Chapter 8 - Properties of Electric Charges-Latest

Physics 1 (PHYS 1200)

Chapter 8

Electric Forces and Electric Fields

Properties of Electric Charges:

1. There two types of charges, positive and negative.

2. Like charges repel and unlike charges attract one another.

3. The electric charge is always conserved.

Charge cannot be created or destroyed, but only exchanged.

Note.

1. The nature’s basic carrier of positive charge is the proton. They do not move from one material to another because they are held firmly in the nucleus. Nature’s basic carrier of negative charge is the electron.

2. Objects are said to be charged if they gain or lose electrons.

3. Objects are said to be negatively charged if they gain electrons and positively

charged if they loose electrons.

4. The SI unit of a charge is the Coulomb (C).

5. The charge of the electron is, e = -1.6 x 10-19, and for the proton is +1.6 x 10-19 C.

6. Charge on a body can be written as an integral multiple of the elementary charge

Q = n e (conservation of charge)

Q - Charge on a body

e - Elementary charge

n - Number of elementary charge.

Problem .1

A body has a charge of -1coulomb. Find the number of (excess) electrons present on it. The charge of an electron e = -1.6 x 10-19C .

Solution: Q = n e

-1C = n x (-1.6 x 10-19C ) , n = 6.25 x 1018

Chapter 8 Electric Forces and Electric Field Zuhair (ZAS)


Materials are classified into three categories in regard of their electrical conductivity:

Conductors Insulators Semiconductors

Conductors are materials in which the electric charges move freely,

Copper, aluminum and silver etc are examples of conductors

They are good conductors electricity

Insulators are materials in which electric charges do not move freely.

Glass and rubber are examples of insulators

Their characteristics are between those of insulators and conductors

Silicon and germanium are semiconductors.

The ways of charging the materials:

1. Charging by Conduction:

A charged object (the rod) is placed in contact with another object (the sphere). Some electrons on the rod can move to the sphere, when the rod is removed, the sphere is left with a charge. The charged object is always left with a charge having the same sign as the object doing the charging

2. Charging by Induction:

When an object is connected to a conducting wire or pipe buried in the earth, it is said to be grounded.

A negatively charged rubber rod is brought near an uncharged sphere. The charges in the sphere are redistributed. Some of the electrons in the sphere are repelled from the electrons in the rod. The region of the sphere nearest the negatively charged rod has an excess of positive charge because of the migration of electrons



away from this location. if a grounded conducting wire is connected to the sphere which allows some of the electrons to move from the sphere to the ground . When the wire to ground is removed, the sphere is left with an excess of induced positive charge. Then the positive charge on the sphere is evenly distributed due to the repulsion between the positive charges

Charging by induction requires no contact with the object inducing the charge.

Coulomb’s Law:

Statement:

The electric force of attraction or repulsion between two point charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance their centers.

The force on q1 is equal in magnitude and opposite in direction to the force on q2 .

Like/ similar charges repels (Fig. a), Unlike/ different charges attracts (Fig. b).

The force is repulsive . The force is attractive.

The mathematical representation of Coulomb's Law is,

ke is called the Coulomb Constant

ke = 8.99 x 109 N m2/C2 = 9X109 N m2/C2 (approximat)

q1,q2 , is the magnitude of charges measured in C , or µC.

F,is the magnitude of the electrostatic force

r,is the distance between the centre’s of the two charges in meters.Chapter 8 Electric Forces and Electric Field Zuhair (ZAS)


Note:

1. Coulombs law is a fundamental law of electrostatic force between between

two stationary point charges.

2. Coulombs law applies only to point charges and to spherical distribution of charges.

3. Coulombs law obeys Newton’s third law, and hence the electric forces F12 andF21 are equal in magnitude but opposite in direction, that is

F12 = -F21

4. The coulombs law is an example of field force, where the force exerted by one object on another without any physical contact between them.

5. The mathematical form of the Coulombs force is the same as that of the gravitational force. The only difference between them is, electric force can be either attractive or repulsive, but the gravitational force is always attractive.

Example 1:

A comb drawn through a person’s hair on a dry day causes 1012 electrons to leave the person’s hair and stick to the comb

(i) Is the force between the comb and the hair attractive or repulsive? (ii) Find the magnitude of this force when the comb is 1.0 m from the person’s

hair.

Solution :

(i) The force is attractive because the charge on the comb is negative and that on the hair is positive. Unlike charges attract.

(ii) From Coulomb’s law

or F = - 2.3 x 10-4 N

where –ve sign means that the force is attractive.



The Electric Field: (E)

(1) It is defined as the space around a charge or a system of charges where another charge will experience a force.

If a charged particle, with charge Q, produces an

electric field in the region of space around it,

then a small test charge, qo, placed in the field, will experience a force

(2) Electric field at a point is defined as-

The electric force acting on a the positive test charge qo placed at that point divided by the magnitude of the test charge qo

Where

E is the Electric field,

r is the distance where the test charge qo is placed.

(3)The direction of the field is defined to be the direction of the electric force that would be exerted on a small positive test charge placed at that point.

The electric field which is produced by a negative

charge is directed toward the charge. If a positive

test charge is placed in this field, then it would be

attracted to the negative source charge. ( Fig.a) ( Fig.a)

The electric field which is produced by a positive

charge is directed away from the charge. If a positive

test charge is placed in this field, then it would be

repelled from the positive source charge ( Fig.b).

( Fig.b)



Note:

1. The test charge qo should be small enough such that it does not disturb the change distribution responsible for the electric field as shown in the figures given below.

2. The electric field is a vector quantity; the direction of E is in the direction of F since we have assumed that acts on a positive test charge.

3. The electric field is a vector quantity

4. The SI unit of electric field is NC-1.

Once the electric field is known at some point, the force on any particle with charge ‘q’ placed at that point can be calculated with the help of equation (1).

F = q E ---------- (2)

Consider a point charge ‘q’ is placed at a distance ‘r’ from a test charge ‘qo’ . According to Coulomb’s law , the magnitude of the force on the test charge is

Since the magnitude of the electric field at the position of the test charge is defined as

E = F / qo ,

the magnitude of the electric field due to the charge ‘q’ at the position of

‘qo

’ is

----------(3)



Fig. 7

5.

Example 2 :

The electron and proton of a hydrogen atom are separated ( on the average) by a distance of about 5.3x10-11m. Compare the electric force and gravitational force that each particle exerts on the other.

Solution : Electric force (Fe) = =

Gravitational force(Fg ) =G =

Now , the gravitational force between the charged atomic particles is

negligible compared with the electric force.

Example 3 :

Find the electric force on a proton placed in an upward electric field of magnitude 2.0 *104N/C.

Solution :

The electric force on the proton is (F) = qp E

= 1.6x10-19 x 2.0x104

= 3.2x10-15N

where the force is upward in the positive y-direction

Example 4 : Chapter 8 Electric Forces and Electric Field Zuhair (ZAS)

If ‘q’ is positive, as in Fig. 7(a) , the field at P due to this charge is radially outward from q

If ‘q’ is negative, as in Fig. 7(b), the field at P due to this charge is radially toward q.


Find the electric field strength at the point P as shown in given figure below, which is 10.0 cm from a charge of + 6.0 µC and 40.0 cm from a charge of - 8.0 µC.

10.0 cm 40.0 cm P

+6.0 µC - 8.0 µC.Solution:

The electric field due to the + 6.0 µC charge (E+ 6.0 µC) = ke (q)/r 2

= ( 9*109)( 6*10-6) / (0.1)2

= 54.0 * 105 N/C

Similarly, the electric field due to the - 8.0 µC charge (E-8.0 µC) = (9*109)(8*10-6)/(0.4)2

= 4.5*105 N/C

The field due to both charges are in the same direction- to the right. Therefore, the resultant has a magnitude of 58.5x105 N/C and is directed toward the right.

Example 5: A water droplet of radius (r) =1.0 m remains stationary in a Millikan oildrop apparatus under the influence of an electric field of intensity 5.1x104 N/C. How many electronic charges does it carry? (qe = 1.6x10-19C , water (density of water) = 103 kg m-3 , g = 9.8 ms-2 )

.

Solution: Droplet stationary under the influence of two forces, its weight acting downwards and the electric force on it acting upwards and these forces must be equal in magnitude . Therefore, Fe = Fg

or q E = m g = (4/3) r3 g [ mass = density * volume ]

q =

=

= 8.0 * 10-19 C



or q = 8.0 *10-19 C

Since e = 1.6 * 10-19 C is the charge on the electron.

It follows that the drop carries = = 5 electronic charges

Electric lines of force: This concept is introduced by Michael Faraday.

Electric field at A, EA B Electric field at B

A

Charge C

Q Electric field at C

OR

The concept of lines of force was introduced by Michael Faraday as an aid in visualizing electric fields.

EA

Field at point A

A B Field at point B

Line of force EB

Fig. 8

Path along which a unit positive test charge would tends to move in an electric field.

Properties of electric field lines:



1. They are imaginary lines starts from positive charge ends at negative charge.

2. Tangent drawn at a point on the field lines represents the direction of the electric field at that point

3. The number of lines per unit area through a surface perpendicular to the lines is proportional to the strength of the electric field in a given region

4. The electric field is stronger where field lines are more crowded weaker where they are far apart.

5. The field lines are racially outward from the positive charge and inward from the negative charge.

6. Field lines extended to infinity.

7. Field lines never intersect each other. if they do so it would mean that at the point of intersection we can draw two tangents ,which represents two direction for the same field at the same time .which is not possible.

Below Fig.9(a) shows some lines of force around a single positive charge and Fig.9(b) shows some lines of force around a single negative charge .

Fig. 9The lines are actually directed radially outward from the positive unit charge in all directions since a positive test charge placed in the field would be repelled by the charge +q. In a similar way , the electric field lines for a single negative point charge are directed toward the charge.

Note : In either case , the lines are extended all the way to infinity. The lines are closer together as they get near the charge, indicates the strength of

the field is increasing. Lines of force never intersect with each other, hence any one point in an electric

field , the field can have but one direction.



Fig. 10

Fig. 11

Fig.12

Rules for drawing electric field lines for any charge distribution are as follows:

(i) The lines must begin on positive charges and terminate on negative charges.

(ii) The number of lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge.

(iii) No two field lines cross each other.

(iv) The electric field intensity at a point is the number of lines of force streaming through per unit area, normal to the direction of the intensity.

8.

Note: Two equal and opposite charge separated by a small distance is called a dipole


Fig.10 shows the beautifully symmetric electric field lines for two point charges of equal magnitude but opposite sign. This charge configuration is called an electric dipole. In this case , the number of lines that begin at the positive charge must

equal the number that terminate at the negative charge. at points very near either charge , the lines are nearly

radial. the high density of lines between the charges indicates a

strong electric field in this region.

Fig.11 shows the electric field lines in the vicinity of two equal positive point charges. In this case, closer to either charge the lines are nearly radial. the same number of lines emerges from each charge because

the charges are equal in magnitude. at large distances from the charges, the field is approximately

equal to that of a single point charge of magnitude 2q. the bulging out of the electric field lines between the charges

shows the repulsive nature of the electric force between like charges.

Fig. 12 is a sketch of the electric field lines associated with the positive charge + 2q and the negative charge -q. In this case

the number of lines leaving charge + 2q is twice the number terminating on the charge - q.

only half of the lines that leave the positive charge end at the negative charge. The remaining half terminate on negative charges that we assume to be located at infinity.


Problems:

Problems:

1. Calculate the force of repulsion between two small positive charges of 10-9C and 2x10-9 C placed 5.0 cm apart in air. [ Ans : 7.2x10-6 N]

2. Find the electric field intensity at a point P distant 10.0 cm from a point charge A of +2x10-8 C in air. [Ans: 18x103 N/C along AP ]

3. A 4.5 × 10–9 C charge is located 3.2 m from a –2.8 × 10–9 C charge. Find the electrostatic force exerted by one charge on the other

[Ans: 1.1x10-8 N attractive ]

4. Two identical conducting spheres are placed with their centers 0.30 m apart. One is given a charge of 12 x 10–9 C and the other a charge of –18 × 10–9 C.

(a) Find the electrostatic force exerted by one sphere on the other.

(b) The spheres are connected by a conducting wire. After equilibrium has occurred, find the electrostatic force between the two. [Ans: (a) 2.2x10-5 N attraction]

[ Ans:(b) 9.0x10-7N repulsion ]

5. Consider two point charges + 1.0 µC and + 2.0 µC separated by a distance of 10.0 m. At what point ( except infinity ) on the line ( or on its extension) joining the charges is the electric field zero?

[ Ans: Between the charges, 41.0 m from the charge +1.0 µC ]

6. (a)The force required to give a 1000.0 kg car an acceleration of 2.5 ms-2 is 2.50x103 N. What would the charge have to be on two small spheres 1.0 m apart to produce the same force between the spheres, if we assume that both spheres have the same charge?

[Ans: 526 µC]

(b) If the charge on each sphere is negative , how many excess electrons are there on each sphere?

[Ans : 3.29x1015]

7. Three charges are arranged as shown in Figure below . Find the magnitude and direction of the electrostatic force on the charge at the origin . [Ans: 1.38 x10-5N at 77.5o below–x axis ]



8. An airplane is flying through a thundercloud at a height of 2000 m. If there are charge concentrations of +40.0 C at height 3000.0 m within the cloud and –40.0 C at height 1000.0 m, what is the electric field E at the aircraft? [Ans :7.20 x 105 N/C downward ]

9. An electron is accelerated by a constant electric field of magnitude 300 N/C. (a) Find the acceleration of the electron. (b) Use the equations of motion with constant acceleration to find the electron’s speed after 1.00 × 10–8 s, assuming it starts from rest.

9. A proton accelerates from rest in a uniform electric field of 640.0 N/C. At some later time, its speed is 1.20 × 106 m/s.

(a) Find the magnitude of the acceleration of the proton. [Ans: (a) 6.12 x 1010 ms-2

(b) How long does it take the proton to reach this speed? (b) 19.6 µs (c) How far has it moved in this interval? (c) 11.8 m (d) What is its kinetic energy at the later time (d) 1/20 x 10-15 J ]

11. A charge of 6.0 x 10-9 C and a charge of – 3.0 x 10-9 C are separated by a distance of

60.0 cm Find the position at which a third charge of 12.0 x10-9 C can be placed so that the net electrostatic force on it is zero.

12. Determine what the mass of a proton would be if the gravitational force between two

of them were equal to the electrical force between them.

13.A 4.5 × 10–9 C charge is located 3.2 m from a –2.8 × 10–9 C charge. Find the electrostatic force exerted by one charge on the other.



14.An alpha particle (charge = +2.0e) is sent at high speed toward a gold nucleus (charge = +79e). What is the electrical force acting on the alpha particle when it is 2.0 × 10–14 m from the gold nucleus?

15.Calculate the magnitude and direction of the Coulomb force on each of the three charges in the Figure .

16.Two small metallic spheres, each of mass 0.20 g, are suspended as pendulums by light strings from a common point as shown in Figure P15.15. The spheres are given the same electric charge, and it is found that the two come to equilibrium when each string is at an angle of 5.0° with the vertical. If each string is 30.0 cm long, what is the magnitude of the charge on each sphere?

17.An electron is accelerated by a constant electric field of magnitude 300 N/C.

(a) Find the acceleration of the electron.

(b) Use the equations of motion with constant acceleration to find the electron’s


Reduced it ¼ of its previous value


speed after 1.00 × 10–8 s, assuming it starts from rest.

,

18.In Figure P15.27, determine the point (other than infinity) at which the total

electric field is zero.




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Chapter 8 - Properties of Electric Charges-Latest