid cylinder)
Region rA < r < rB ( the space separating the two concentric
cylinders), and
iii. Region r > rB ( the region outside the outer cylinder).
1. Region r < rA rA S1 S3

58. Application of Gauss’s law to the inside of the solid cylinder leads us the equation
where E is the electric field inside the cylinder, Qenclosed is the total charge enclosed by a length L of the cylinder, S1, S2 are the endcap surfaces, and S3 is the curved surface.
Now, the electric field is parallel to the endcap surfaces and it is everywhere constant and normal on the curved surface. Therefore, for the two endcap surfaces, we have

59. and for the curved surface we have
Therefore, the Gauss’s law expression above reduces to the form or

60. For a dc or a single-phase circuit, the solid cylinder will be carrying the outgoing current +Q and the hollow cylinder will be carrying the return current, -Q. Thus, we can write, for the solid cylinder
where  is the linear charge density and

61. For the two endcap surfaces, we have
2 Region rA < r < rB
Application of Gauss’s law to Gaussian surface between rA and rB leads us the equation rB rA S4 S5 S6
For the two endcap surfaces, we have
and for the curved surface we have

62. Therefore, the Gauss’s law expression above reduces to the form.
Putting
into the above expression, we hence obtain
for
where we have used the relation

63. 3 Region r > rB rB S7 S9 S8
Application of Gauss’s law to Gaussian surface with r = rB leads us the equation
where E is the electric field inside the cylinder, and Qenclosed is the total charge enclosed by the Gaussian surface.

64. Now for r = rB, the Gaussian surface encloses both the sending current conductor and the returning current conductor. Hence
giving
Hence E = 0 for r  rB.

65. Worked Example
The figure below is a section of a coaxial electrical cable of radius R1 = 1.30 mm and length L = m inside a thin-walled coaxial conducting cylindrical shell of radius R2 = 13 mm and the same length L. The net charge on the inner conductor is Q1 = 3.40 x C; that on the shell is Q2 = x10-12 C. What are the
(a) magnitude E, and
(b) direction (radially inward or outward) of the electric field at radial
distance (i) r = 1.3 mm, (ii) r = 5 mm, and (iii) r = 12.9 mm.

66. (a) At r = 1.3 mm (at the surface of the inner conductor) we have
Solution
(a) At r = 1.3 mm (at the surface of the inner conductor) we have
V/m
(b) At r = 5 mm (in the region between the two conductors) we have
V/m
At r = 12.9 mm (in the region between the two conductors) we
have
V/m

67. END