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E L E C T R I C F I E L D

1.0 INTRODUCTION

You are already familiar with many electrical phenomena such as lightning. You can see a brief spark, but the effect does not persist. Such events are due to what we call static electricity or electrostatic.

1.0.1 ELECTROSTATICS

We will begin our discussion of electricity by studying electrostatics, the science of stationary electric charges What is static electricity? Static electricity is defined as an electrical charge caused by an imbalance of electrons on the surface of a material. Put in simple terms, static electricity is when electrons (not protons) are moved, usually by rubbing or brushing. The study of static electricity, electrostatics investigates every aspect of static charge. There is also one more way to define static electricity, by defining and comparing its opposite. Current electricity is the opposite of static electricity, the difference between the two is static is when electrons are moved, as stated above, and current is when electrons flow inside a conductor. So just remember, static electricity is a build up of charge, positive or negative, in one place.

Electric Charge

The structure of atoms can be described in terms of three particles that are electron, proton and neutron. Proton and electron are charged particle but neutron is the uncharged particle. There are two types of electric charged that is positive and negative charges. The positive charge is called proton and the negative charge is called electron. Basically, if an object has an equal number of proton and electron, the object is called neutral charged. If the object has the number of electron more than the number of proton the object is said as negatively charged and if the object has deficiency of electron it will become as positively charged

The unit of charge used in the SI system is called the Coulomb (C) *. 1 Coulomb charges = 6.242 x 1018 number of electrons, therefore 1 electron charge = 1.602 x 10-19 C. Similarly, the proton charge is 1.602 x 10-19 C. The sign of charge is labeled as (+ve) for proton and (-ve) for an electron.

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* Charles Augustin de Coulomb was a French physicist. He found the mathematic theory of interaction of electric charges.

The formula for charge is 'Q = ne'. In the equation Q stands for charge of an object in coulombs, n is the number of elementary charges, and e is the elementary charge, to be 1.60 x 10-19C.

Exercise 1.1

A person has received a charge of 8 coulombs how many elementary charges does the person have?

Test your understanding

What is the mass of the electron, proton and neutron?

In electrostatics, we are studying three important quantities, those are (i) Electric Force (vector quantity)(ii) Electric Field (vector quantity)(iii) Electric Potential (scalar quantity)

1.1 ELECTRIC FORCE AND COULOMB’S LAW

Electric Force between two point charges.

We now know the properties of charges, but we still do not know how to measure charges interactions. When two charges interact, there are two possibilities the charges can either attract or repel . In both of these possibilities, the charges feel either an attractive or repulsive force– we know that this force exists, but how can we measure it? Charles-Augustin de Coulomb asked this same question in 1784. Coulomb was a great French physicist who was responsible for discovering a relationship between the force of two charges and the square of its distance. He compiled this relationship into an equation called Coulomb's Law. Electric charge and a basic rule describing their behavior is- like electric charges repel each other, and unlike charges attract each other. If two point charges are arranged as shown in Figure 1.1, each point charge will experienced the force due to other charge as indicated in figure. The direction of electric force on each point charge is depend on the charge itself.

2

21

+

++

Q1 Q2

12 21

12

12

21

Figure 1.1

Coulomb’s Law

Charles Augustin de Coulomb was the first to measure electrical attractions and repulsion quantitatively. As a result of his experimentation, he deduced the law that governs these forces. It is called Coulomb’s Law, The force between two stationary point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

F Q1Q2 and F

Coulomb’s law can be expressed algebraically as

F = or F =

F = magnitude of force experienced by each of two charges.

Q1 and Q2 = magnitude of charges

r = distance between charges

k = constant of proportionality ( 9 x 109 Nm2/C2 )0 = permittivity of free space ( 8.854 x 10-12 C2/Nm2 )

Thus, the magnitude of the force between two point charges is

=

Example 1.1

Two isolated small objects have charges of 0.04 C and –0.06 C and are 5cm apart as Figure 1.2. What will be the magnitude of electrostatic force acting on each object?

Solution

= =

= 0.00864 N

3

Q1 = +0.04C Q2 = -0.06C

50.0 cm Figure 1.2

= When dealing with Coulomb’s law, you must remember that the force is vector quantity and must be treated accordingly

=

= + +

Exercise 1.2

Refer to an example 1.1 above. Identify the direction of electric force and write the vector component on each charge.

Superposition of Electric Forces

As we have stated in Coulomb’s Law describes that the only interaction of two point charges. Experiments show that when two charges exert force simultaneously on a third charge, the total force acting that charge is the vector sum of the forces that the two charges would exert individually. This independence of the forces is known as the principle of superposition of Electric Forces. We can determine the net force on any one charge by summing up the individual contributions to the force due to each of the other charge.

= + + + ……

Example 1.2

Calculate the resultant force on the charge Q3 due to other two charges located as shown in Figure 1.3. Identify the direction of resultant/ net force.

PROBLEM-SOLVING HINTS.

Step 1: Determine the directions of the individual forces on Q3 due to charge Q1 and Q2 and indicate it in a sketch.

Step 2: Find the magnitude of forces on Q3 due to charge Q1 and Q2 that is │F31│& │F32│. Please make sure these values always positive.

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1.0 cmFigure 1.3

2.0 cm

Q1 = -4.2C Q2 = +1.3C Q3 = +1.1C

Step 3: Find the vector components for 31 & 32 by using , , form (normally in or only in one dimension case ).

Step 4: Find the resultant forces vector net = 31 + 32

Step 5: Identify the direction and magnitude of resultant force from net

Solution

(1) 31 32

(2) = = = 103.95 N

= = = 128.70 N

(3) 31 = - 103.95

32 = + 128.70

(4) net = 31 + 32 = + 24.75

The magnitude of electric force are 24.75 N and in positive x axis direction. Exercise 1.3

Using the same method as above to find the resultant (net) force on Q1 due to charges Q2 and Q3. Identify the direction for resultant electric force?

Example 1.3

Three point charges QA, QB, and QC of +12µ , -16µ and + 20C respectively, are arranged as shown in Fig 1.4 Find the magnitude and direction of the net force on charge QA.

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QC = +20CQB = -16C

QA =+12C

4m

3m

Figure 1.4

PROBLEM-SOLVING HINTS.

Step 1: Determine the directions of the individual forces on QA due to charge QB and QC and indicate it in a sketch.

(1)

Step 2: Find the magnitude of forces on QA due to charge QB and QC that is │FAB│& │FAC│..

(2) = = = 0.069N

= = = 0.135 N

Step 3: Find the vector components for AB & AC by using , , form.

(3) AB = │FAB│cos + │FAB│sin

AB = 0.069 cos 233.130 + 0.069 sin 233.130

AB = -0.041 + ( -0.055 )

AC = +│FAC│

AC = +0.135

net = AB + AC

Step 4: Find the resultant forces vector net = AB + AC

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AC

AB

= 53.130

= 53.130

(4) net = -0.041 + -0.055 + 0.135

net = -0.041 + 0.08

Step 5: Find the magnitude of resultant force on │ net │by using the formula.

(5) = = 0.089N

Step 6: Find the direction of resultant force on θ by using the formula.

(6) Direction θ = tan –1 = 63.430

Step 7: Check the quadrant for direction of the force.

(7) 63.430 (2 nd quarter) or 63.43 above –ve x-axis

Exsercise 1. 4

Using the same methods as above calculate the resultant force on QC due to the charges QA and QB.

Tutorial 1.1

1. How many electron must be removed from an electrically neutral silver ball to give it a charge of +2.4μC ? (Ans : 1.5 x 1013 electrons)

2. Excess electrons are placed on a small lead sphere with mass 8.00g so that the net charge is –3.20 x 10-9C. Find the number of excess electrons on the sphere(Ans : 2.0 X 1010 numbers of electron)

3. Two isolated small objects have charges of 1.0 C and –2.0 C and are 50cm apart as Figure 1. What will be the magnitude of electrostatic force acting on each object? Write the electrostatic force vector component for each object.

(Ans : 0.072N , , )

4. Three point charges are arranged as Figure 2. Find the resultant electric force vector component on the charge q3 due to charge q2 and q1.

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1.5 m 2.0 m

q1 = +5C q2 = -10 C q3 = +20C

Figure 2

Q1 = +1.0C Q2 = -2.0C

50.0 cm Figure 1

(Ans : )

5. Three point charges Q1 , Q2 and Q3 are arranged as shown in Figure 3 (i) Write the vector component for resultant force on charge Q3 due to other charges (ii) Calculate the resultant force on charge Q3.

[ Ans : i. , ii. 2.38N , 79.10 (1st quarter)]

6. Three point charges Q1 , Q2 and Q3 are arranged as shown in Figure 4.(i) Find the vector component for resultant force on charge Q2 = + 200C due to other

two charges.(ii) Calculate the resultant force on charge Q2. .

(Ans: i. 4.5 x 105 N i – 6 x 105 N j, ii. 7.5 x 105 N, 530, 4th quarter)

7. Two point charges Q1 and Q2 are 3m apart and repel each other with a force 0.075N. If Q1 + Q2 = 20C, what is the charge on Q1 and Q2 . ( Ans: 5C , 15C )

8. Four point charges Q1 , Q2 , Q3 and Q4 are arranged as shown in Figure 5, (a) find the vector component for the force acting on Q2 due to Q1.(b) find the vector component for the force acting on Q2 due to Q3.(c) find the vector component for the force acting on Q2 due to Q4

(d) find the force acting on Q2

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5cm

Q1 = +2C Q2 = +1C

Q3 = +1C

10cm

10cm

10cmFigure 3

Q1= +400C Q2= +200C

Q3= -300C

3 cm

4 cm

Figure 4

Q1 = +20C

Q2 = -30C Q3 = +40C

5cm Figure 5

Q4 = -30C

(Ans : a. +2160N j , b. +4320N i , c –1145.5 i – 1145.5 j , d. 3332.21 N, = 17.720 (1st

quarter) @ 17.72 above posite x axis.)9. Three stationary charges are arranged at the corners of a triangle as shown in Figure 6. Given

Q1 = + 5 C, Q2 = - 3 C and Q3 = - 3 C

Figure 6

a) Sketch the direction of the forces exerted on Q1 due to the other two chargesb) Calculate the magnitude and direction of the resultant electrostatics force on Q1

(Ans : 20.72 N, - 62.21o (4th quarter @ below +ve x-direction)

10. Three point charges Q1, Q2 and Q3 are arranged as shown in Figure 7. Calculate the net force experienced by charge Q3 due to charges Q1 and Q2.

Figure 7

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Q3 = +4C Q2 = -3C

Q1 = +2C

3 cm

4 cm

1.2 ELECTRIC FIELD

An electric field is said to exist at a point if a small positive charge placed at that point experiences an electric force on it.

The electric field E has magnitude and direction (vector quantity). The electric field intensity (or strength) E at any point is defined as the force per unit positive charge acting on a charged body at that point.

E =

Just as charges cannot be seen with the eye, electric fields are also not visible. To diagram electric fields, we use electric field lines. Electric field lines are drawn out of or into a charge. Each line represents the path a test charge (a very small positive charge) would take if it came into contact with the given charge.

When multiple lines are drawn, an electric field diagram is constructed. Also the number of lines and there spacing are not drawn in a random or artistic manner instead they are representative. The number of lines drawn around a given charge represents the magnitude of that charge while the space of the drawn lines represents the approximate strength of the electric field. So a diagram with many lines drawn close together would represent a powerful charge with a strong electric field.

The direction of the electric field at a point is defined as the direction of the force on a positive charge (test charge q) placed at that point.

The field points radially away from an isolated positive charge (Q> 0C), and radially toward a negative charge (Q< 0C) as shown in Figure 1.5 and Figure 1.6.

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Figure 1.5 the electric field lines around a single charged

The magnitude of electric field created by the positive charge Q is

│Eon Q│ = =

The SI unit of field intensity is the Newton per coulomb (N/C).

Thus the electric field of the positive charge Q at a distance r from Q can be written in vector form as

=

= + +

Example 1.4(a) Calculate the magnitude of electric fields at point P in Figure 1.7 which is 30cm to the right of a

point charge Q = -3.0 x 10-6 C. (b) Identify the direction of electric field at point P and write the vector component of electric field at

point P.

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Figure 1.6 the electric field lines around (a) & (b) two unlike charges (c) two like positive charges

30 cm

PQ = -3.0 x 10-6 C

Figure 1.7

The magnitude of electric fields is 3.0 x 105 N/C and in negative x axis direction

Superposition of Electric Fields

The electric field vector of a static arrangement of charges obeys a principle of linear superposition. The total field at any point is the vector sum of the individual fields produced by each charge acting alone :

= + + + ………

In this way, if a charge q, placed at a point in space where the electric field is exist, the charge will be experiences a force that is

q = q

Example 1.5A distance of 10.0 cm as shown in Figure 1.8 separates two point charges. Find the electric fields at point P. (6.3 x 108 N/C)

Solution

(1)

(2) = = = 56.25 x 107 N/C

Solution

(a) │EP│ = = = 3.0 x 105 N/C

(b)

= - 3.0 x 105 N/C

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P

r2 = 8.0cmr1 = 2.0cm

Q1 = +25 μC P Q2 = +50 μC

Figure 1.8

= = = 7.03 X 107 N/C

(3) = + 56.25 X 107

= - 7.03 X 107

(4) = + = + 49.22 X 107

The magnitude of electric fields are 49.22 x 107 N/C and in positive x axis direction

Exercise 1.4Two point charges are arranged as in a Figure 1.9. Find the electric field at point P.

Example 1.6

1. Two point charges are arranged as in a Figure 1.10. Calculate the electric field at point G due two both charges Q1 and Q2. Identify the direction for resultant electric field.

2. Determine the force on a charge q = -2.0μC if placed at point G.

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r1 = 2.0cm

r2 = 8.0cm

P

Q1 = -25 μC

Q2 = -50 μC

Figure 1.9

x

30 cm

26 cm26 cm

G

y

Q2 = -40 μCQ1 = +50 μC

Figure 1.10

Solution

1. (i)

(ii) = = = 2.86 X 106 N/C

= = = 2.28 X 106 N/C

(iii) = │EGQ1│cos + │EGQ1│sin

= 2.86 X 106 cos 49.10 + 2.86 X 106 sin 49.10

= 1.87 X 106 + 2.16 X 106

= │EGQ2│cos + │EGQ2│sin

= 2.28 X 106 cos (40.90 + 2700) + 2.28 X 106 sin (40.90 + 2700 )

= 1.49 X 106 + (-1.7 X 106 )

(iv) = + = (1.87 X 106 + 2.16 X 106 ) +

(1.49 X 106 - 1.70 X 106 )

net = 3.36 X 106 + 0.46 X 106

(v) The magnitude = = 3.39 X 106 N/C

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= 49.10

(vi) Direction θ = tan –1 = 7.50

(vii) The magnitude of electric fields are 3.65 x 106 N/C and 5.70 ( 1st quarter) @ 5.70 above positive x axis.

2. qG = q

= (-2.0 X 10-6 ) ( 3.36 X 106 + 0.46 X 106 )= (-6.72 - 0.92 )

= = 6.78 N

Direction θ = tan –1 = 7.790

= 7.790 (3 rd quarters)

The magnitudes of electric force are 6.78 N and 7.790 (3 rd querters) @ 7.790 below negative x-axis.

Exercise 1.51. If a charge Q3 = - 10μC are placed at origin in Figure 1.10, find the magnitude and direction for

resultance electric fields at point G.

2. Determine the force on a charge q = +2.0μC placed at point G now. What is the magnitude of the force on this charge and sketch the direction of that force?

1.3 GAUSS’S LAW

Gauss Law states that

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Where ФE = electric flux A = surface area θ = angle between E and normal to the surface

Electric fluxis the number of field lines that pass through a given surface.

The net flux through a closed surface is equal to the amount of charge enclosed by the surface per εo

εo is the permittivity of free space, and the above equations hold true for charges in vacuum only. The net number of field lines through the surface is proportional to the charge enclosed, and also to the flux. Gauss Law can be used in two ways : 1) to obtain the amount of charge enclosed when the flux / electric field is known 2) to obtain the electric field if the amount of enclosed charge is known.

Application of Gauss’s Law

Uniformly charged spherical shellIf the conductor contains a cavity, as in a spherical shell (Figure 1.11), the amount of charge in the cavity can be obtained by drawing a Gaussian surface inside the conductor close to the surface of the cavity. Because E is zero inside the conductor, there is no flux through this Gaussian surface. Therefore there is no net charge on the inside wall of the cavity.

rR

Gaussian surface

Figure 1.11

The Gaussian surface is a sphere of radius r (See Figure 1.11). By Gauss' law, the electric field is given by

Case 1 : when R > r (outside the surface)

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Case 2 : when R = r (at the surface)

Case 3 : when R < r (inside the surface)

Example 1.71. What is the electric flux passing through a Gaussian surface that surrounds a +0.075 C point

charge?

2. A sphere 8 cm in diameter has a charge of 4 µC place on its surface. What is the electric field intensity

a) at the surface b) 2 cm outside the surface c) 2 cm inside the surface

Exercise 1.61. A hollow conducting sphere of radius 15 cm has a net charge of -3 µC on its surface.

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a) What is the electric field 10 cm from its center? b) What is the electric field 20 cm from its center? (Answer = a) 0 N/C b) 6.75 x 105 N/C pointing towards the center of the sphere)

2. A charge of +5 nC is placed on the surface of a hollow metal sphere whose radius is 3 cm.a) Use Gauss’s Law to find the electric field intensity at a distance of 1 cm from the surface of

the sphere. b) What is the electric field at a point 1 cm inside the surface (Answer = a) 2.81 x 104 N/C b) 0 N/C)

3. At a distance of 0.20 cm from the center of a charged conducting sphere with radius 0.10 cm, the electric field is 480 N C-1. What is the electric field 0.6 cm from the center of the sphere?

(Answer = 53 N/C)

Tutorial 1.21. A point charge q1 = 2 C is placed at the origin. Find the electric fields at point A (3,0) cm from

origin . (Ans : +2.0 X 107 N/C i )

2. Find the total electric field along the line of the two point charges shown in Figure 1.18 below at the midpoint between them.

(Ans : -54800 N/C i )

3. Three point charges are arranged as in a Figure 1.19. Determine the electric fields at point R due to other charges.

(Ans : -4.16 X 108 NC-1 i)

4. A point charge Q1 = 2C is placed at the origin and charge Q2 = -3C at (3,0) cm as Figure 1.20, (a) Sketch electric field vector E1 and E2 at point P due to charge Q1 and Q2.(b) write the vector component for field at point P due to Q1 charge.(c) write the vector component for field at point P due to Q2 charge.(d) write the vector component for resultant electric field at point P(e) find the magnitude and direction for resultant electric field at point P(f) if a point charge - 5C is placed at point P, determine the force experience by this

charge due to electric field at point P.

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9.0μC-4.7μC Figure 1.18

2cm 2cm 2cm

Q1 =+ 6C Q2 = - 8C Q3 = +12CR

Figure 1.19

Figure 1.20

Q1 Q2

x

y

3 cm

2 cm

P

{Ans : b) 1.148 x 107 N/Ci + 0.766 x 107 N/C j c)-6.75 x 107 N/C j d) 1.148 x 107 N/Ci - 5.98 x 107 N/C j e) 6.1 X 107 N/C , 2810 @ (79.10 4th

quarter) f) 305N , 110 ( 2nd quarter) }

6. A small 2g plastic ball is suspended by a 20cm long string in a uniform electric field (Figure 1.22). If the ball is in equilibrium when the string makes a 150 angle with the vertical as indicated, what is the net charge on the ball? (g = 9.80 ms-2).(Ans : 5.25 x 10-6 C)

7. Two positive charges Q1 located at (-3,0) m and Q2 at (0,2) m. Calculate the magnitude and direction of the net electric field at the origin due to these two point charges if Q 1 = 7 x 10-9 C and Q2 = 6 x 10-9 C.

(Ans : 15.2 N/C, θ= -62.6o (2nd quadrant or above negative x-axis)

5. Two point charges Q1 = +5C and Q2 = +15C lie 5m apart on a straight line in a vacuum as shown in Figure 1.21. Find the point between of two charges where the resultant electric fields intensity is zero.

(Ans : 1.83m from Q1)

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Q1 =+ 5C Q2 =+ 15C

5 m

Figure 1.21

Figure 1.2220cm

E = 103i N/C

m = 2g