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16.1 Coulomb’s Law 16.2 Electric Field 16.3 Charge in a uniform electric field 16.4 Electric Potential

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## 16.1 Coulomb’s Law 16.2 Electric Field 16.3 Charge in a uniform electric field 16.4 Electric Potential

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**UNIT 16 : ELECTROSTATICS**(4 hours) 16.1 Coulomb’s Law 16.2 Electric Field 16.3 Charge in a uniform electric field 16.4 Electric Potential Electrostatics - the study of electric charges at rest, the forces between them and the electric fields associated with them.**SUBTOPIC :**16.1 Coulomb’s Law (1 hour) LEARNING OUTCOMES : At the end of this lesson, the students should be able to : • State Coulomb’s law, • b) Sketch the force diagram and apply Coulomb’s law • for a system of point charges.**16.1 Coulomb’s Law**• Coulomb’s law states that the force,F between two point • charges q1 and q2 is directly proportional to the product • of their magnitudes and inversely proportional to the • square of the distance separating them, r. • In equation form : attractive force repulsive force electrostatic force Coulomb’s law equation • Unit : N(Newton) • Vector quantity**+**+ - + 16.1 Coulomb’s Law F12 F21 F12 F21 repulsive force attractive force F12 : the force on charge q1 due to charge q2. F21 : the force on charge q2 due to charge q1.**16.1 Coulomb’s Law**Graphs above show the variation of electrostatic force with the distance between two charges. Note: • The sign of the charge can be ignored when substituting into the Coulomb’s law equation. • The sign of the charges is important in distinguishing the • direction of the electric force.**+**- 16.1 Coulomb’s Law Example 16.1 Two point charges, q1=-20 nC and q2=90 nC, are separated by a distance of 4.0 cm as shown in figure below. Find the magnitude and direction of a) the electric force on q1 due to q2. b) the electric force on q2 due to q1. c) the electric force on each charge. (Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)**-**16.1 Coulomb’s Law Solution 16.1 q1= 2.0 x 10-8C, q2= 9.0 x 10-8 C, r = 4.0 x 10-2 m + a) b)**+**+ - 16.1 Coulomb’s Law Example 16.2 Three point charges lie along the x-axis as shown in figure below. Calculate the magnitude and direction of the net electric force exerted on q2. (Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)**+**+ - 16.1 Coulomb’s Law Solution 16.2 r21= 3.0 x 10-2 m, r23= 5.0 x 10-2 m**+**- - 16.1 Coulomb’s Law Example 16.3 Figure below shows the three point charges are placed in the shape of triangular. Determine the magnitude and direction of the resultant electric force exerted on q1. Given q1=-1.2 C, q2=+3.7 C, q3 =-2.3 C, r12=15 cm, r13=10 cm, =32 and k = 9.0 x 109 N m2 C-2.**+**- - 16.1 Coulomb’s Law Solution 16.3 q1= 1.2x10-6 C, q2= 3.7x10-6 C, q3= 2.3x10-6 C, r12= 15x10-2 m, r13= 10x10-2 m**+**- - 16.1 Coulomb’s Law**+**+ + 4m - + - 3m 16.1 Coulomb’s Law Exercise 1. Two point charges of -1.0 x 10 -6 C and +2.0 x 10 -6 C are separated by a distance of 0.30 m. What is the electrostatics force on each particle ? (0.20 N , directed to one another) 2. Calculate the net electrostatics force on charge i ) q1 ii) q2 iii) q3 (28.7 N , to the left , 116.25 N to the right , 87.55 N to the right) 3. Calculate the net electrostatics force on charge q2. (4.11 x10-3 N , 18.4 o below the x-axis)**+**+ + + + + + 16.1 Coulomb’s Law 4. What is the electrostatics force on charge q3? (0,+0.30 m) (0.43 N in the +x direction) (0,-0.30 m) 5. Four identical point charges (q = +10.0 C) are located on the corners of a rectangle as shown in figure below. The dimension of the rectangle are l = 60.0 cm and w = 15.0 cm. Calculate the magnitude and direction of the resultant electric force exerted on the charge at the lower left corner by the other three charges. (40.9 N at 263)**SUBTOPIC :**16.2 Electric field (1 hour) LEARNING OUTCOMES : At the end of this lesson, the students should be able to : • Define electric field. • Define and use electric field strength, . • Sketch the electric field lines of isolated point charge, • two charges and uniformly charged parallel plates. • d) Sketch the electric field strength diagram and determine electric field strength for a system of charges.**16.2 Electric field**Definition of electric field : A ………….....of space around isolated charge where an …..……..………..……. is experienced if a ………………………………. placed in the region. • Electric field around charges can be represented by drawing a series of lines. These lines are called electric field lines (lines of force). • The direction of electric field is tangent to the electric field line at each point.**Field direction**-q +q 16.2 Electric field • Figures below show the electric field patterns around the charge. a. Single positive charge b. Single negative charge the lines point radially ……….……… from the charge the lines point radially ………………… toward the charge**X**16.2 Electric field • c. Two equal point charges of opposite sign, +q and -q Field direction +q -q the lines are curved and they are directed from the positive charge to the negative charge. d. Two equal positive charges, +q and +q • (point X is neutral point ) • Is defined as a point (region) where the resultant electric force is zero. • It lies along the vertical dash line. Field direction +q +q**Field direction**-q +2q • e. Two opposite unequal charges, +2q and -q 16.2 Electric field • note that twice as many lines leave +2q as there are lines entering –q. • number of lines is proportional to magnitude of charge. • f. Two opposite charged parallel metal plates • The electric field lines are perpendicular to the surface of the metal plates. • The lines go directly from positive plate to the negative plate. • The field lines are parallel and equally spaced in the central region.Thus, in the central region, the electric field has the same magnitude at all points (uniform) except near the edges.**16.2 Electric field**The properties of electric field lines: • The field lines indicate the direction of the electric field (the field points in the direction ………………..to the field line at any point). • The lines are drawn so that the magnitude of electric field is proportional to the number of lines crossing unit area perpendicular to the lines. The closer the lines, the ……………………… the field. • Electric field lines start on …………….. charges and end on ……………… charges, and the number starting or ending is proportional to the magnitude of the charge. • The field lines never …………. because the electric field does not have two values at the same point.**16.2 Electric field**Electric field strength • The electric field strength at a point is defined as • the electric force (electrostatic) per unit positive • test charge that acts at that point in the same direction • as the force. where • It is a vector quantity. • The units of electric field strength is N C-1 or V m-1. • The direction of the electric field strength, E depends • on the sign of isolated charge.**q**q 16.2 Electric field From and A positive isolated point charge. thus • In the calculation of magnitude E, substitute the MAGNITUDE of the charge only. A negative isolated point charge.**16.2 Electric field**Example 16.5 2.0 cm + A Calculate the electric field strength at point A, 2.0 cm from a point charge q1. (Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2) Solution**16.2 Electric field**Example 16.6 2.0 cm - A Calculate the electric field strength at point A, 2.0 cm from a point charge q1. Solution**-**+ 16.2 Electric field Example 16.6 Two point charges, q1=1.0 C and q2=-4.0 C, are placed 2.0 cm and 3.0 cm from the point A respectively as shown in figure above. Find a) the magnitude and direction of the electric field intensity at point A. b) the resultant electric force exerted on q0=4.0 C if it is placed at point A. (Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)**+**16.2 Electric field Solution 16.6 - a) The electric field strength at point A due to the charges is given by**+**16.2 Electric field Solution 16.6 - b)**-**- 16.2 Electric field Example 16.7 Two point charges, q1=-1.0 C and q2=-4.0 C, are placed 2.0 cm and 3.0 cm from the point A respectively as shown in figure above. Find a) the magnitude and direction of the electric field intensity at point A. b) the resultant electric force exerted on q0=4.0 C if it is placed at point A. (Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)**-**16.2 Electric field Solution 16.7 - a) Direction : to the left (q1) Direction : to the right (q2) The electric field strength at point A due to the charges is given by Direction : to the right (q2)**-**16.2 Electric field Solution 16.7 - b) Direction : to the right (q2)**+**- 16.2 Electric field Example 16.8 - Three charges are placed on three corners of a square, as shown above. Each side of the square is 30.0 cm. Calculate the electric field strength at point A. What would be the force on a 6.00 µC charge placed at the point A?**+**- 16.2 Electric field Solution16.8 30.0 cm - 30.0 cm EA2 30.0 cm 42.4 cm EA3 EA1**+**- 16.2 Electric field Solution 16.8 30.0 cm - 30.0 cm EA2 30.0 cm 42.4 cm 45o EA3 E EA1**+**16.2 Electric field Exercise • Determine • a) the electric field strength at a point X at a distance 20 cm from a • point charge Q = + 6µC. (1.4 x 10 6 N/C) • b) the electric force that acts on a point charge q= -0.20 µC placed at • point X. (0.28 N towards Q) • 2. - • Two point charges, q1 = +2.0 C and q2 = -3.0 C, are separeted by a distance of 40 cm, as shown in figure above. Determine • The resultant electric field strength at point X. • (1.13 x 103kN C-1 towards q2) • b) The electric force that acts on a point charge q = 0.50 µC placed at X. • (0.57 N)**16.2 Electric field**Exercise • 3. Find the magnitude of the electric field at point P due to the four point charges as shown in the figure below if q=1 nC and d=1 cm. • 4. Find the magnitude and direction • of the electric field at the centre of • the square in figure below if • q=1.0x10-8 C and a= 5cm. (Given 0=8.85 x 10-12 C2 N-1 m-2) (HRW. pg. 540.11) Ans. : zero. (Given 0=8.85 x 10-12 C2 N-1 m-2) (HRW. pg. 540.13) Ans. : 1.02x105 N C-1,upwards.**SUBTOPIC :**16.3 Charge in a uniform electric field (1 hour) LEARNING OUTCOMES : At the end of this lesson, the students should be able to : • Explain quantitatively with the aid of a diagram the motion of a charge in a uniform electric field. Cases : 1. stationary charge 2. charge moving perpendicularly to the field 3. charge moving parallel to the field 4. charge in dynamic equilibrium**Figure a : Case 1**16.3 Charge in a uniform electric field Case 1 : Stationary charge Case 4 : Charge in dynamic equilibrium (moves in straight line, consider the weigh, W of the particle) vo v=0, a=0 Figure b : Case 4 • Figure a and Figure b show a particle with positive charge q is held stationary • and moves at constant speed respectively, in a uniform electric field, E . • The forces acted on the particle are electrostatic force (upwards) • and weight (downwards). • For the particle in static equilibrium (Figure a) and dynamic • equilibrium : moves horizontally in straight line (Figure b), • electrostatic force = weight • FE= W • qE = mg**16.3 Charge in a uniform electric field**Case 2 : Charge moving perpendicularly to the field - vx FE θ vy Figure above shows the path of an electron qo which enters the uniform electric field between two parallel metal plates with a velocity vo. • When the electron enters the electric field , E the only • force that acts on the electron is electrostatic force , FE=qE.**16.3 Charge in a uniform electric field**• This causes the electron of mass m to accelerate downwards • with an acceleration a. • Since the horizontal component of the velocity of the • electron remains unchanged as vo , the path of the electron • in the uniform electric field is a parabola. • The time taken for the electron to tranverse the electric field • is given by**16.3 Charge in a uniform electric field**• The vertical component of the velocity vy, when the electron • emerges from the electric field is given by • After emerging from the electric field, the electron travels • with constant velocity v, where**16.3 Charge in a uniform electric field**• The direction of the velocity v is at an angle • The position of the electron at time t is or**16.3 Charge in a uniform electric field**Example 16.9 + 20 mm uo = 1.5 x 107m/s - 60 mm • An electron travelling at speed of 1.5 x 107 m/s enters the • space between two parallel metal plates 60 mm long. The electric field between the plates is 4.0 x 103 V/m. • Sketch the path of the electron in between plates, and • after emerging from the space between the plates. • Find the magnitude and direction of the acceleration of the electron in between the plates. • Calculate the vertical and horizontal components of the electron velocity when it emerges from the space between the plates. • Find the angle of deflection of the electron beam.**16.3 Charge in a uniform electric field**Solution 16.9 + 20 mm uo = 1.5 x 107m/s - 60 mm • DIY**16.3 Charge in a uniform electric field**Solution 16.9 c) Vertical component of velocity, Horizontal component of velocity, d) Angle of deflection,**Figure 1**Figure 2 16.3 Charge in a uniform electric field Case 3 : Charge moving parallel to the field Consider 2 cases as shown in figure 1 and figure 2**16.3 Charge in a uniform electric field**• Figure 1 • A particle with positive charge q moves with initial velocity vo • towards negative plate (upwards). • For a positive charge, its acceleration a is in • the direction of the electric field is given by • Since the direction of electric force is upward • (same direction and parallel to the direction of motion of • particle), the particle is then accelerated in straight line • towards negative plate with speed v, where v > vo .**16.3 Charge in a uniform electric field**• The velocity v of the particle after time t is given by • The displacement s after time t is given by • The displacement s in terms of velocity v is given by**16.3 Charge in a uniform electric field**• Figure 2 • A particle with negative charge q moves with initial velocity vo • towards positive plate (downwards). • For a negative charge, its acceleration is in • the directionopposite the electric field is given by • Since the direction of electric force is downwards, • (same direction and parallel to the direction of motion of • particle), the particle is then accelerated in straight line • towards positive plate with speed v, where v > vo**16.3 Charge in a uniform electric field**• The velocity v of the particle after time t is given by • The displacement s after time t is given by • The displacement s in terms of velocity v is given by

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