1. ELECTROSTATIC POTENTIAL
Lecture 3
ELECTROSTATIC
POTENTIAL

2. After completing this lesson you will be able to:
LEARNING OUTCOMES
After completing this lesson you will be able to:
Calculate work done by an electric field
Define electrostatic potential
Calculate potential difference between two points in an electric field
Define absolute potential

3. WORK DONE BY AN ELECTRIC FIELD
Suppose the coulomb force F moves a positive test charge Q1 from point A to point B in an electric field as shown in the figure below. How much work does this involve? Q
Electric field region of charge Q A B E Q1 F

4. WORK DONE BY AN ELECTRIC FIELD
In the figure shown the line joining A and B represents the path along which a test charge Q1 is moved in the field of point charge Q. If both charges are positive, work is done by the repulsive electrostatic force F as the test charge moves from A to B. The work done along the infinitesimal section dl of the path is
dW = F.dl
Since the force acting on the test charge is the Coulomb force, F = Q1E and therefore
The unit of force in the newton and the unit of work is the joule.

5. WORK DONE BY AN ELECTRIC FIELD (CONTD.)
The total work done when the test charge is moved from A to B is thus approximated as an infinite sum of incremental amounts of work, each being given by the preceding equation:
(J/C)
where Ek is the electric field intensity at the k-th point on the path. If we let N approach infinity in such a way that lk approaches zero for any k, the total work done in moving Q1 from A to B in the field E is given by the line integral

6. WORK DONE BY AN ELECTRIC FIELD (CONTD.)
Now, in the cylindrical coordinate system the electric field at position vector r is Q
Electric field region of charge Q A B E Q1 F rA rB
and for A, B on the same radial, dl = drr the integral reduces to the form
Therefore,

7. Note that the work done by the electrostatic force is positive if the charge is moved in the direction of the electric field and negative when moved against it. x y z
P(1,1,0)
10 nC O
Path travelled
Example
Calculate the work to move a 10 nC charge from the origin to point P(1,1,0) against the static field E = 5 V/m.

8. In the rectangular coordinate system, we have
Solution
In the rectangular coordinate system, we have
Thus, work done by the field to move the 10 nC from the origin to point P(1,1,0) in the region where the electric field is E = 5 V/m is
Next, using the identities
î · î = ĵ · ĵ = k̂ · k̂ = (1)(1)(cos 0°) = 1
î · ĵ = ĵ · k̂ = k̂ · î = (1)(1)(cos 90°) = 0

9. we obtain
Since work done is positive, the field is doing 50 nJ of work.

10. THE ELECTRIC POTENTIAL
If E is a static electric field, the integral
is independent of the path taken from A to B; it only depends on the end points. In other words, the electric force F = Q1E is a conservative force if E is static field. For a conservative field, work done in moving round a closed path is zero, that is A B
Path x
Path y
Path z

11. THE ELECTRIC POTENTIAL
We can associate a potential energy function U with the force F. Thus, at point A we can associate it with the potential energy function UA, and at point B we can associate it with a potential energy function UB.
By the work-energy principle, the work done by a conservative force equals the negative of the difference in potential energy between the final position and the initial position. Thus, we can write A B
Path x
Path y
Path z

12. THE ELECTRIC POTENTIAL (CONTD.)
Let us define the electric potential V as the potential energy per unit charge; that is, P
Unit positive test charge, Q B A
Electric field region of charge P dl
Thus, we have or E
This can be rewritten as

13. THE ELECTRIC POTENTIAL (CONTD.)
Thus, we have x z P
Unit positive test charge, Q y B A
Electric field region of charge P rB rA r dl or
This can be rewritten as

14. THE ELECTRIC POTENTIAL (CONTD.)
For A,B on the same radial
and
Therefore,
Also, using V= U/Q, we obtain or

15. THE ELECTRIC POTENTIAL (CONTD.)
The quantity, denoted by VAB, is known as the potential difference between points A and B. The quantities denoted by VA, VB are known as the absolute potentials at points A and B, respectively.
Note that
In determining VAB, A is the initial point while B is the final point.
If VAB is positive, there is loss in potential energy in moving from A to A; this implies that the work is being done by the field. However, if VAB is negative, there is a gain in potential energy in the movement; an external agent performs the work.
VAB is independent of the path taken.
VAB is measured in joules per coulomb, commonly referred to as volts (V).

16. SI UNITS FOR POTENTIAL
By definition, electric potential is potential energy per unit charge. So,
The SI unit for electric potential is the volts. Both potential and its unit are notated by the capital letter “V.” Based on the definition above, a volt is defined as joule per coulomb:
Example: If an object with a 10 C charge is placed at a certain point in an electric field so that its potential energy is 50 J, every coulomb of charge in the object contributes to 5 J of its energy, and its potential is 5 J/C, that is, 5 V.

17. This parasailer landed on a 138,000-volt power line
This parasailer landed on a 138,000-volt power line. Why didn’t he get electrocuted?
He touches only 1 line – there’s no potential differences & hence no energy transfer involved.

18. UNITS FOR ENERGY Example
There is an additional unit that is used for energy in addition to that of joules
Example
A particle having the charge of e (1.6 x C) that is moved through a potential difference of 1 Volt has an increase in energy that is given by

19. ABSOLUTE POTENTIAL
Instead of speaking continually of potential difference between pairs of points, we may speak of the potential at a single point- provided we always refer to some other, agreed, reference point.
For practical purposes, we generally choose as our reference point the electric potential of the surface of the earth. P A
Electric field region of charge P B rA rB
Earth

20. Earth P A
Electric field region of charge P B rB rA
Suppose we define VB = 0 V at an infinite radius. Then, potential of point A relative to the zero reference (or, the absolute potential at point A) is
If the point charge Q is not located at the origin but a point whose position vector is r’, the potential V(x,y,z) or simply V(r) at r becomes

21. Some choices of zero potential
THE ZERO OF POTENTIAL
Some choices of zero potential
Power systems / Circuits
Earth ( Ground )
Automobile electric systems
Car’s body
Isolated charges
Infinity

22. Example
Let the positive charge Q in the figure below be equal to 223 pC. Also let rB = 400 mm and rA = 100 mm. The medium is air. Find the absolute potential VB at B, the absolute potential VA at A, and the potential rise VBA. Q A
Electric field region of charge Q B rB rA E

23. Solution Q A
Electric field region of charge Q B rB rA E
5 V
20 V

24. Example
A point charge of 5 nC is located at the origin. If V = 2 V at A(0, 6, -8), find
The potential at B(-3, 2, 6)
The potential at C(1, 5, 7)
The potential difference VBC
A(0,6,-8)
5 nC
B( -3, 2, 6)
C(1, 5, 7) rA rB rC x z y

25. Solution
Given
Q = 5 nC
VA = 2 V
rA = (0, 6, - 8)
εo = 8.85 x 10-12
and
(b) Given
rC = ( 1, 5, 7)
Therefore,
Therefore,
rA = |rA| =10
rC = |rC| = 8.66
and
(a) Given
rB = ( -3, 2, 6)
(c) Potential difference
Therefore,
VB – VC = VBC = V
rB = |rB| = 7

26. Example
A 15-nC point charge is at the origin in free space. Calculate VA if point A is located at A(- 2, 3, -1) and (a): VB = 0 at (6, 5, 4); (b) VB = 0 at infinity; (c) VB = 5 V at B(2, 0, 4).
Solution
A = (-2, 3, -1); B = (6, 5, 4)
rA = (-2, 3, -1)
rB = (6, 5, 4)
Therefore,

27. Solution
(b) A = (-2, 3, -1); B = (, , )
rA = (-2, 3, -1)
rB = (, , )
Therefore,

28. Solution
(c) A = (-2, 3, -1); B = (2, 0, 4)
rA = (-2, 3, -1)
rB = (2, 0, 4)
Therefore,
Now VAB = VA – VB
Therefore,
VA = VAB + VB = = V

29. POTENTIAL DUE TO N POINT CHARGES
For a system consisting of N point charges, the total electric potential at a point is the algebraic sum of the individual component potentials.
Thus, for N point charges Q1, Q2, …., QN located at points with position vectors r1, r2, ….., rN, the potential at r is or P Q1 Q2 QN z y x r1 r2 rN r

30. Example
Two point charges – 4 µC and 5 µC are located at (2, -1, 3) and (0, 4, -2) respectively. Find the potential (1, 0, 1) assuming zero potential at infinity.
(0,4,-2)
5 µC
- 4 µC
(2, -1, 3)
(1,0,1) r2 r1 r x z y
Solution
Let Q1 = - 4 µC; Q2 = 5 µC
r1 = (2, - 1, 3)
r2 = (0, 4, -2)
r = (1, 0, 1)

31. Solution (continued) Therefore, r - r1 = (1, 0, 1) - (2, - 1, 3) =
(0,4,-2)
5 µC
- 4 µC
(2, -1, 3)
(1,0,1) r2 r1 r x z y

32. EQUIPOTENTIAL SURFACES: POSITIVE POINT CHARGE
Imagine a positive test charge, q, approaching an isolated, positive, point-like field charge, Q. The closer q approaches, the more potential energy it has. So, potential increases as distance decreases.
We have found earlier that the formula for absolute potential at radial distance r due to a point charge Q is given by
This means, for a point charge,
the surface of constant potential
is a sphere of radius r centered at
Q. And because of the circular
symmetry, the equipotentials are
everywhere perpendicular to the field
lines. +

33. The equipotentials have units of volts
The equipotentials have units of volts. Here the surfaces could be labeled from the inside out: 100V, 90 V, 80 V, and 70 V. Every 10 V step is bigger than the previous, since the field is getting weaker with distance.
The gap between the 50 V and 40 V surfaces would be very large, and the gap between 10 V and 0 V would be infinite. +
70 V
80 V
90 V
100 V A B D C

34. Picture of equipotential surfaces surrounding a positive charge:
Distance, r
Electric Potential, V
100 V
70 V
50 V
20 V

35. PROPERTIES OF EQUIPOTENTIAL LINES:
Equipotentials are always perpendicular to the field lines.
Equipotentials never intersect one another.
The potential is large & positive near a positive charge,
large & negative near a negative charge, and near zero far from all the charges.
Equipotentials are close together where potential energy changes quickly (close to charges).
The electric field does no work as a charge is moved along an equipotential surface.
Since no work is done, there is no force, qE, along the direction of motion

36. EXAMPLES OF EQUIPOTENTIAL SURFACES
Point Charge
Two Positive Charges

37. E A B x
x +δx V
V + δV δx
Positive charge Q
Distance, r
Electric Potential, V
100 V
70 V
50 V
20 V
POTENTIAL GRADIENT
Consider a region in the vicinity of a positive point charge. Suppose A and B are two neighbouring points on a line of force, so close together that the electric field intensity between them is constant and equal to E. If V is the potential at point A, (V + δV) is the potential at B; and the respective distances of A and B from the origin are x and (x + δx), then
VAB = potential difference between
A and B
= VA – VB
= V – (V + δV)
= - δV

38. The work done in taking a unit charge from B to A = force x distance
= VAB
= - δx
Hence, E A B x
x +δx V
V + δV δx
Positive charge Q
Or, in the limit as δx approaches zero,

39. Electric potential, V
Distance, x dV dx
Slope of line is potential gradient
The quantity dV/dx is the rate at which the potential rises with distance, and is called the potential gradient.
The equation
shows that the strength of the electric field is equal to the negative of the potential gradient. Strong fields are represented by a steeper slope and weaker fields by a gentler slope.
Electric potential, V
Distance, x
Slope of line is potential gradient
Strong field
Weaker field

40. Worked Example
The graph shows the variation of potential with distance from the charged dome of a van de Graaff generator. Use the graph, together with the equation
160
140
120
100 80 60 40 20
V (kV)
r (m)
to find the electric field strength at a distance of 0.3 m from the dome.

41. Solution
The field strength will be the negative of the gradient at r = 0.3 m. This should give a value of around 1.7 x 105 N/C.
To calculate this we note that at 0.3 m the potential is 50 kV. Now
V = kQ/r
and we note that E = kQ/r2, so numerically, we can see that E = V/r. In other words, in this case
E = 50,000 / 0.3 = 1.7 x 105 N/C.

42. ELECTRIC FIELD AS GRADIENT OF POTENTIAL
One of the values of calculating the scalar electric potential (voltage) is that the electric field can be calculated from it. The component of electric field in any direction is the negative of rate of change of the potential in that direction.
If the differential voltage change is calculated along a direction ds, then it is seen to be equal to the electric field component in that direction times the distance ds.
The electric field can then be expressed as
For rectangular coordinates, the components of the electric field are

43. In general, the electric potential is a function of all three dimensions
Given V (x, y, z) you can find Ex, Ey and Ez as partial derivatives
The expression of electric field in terms of voltage can be expressed in the vector form
This collection of partial derivatives is called the gradient, and is represented by the symbol ∇. The electric field can then be written as

44. Example
Suppose that in the figure below the potential decreases by 2 V/m in the x direction and by 1 V/m in the y direction. Find the electric field E. y Ey E x P Ex
V = 102 V
V = 103 V
V = 101 V
V = 104 V
V = 105 V

45. Solution
and
Therefore, E has a magnitude of 2.24 V/m and is directed at an angle of 26.6o with respect to the positive x axis.

46. FORCE AS A DIFFERENTIAL OF THE POTENTIAL FUNCTION
A positive charge Q placed in an electric field will experience a force given by
But E is also given by
Therefore
Since Q is positive, the force F points in the direction opposite to increasing potential or in the direction of decreasing potential. 46

47. POTENTIAL ENERGY
Looking at the work done we notice that there is the same functional at points A and B and that we are taking the difference
We define this functional to be the potential energy
The potential energy is taken to be zero when the two charges are infinitely separated.
The signs of the charges are included in the calculation.

48. POTENTIAL ENERGY OF A SYSTEM OF POINT CHARGES
Suppose we have more than two charges.
Have to be careful of the question being asked.
Two possible questions:
1) Total Potential energy of one of the charges with respect to
remaining charges or
2) Total Potential Energy of the System

49. CASE 1: POTENTIAL ENERGY OF ONE CHARGE WITH RESPECT TO OTHERS
Given several charges, Q1…Qn, in place Q1
Now a test charge, Q0, is brought into position Q2
Work must be done against the electric fields of the original charges Q3 Q0
This work goes into the potential energy of Q0
We calculate the potential energy of Q0 with respect to each of the other charges and then just sum the individual potential energies
Remember - Potential Energy is a Scalar

50. CASE 2: POTENTIAL ENERGY OF A SYSTEM OF CHARGES
Start by putting first charge in position
No work is necessary to do this
Next bring second charge into place
Now work is done by the electric field of the first charge. This work goes into the potential energy between these two charges.
Now the third charge is put into place
Work is done by the electric fields of the two previous charges. There are two potential energy terms for this step.
We continue in this manner until all the charges are in place
The total potential is then given by

51. Example 1
Two test charges are brought separately to the vicinity of a positive charge Q
Charge +q is brought to pt A, a distance r from Q
Charge +2q is brought to pt B, a distance 2r from Q
Compare the potential energy of q (UA) to that of 2q (UB)
(a) UA < UB A q r Q
(b) UA = UB
(c) UA > UB B Q 2q 2r 51

52. Example 2
Two test charges are brought separately to the vicinity of a positive charge Q
Charge +q is brought to pt A, a distance r from Q
Charge +2q is brought to pt B, a distance 2r from Q
Compare the potential energy of q (UA) to that of 2q (UB)
(a) UA < UB A q r Q B 2q 2r
(b) UA = UB
(c) UA > UB
The potential energy of q is proportional to Qq/r
The potential energy of 2q is proportional to Q(2q)/(2r) = Qq/r
Therefore, the potential energies UA and UB are EQUAL!!! 52

53. Solution The potential energy of q is proportional to Qq/r
The potential energy of 2q is proportional to Q(2q)/(2r) = Qq/r
Therefore, the potential energies UA and UB are EQUAL!!! A q r Q B 2q 2r 53

54. END