ELECTRICITY and MAGNETISM

NỘI DUNG MÔN HỌC Chương 1. Trường tĩnh điện Chương 2. Vật dẫn trong điện trường Chương 3. Dòng điện không đổi Chương 4. Từ trường tĩnh Chương 5. Cảm ứng điện từ

TÍNH ĐIỂM MÔN HỌC Điểm thường kỳ: là tổ hợp của: Bài kiểm tra trên lớp. Chuyên cần. Thái độ học tập. Điểm chấp nhận phải từ 3 trở lên. Lưu ý: không nghỉ học buổi nào, thái độ học tập tốt, không vi phạm qui chế sẽ được cộng thêm từ 1 đến 1,5 điểm TK và nếu điểm giữa kỳ nhỏ hơn 4, lớn hơn hoặc bằng 3 vẫn được dự thi cuối kỳ.

Thi giữa kỳ: Nội dung thi: Chương1,2. Hình thức thi: Trắc nghiệm và tự luận, thời gian 45’. Thời gian thi: Dự kiến vào tuần thứ 7 (thực học). Điểm đạt 4 trở lên (thi một lần duy nhất) Thi cuối kỳ: Nội dung thi: Chương 3,4 Hình thức thi: Trắc nghiệm và tự luận, thời gian 60’ Điểm đạt 4 trở lên (thi một lần duy nhất)

Điều kiện để được thi cuối kỳ: Mắc phải một trong những điều sau đây sẽ không được dự thi cuối kỳ: Nghỉ quá 20% tổng số tiết học ( quá 6 tiết). Điểm thi giữa kỳ nhỏ hơn 3. Điểm thường kỳ dưới 3.

Chapter 1 ELECTRIC FIELDS

1 – Properties of electric charges 2 – Coulomb’s Law 3 – The electric field 4 – Electric field lines

1 – Properties of electric charges -There are two kinds of charges in nature : charge positive and charge negative.

- Charges of the same sign repel one another and charges with opposite signs attract one another.

- Total charge in an isolated sytems is conserved - Electric charge has the smallest value is call element charge. q = Ne where N is some interger e = -1, C m e = 9, kg

2 – Coulomb’s Law

Electric force between two stationary charged particles:

k = (Nm 2 /C 2 ) Coulomb constant r : space between two point charges (m) is known as the permitivity of free space

Exercise There are two point electric charges: q 1 = C at A, q 2 = C at B, for from AB = 10cm. Find the electric force exerted by q 1 on q 2 ? The electric force exerted by q 1 on q 2 is:

To calculate the electric force due to a group of point charges, the vector sum of the electric force of all the charges exerted by q is:

Plus two vector:

Exercise Consider three point charges located at the corners of a right triangle as shown in figure, where q 1 = q 3 = 5μC, q 2 = -2μC, and a = 0,1m. Find resultant force exerted on q 3. a a q2q2 q1q1 q3q3 The resultant force exerted on q 3 is:

Exercise Three point charges lie along the x axis. The positive charge q 1 = 15μC is at x = 2m, the positive charge q 2 = 6μC is at the origin, and the net force acting on q 3 is zero. What is the x coordinate of q 3. + x q2q2 + q1q1 x = 2m O The x coordinate of q 3 is:

3 – The electric field Electric fields is the region of space around a charged object the source charge. When another charged obejct the test charge enters this electric field, an electric force acts on it.

The electric field vector at a point in space is defined as the electric force F acting on a positive test charge q 0 placed at that point divided by the test charge:

If the source charge q positive, the situation with the test charge removed the source charge sets up an electric field at point P, directed away from q. + P * The electric field created by q is positive : The force is in the same direction as the field

- P 0q  If q is negative, the force on the test charge is toward the sourse charge, so the electric field at P is directed toward the sourse charge. * The electric field created by q is negative: The force and the field are in opposite directions.

r : is the space from point P to the source charge q (m) + P

* The electric field at point due to a group of sourse charges can be expressed as the vetor sum: r i is the distance from the i th sourse charge q i to the point P (m). * The charge distribution is modeled as continuous, the total field at P in the lim it Δq i → 0 :

. If a charge q is unifromly distributed throughout a volume V, the volume charge density ρ is defined by: ρ = q/V (C/m 3 ). If a charge q is unifromly distributed throughout on a surface of area A, the surface charge density σ is defined by: σ = q/A (C/m 2 ). If a charge q is unifromly distributed a long a line of length ℓ, the linear charge density σ is defined by: ℓ = q/λ (C/m 2 )

. If the charge q is nonuniformly distributed over a volume, surface, or line, throughout a volume V, the of charge dq in a small volume, surface, or length element are:

In the air, there are a point electric charges Q = C. Caculate the electric field at the point P to electric charge Q is 30cm ? Exercise The electric field at the point P to electric charge Q is 30cm:

Exercise There are two point electric charges: q 1 = C at A, q 2 = C at B, for from AB = 10cm. At P, with PA = PB = 5cm. Find the electric field at the point P? r 1 = r 2 = 0,05m q1q1 q2q2 B - + A P

Exercise: There are two point electric charges: q 1 = 8µC at A, q 2 = -6µC at B, for from AB = 10cm. At P, with PA = 8cm, PB = 6cm. Find the electric field at P ? + - q1q1 q2q2 P r1r1 r2r2 A B

Exercise A point charge of -4nC is located at (0, 1)m. What is the x component of the electric field due to the point charge at (4,-2)m?

Exercise 1. Put a point electric charge Q = - 5μC is at the origin. Caculate the electric field at point P located (4,-3)m? 2. There are two point electric charges: q 1 = C at A, q 2 = C at B, for from AB = 20cm. At point P is a centre of AB. Caculate : a. The electric field at point P. Draw a figure. b. The electric force affecting on electron put at P.

Exercise Four charged particles are at the corners of a square of side a. Determine: a.The total electric field at the centre of square. b.The total electric force exerted on electron at the centre of square q 3q 4q q

4 – Electric field lines The electric field vector E is tangent to the electric field line at each point. The line has a direction, indicated by an arrowhead, that is the same as that of the electric vector.

The number of lines per unit area through a surface perpendicular to the lines is proportional to the magnitude of the electric field in the region. Thus, the field lines are close together where the electric field is strong and far apart where the field is weak. A B E A > E B

The rules for drawing electric field lines are as follows:. The lines must begin on a positive charge and terminate on a negative charge. In the case of an excess of one type of charge, some lines will begin or end infinitely far away.. The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge.. No two field lines can cross.

Electric field lines

Chapter 2 GAUSS’S LAW

1 – Electric Flux 2 – Gauss’s Law 3 – Application of Gauss’s Law to various charge distributions 4 – Conductors in electrostatic equilibrium

1. Electric Flux Φ E (Vm) is proportional to the number of electric field lines that penetrate a surface.

θ dA A is area (m 2 ) If the electric field is uniform and makes an angle θ with the normal to a surface of area A, the electric flux through the surface is:

Exercise. In the electric field is uniform E = V/m has A = 20cm 2 is area of suface and. Find the electric flux send to a area A.

Exercise. A flat surface of area 3,2m 2 is rotated in a unifrom electric field of magnitude E = 6, V/m. Determine the electric flux through this area : a. When the electric field is perpendicular to the surface. b. When the electric field is parallel to the surface.

2. Gauss’s Law Electric flux Φ E through any closed gaussian surface is equal to the net charge q in inside the surface divided by ɛ 0 :

Exercise. There are three point electric charges: q 1 = C, q 2 = C, q 3 = C put inside the sphere, radius 30cm. Find the electric flux send to plane of sphere it. The electric flux send to plane of sphere is:

Exercise. There are three point electric charges: q 1 = C at A, q 2 = C at B, q 3 = C at C. For AB = 20cm and BC = 30cm. Caculate the electric flux send to plane of sphere has centre B, radius 25cm A B C q1q1 q2q2 q3q3 The electric flux send to plane of sphere has the radius 25cm is:

Exercise. Give a right triangle ABC, AB = BC = 10cm, put electric charge q 1 = C at A and electric charge q 2 = C at C. Caculate : a. The electric field at point B. b. The electric force affecting on electron put at B. c. The electric flux send to plane of sphere has centre C, radius 5cm

Chapter 3 ELECTRIC POTENTIAL

1 – Potential difference and electric potential 2 – Potential difference in a uniform electric field 3 – Electric potential and potential energy due to point charge 4 – Obtaining the value of the electric field form the electric potential

+ + q Q r A B 1 – Potential difference and electric potential When a positive test charge q is moved between points A and B in an electric field E of electric charge Q, the potential energy (J) of the charge field system is: rArA rBrB

+ + q Q r A B rArA rBrB r A is the space from point A to the source charge Q (m) r B is the space from point B to the source charge Q (m)

On the other side: U AB : is the potential difference between points A & B (V) V A & V B : is the electric potential at point A & B (V) = V A - V B

Comment If the first point A coincide the final point B (A≡ B) then W = 0 and U AB = 0 → V A = V B The electric energy a charge – field sytems do not depend on the shape way, only depend on the first point and the final point.

- + q B A Q C Exercise there are two point electric charges: Q = C at A, q = C at B, for from AB = 90cm. Caculate the potential energy of the charge – field system when charge q is moved from B to C ? BC = 10cm r B = AB, r C = AC

If we choose V = 0 at r = ∞ (V ∞ = 0 ), the electric potential due to a point charge at any distance r from the charge is: r: is the space from point caculate the electric potential to the source charge Q (m) * The electric potential at a point due to a point charge Q: Notice

The total electric potential at some point due to several point charges is the sum of the potential due to the individual charges. For a group of point charges, we can write the total electric potential : r i : is the distance from the point caculate the electric potential to the source charge Q i (m) * The electric potential at a point due to several point charges:

Exercise. There are three point electric charges: q 1 = C, q 2 = C, q 3 = C put at three the tops of regular triangle, side a = 20cm. Choose V ∞ = 0. Caculate: a. Electric potental at centre triangle. b. The potential energy of the charge – field system when a electron is moved from centre triangle far to endless.

q 1 = C, q 2 = C, q 3 = C, a = 20cm. a. Electric potental at centre triangle is: O r r r

b. The potential energy of the charge – field system when a electron is moved from centre triangle far to endless is: O r r r

Exercise. Put two point electric charges q 1 = C at A and q 2 = C at B, for AB = 10cm. Point C consistant with AB a right triangle at C. Choose V ∞ = 0. Caculate 1. The electric potential at centre point of AB. 2. The potential difference U MC = ?

Problem. Put two point electric charges q 1 = C at A and q 2 = C at B, for AB = 10cm. Point C is centre of AB. Choose V ∞ = 0. Caculate : a. The electric force exerted by q 1 on q 2 ? b. The electric field at C. c. The electric flux send to plane of sphere has centre B, radius 15cm. d. The electric potential at point C. e. The potential energy of the charge – field system when a electric charges q 0 = C moved from point C far to endless.

Because E = constant where r = |s| is space of the electric charge q moved between points A and B (m) 2 – Potential difference in a uniform electric field

A B V A > V B ; E A > E B A B V A E B A B C V A = V B > V C E A = E B = E C Electric field lines always point in the direction of decreasing electric potential

U AB = E.AC = - 240V U BA = -U AC = 240 V U CB =V C – V B = 0 BC = 3cm, AC = 4cm A C B Exercise

71 1 – Sơn tĩnh điện: MỘT SỐ ỨNG DỤNG CỦA TĨNH ĐIỆN Negatively charged paint adheres to positively charged metal. Fine mist of negatively charged gold particles adhere to positively charged protein on fingerprint.

72 2 – Làm sạch không khí: MỘT SỐ ỨNG DỤNG CỦA TĨNH ĐIỆN

73 3 – Băng dính: MỘT SỐ ỨNG DỤNG CỦA TĨNH ĐIỆN

74 4 – Kỹ thuật photocopy: MỘT SỐ ỨNG DỤNG CỦA TĨNH ĐIỆN The drum is an aluminum cylinder coated with a thin layer of Selenium. Aluminum is a conductor. Selenium is a photoconductor, it is an insulator in the dark and a conductor when exposed to lightlight. So, a positive charge deposited on the Selenium layer will stay there. However, when the drum is esposed to light, electrons from the aluminum will pass through the conducting selenium and neutralize the positive charge. 1. Charging the drum 2. Imaging the document on the drum 3. Fixing the toner 4. Transferring the toner to the paper.

75 5 – Kỹ thuật in phun: MỘT SỐ ỨNG DỤNG CỦA TĨNH ĐIỆN

76 6 – Kỹ thuật in Laser: MỘT SỐ ỨNG DỤNG CỦA TĨNH ĐIỆN

77 7 – Đèn hình TV: MỘT SỐ ỨNG DỤNG CỦA TĨNH ĐIỆN

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