Section 1 Static Electricity: Electric Charge: Attraction and Repulsion Law:

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Chapter 21Section 1 Static electricity: Electric Charge: Attraction and Repulsion Law: Unlike charges attract and like charges repel. The law of conservation of electric charge: The net amount of electric charge produced in any process is zero. Section 2 Electric charge in the atom: Q1 Describe the structure of neutral atom. Q2 Why a plastic ruler becomes negatively charged by rubbing with a wool towel? Because the electrons are transferred from the wool towel to the plastic ruler by rubbing and the plastic becomes negative. Q3 Why a glass rod becomes positively charged by rubbing with a silk towel? Because the electrons are transferred from the glass rod to the silk towel by rubbing and the glass rod become negative. Section 3 and 4 Insulators and conductors: Methods of charging objects: 1- by rubbing 2- by contact 3- by induction The electroscope Section 5 Coulombs Law: it gives the force between two point charges Q1 and Q2, a distance r apart by the following equation: F=kQ1Q2/R2Q1 r Q2

Where k is the constant in this equation in the case of the air and equal to 9x109 N.m2/c2 Also K in term of the constant E permittivity of free space is equal to 1/4 E where -12 C2/N2 E 1/4= K= 8.8 x 10 Examples: 1,2,3,4 pages 551 and 552 Section 6 The Electric Field: Is the space surrounding a charge or group of charges when Electric forces exist due to it or them.

The Electric Field E means the strength of the electric field, at any point in a space is defined as the force per unit charge that would act on a test charge q placed at that point (a) where: + a E r E=F/q N/C Or E=KQ/R2 N/C +Q r+q A F

If the field is due to more than one charge, the total field at any point vectorially: E=E1+E2+E3+. Where E is a vector quantity whose direction is the direction of the force acting on a positive test charge. Examples 7& 8 page 556 Section 8 Field lines or lines of force: In order to visualizes the electric field, we draw a series of lines to indicate the direction of the electric field at various points in space. Toes lines are drown to indicate the path of a positive test charge moving freely in the field as shown in the following figures:

The properties of field lines: 1- The field lines indicate the direction of the electric field. 2- The direction of the tangent at any point on the field line points to the direction of the field. 3- The lines are drawn so that the magnitude of the electric field E is propotional to the number of lines crossing unit area perpendicular to the lines. 4- The closer the lines the stronger the field. 5- The field lines never cross, because the electric field has only one value at the same point. 6- The electric field lines start on positive charges and end on negative charges. Section 10 Motion of a charged particle in a uniform electric field Examples: 14 page 564 + + + + + + + +

Home work 11- a positively charged rod is bought close to a neutral piece of paper which attracts it. Draw a diagram that shows the separation of charge and explain why attraction occurs?

-

+

+ + +

The attraction occurs due to the separation of charge as shown in the figure where the electrons attracted toward to the positive charged rod and the positive charges repel to the far end of the piece of the paper which results the attraction force to be bigger than the repulsion force and there for attraction occurs. 2- When a charged ruler attracts small pieces of paper sometimes a piece jumps quickly away after touching the ruler why? In the first question we explained why the neutral piece of paper attracted to charged rod. Now when the piece of paper touches the charged ruler some charges transfer from the ruler to the piece of paper which became charged by touch with the same kind of charges. Therefore this piece jumps quickly away by the repulsion force. 3-make use with the following figure: 10CMQ2=-2x10-6

10CMq1=+2x10 -6 q3=+5x10 -6

Compute: F31 a) the force acting on q3 due to q1 (F13) F32 b) the force acting on q3 due to q2 (F12) c) the net force acting on q3 (F net) F31 =k q1 q3/r2= 9x 109 x210-6 x5x10-6 /0.1=9N (repulsion) F32= k q2 q3/r2= 9x 109x210-6 x5x10-6 /0.2=2.25N (attraction) F net = F31- F32= 9- 2.25= 6.75N (repulsion)

F(net)

Chapter 22Electric Flux The electric flux through a planar area is defined as the flux field times the component of the area perpendicular to the field. When the area is used in a vector operation like this, it is understood that the magnitude of the vector is equal to the area and the direction of the vector is perpendicular to the area. Note: The previous figure shows the general case where the area A make an angle uniform electric field E. =E A cos If If N/C. m2 = 0 the area A becomes perpendicular to E, hence = E A N/C. m2 = 90 the area A becomes parallel to E and cos 90=0, =0 with the

Gausss Law: The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The total electric flux through closed surface is directly proportional to the total charge exist inside the close surface.

E^ = E^A

, =Q/E Q

Sum of flux proportional to total charge enclosed

Home work 2Answer the following questions: 1- define the following concepts in term of flux: It is defined as the electric field times the component of the area perpendicular to the field. 2- a uniform electric field E= 200N/C passing through the surface of the rectangle ( its dimensions 10Cm, 20Cm) calculate the electric flux in the following cases: a) When the rectangle is perpendicular to the electric field? = E A= 200 x (0.1x 0.2)= 4N/C m2 b) When the rectangle makes an angle of = 30 with E?

=E A cos = 200 x (0.1x0.2) cos 30=3.46N/C m2 c) When the rectangle is parallel to the electric field E? =zero 3- define the Gausss Law and then write the conclusion of its math expression? Gausss Law the total electric flux through closed surface is directly proportional to the total charge exist inside the closed surface. = Q/ E 4- a 3m straight wire passes uniform positive charge 5x 10-11 C. Calculate the electric field at point 2 Cm above the wire far from the ends. = EA= Q/ E A= 2 r x l = 3.77 m2 E .E.A = Q E=Q/A. E 5 = x 10-11/ ( 8.85x10-12 x 3.77)= 1.49x 10-3 Er=2cm

r

+ + +_+ + + + + +L=3Cm

Chapter 23 The Electric Potential1)As shown in the following figure, if we placed test charge +q at (a) it acquires Potential Energy Ua and at (b) Ub, therefore we can define Potential Energy as the capacity for doing work which arises from the position at (a) or (b) this work is equal to the electric force times the distance ( r) required for moving test charge +q from position to another.

a b rm QC +qc

2) Electric potential (v) is defined as potential per unit charge. V=U/ q J/C

3) Electric Potential Difference between two points a and b is defined as the difference in potential energy of a test charge q placed at those two points divided by the charge q. Va- Vb= W/q J/C or volt a 4) Electric potential V due to point charge Q+: a +Q rm Va= +KQ/r volt If the point charge was Q Va= -KQ/r volt -Q rm 5) Electric Potential V due to several point charges is equal to algebraic sum: V net =-+V1-+ V2-+ V3 -+.. 6) The electric potential for charged conducting sphere Example 4 page 596 Va= +KQ/r=Vb Vc= +KQ/r volt Vd=+KQ/r volt 7) Equipotential surface: Is one on which all points are at the same potential. The potential difference between any two points on equipotential surface is zero and no work is required to move a charge from one point to the other. An equipotential surface should be perpendicular on the electric field.

Home Work 3Answer the following questions: 1- Define the following concepts: a)The potential energy: we can define Potential Energy as the capacity for doing workwhich arises from the position at (a) or (b) this work is equal to the electric force times the distance ( r) required for moving test charge +q from position to another.

b) The electric potential: is defined as potential per unit charge. V=U/ q J/C c) The electric potential difference between two points a and b is defined as the difference in potential energy of a test charge q placed at those two points divided by the charge q. d) The equipotential surface: Is one on which all points are at the same potential. 2) Draw the diagram for E and V versus due to a charge conducting sphere.

3) Make use with the following figure: C

10Cm

10Cm

A+q=2oMc

10Cm

B-q=2oMc

10Cm

D

a) Calculate the electric potential at the points D and C. At C v total= kq1/r + (- kq2/r)= 1.8 x 106 - 1.8 x 106 =zero At D v total= kq1/r + (- kq2/r)= 9 x 109 x 20 x 10-6 / 0.2 - 9 x 109 x 20 x 10-6 / 0.1 = -0.9 x 106 Volt b) Find the electric potential difference between D and C. VCD= VC-VD= zero (-900 000 )= 900 000 Volt c) Calculate the work that is required to bring a charge q= +20 mC from the point C to D. W= VCD x q = 20 x 900000= 18 000 000 Joule 3) Draw the method to produce a uniform electric field and right the relation you use to find the E.

+ + + d

E=V/d V/ m or N/C

+ EMF

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