VIEWS: 65 PAGES: 95 POSTED ON: 10/4/2012 Public Domain
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. 1 SUBTOPIC : 16.1 Coulomb’s Law (1 hour) LEARNING OUTCOMES : At the end of this lesson, the students should be able to : Qq kQq a) State Coulomb’s law, F 2 4o r 2 r b) Sketch the force diagram and apply Coulomb’s law for a system of point charges. 2 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. F q1q2 1 F r2 kq1q2 attractive force • In equation form : F repulsive force r2 electrostatic force Coulomb’s law equation • Unit : N(Newton) • Vector quantity 3 16.1 Coulomb’s Law kq1q2 F k 1 r2 4 0 0 : permittivi ty of free space (called proportion ality constant (vacuum or air) or Coulomb' s constant) ( 0 8.85x1012 C2 N1 m2 ) k 9.0 x 10 9 N m 2 C 2 q1 F12 F21 q2 F12 q1 q2 F21 + - + + r r attractive force repulsive force F12 : the force on charge q1 due to charge q2. F21 : the force on charge q2 due to charge q1. kq1q2 F12 F21 r2 4 Example of force diagram 5 16.1 Coulomb’s Law kq1q2 F F 2 F r Gradient, M kq1q2 1 0 r 0 r2 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. 6 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. q1 - + q2 4.0 cm 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) 7 16.1 Coulomb’s Law Solution 16.1 q1= 2.0 x 10-8 C, q2= 9.0 x 10-8 C, r = 4.0 x 10-2 m F12 F21 q1 - + q2 kq1q2 4.0 cm a) F12 2 r kq2 q1 b) F21 2 r 8 16.1 Coulomb’s Law Example 16.2 Three point charges lie along the x-axis as shown in figure below. q1 2 C q2 4 C q3 6 C + + - 3.0 cm 5.0 cm 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) 9 16.1 Coulomb’s Law Solution 16.2 r21= 3.0 x 10-2 m, r23= 5.0 x 10-2 m q1 2 C q2 4 C F23 q3 6 C + + F21 - 3.0 cm 5.0 cm kq2 q1 F21 2 r21 F21 kq2 q3 F23 2 r23 F23 10 16.1 Coulomb’s Law Solution 16.2 r21= 3.0 x 10-2 m, r23= 5.0 x 10-2 m q1 2 C q2 4 C F23 q3 6 C + + F21 - 3.0 cm 5.0 cm Magnitude ; F2 F21 F23 F2 F21 F23 11 16.1 Coulomb’s Law Example 16.3 Three point charges lie along the x-axis as shown in figure below. q1 2 C q2 4 C q3 6 C + - - 3.0 cm 5.0 cm 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) 12 16.1 Coulomb’s Law Solution 16.3 r21= 3.0 x 10-2 m, r23= 5.0 x 10-2 m q1 2 C 23 q2 4 C F q3 6 C + F21 - - 3.0 cm 5.0 cm kq2 q1 F21 2 r21 Magnitude ; F2 F21 F23 F2 F21 F23 kq2 q3 F23 2 r23 13 16.1 Coulomb’s Law Example 16.4 Three point charges lie along the x-axis as shown in figure below. q1 2 C q2 4 C q3 6 C + + - 3.0 cm 5.0 cm Calculate the magnitude and direction of the net electric force exerted on q1. (Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2) 14 16.1 Coulomb’s Law Solution 16.4 r21= 3.0 x 10-2 m, r23= 5.0 x 10-2 m q 4C q1 2 C F 2 q3 6 C + 13 + - F12 3.0 cm 5.0 cm kq1q2 F12 2 Magnitude ; r12 F1 F12 F13 F1 F12 F13 kq1q3 F13 2 r13 15 16.1 Coulomb’s Law Example 16.5 Figure below shows the three point charges are placed in the shape of triangular. q3 - r13 - r12 + q1 q2 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, and k = 9.0 x 109 N m2 C-2. 16 16.1 Coulomb’s Law Solution 16.5 q1= 1.2x10-6 C, q2= 3.7x10-6 C, q3= 2.3x10-6 C, r12= 15x10-2 m, r13= 10x10-2 m kq1q2 F12 2 r12 q3 - r13 kq1q3 - r12 + F13 2 q1 F12 q2 r13 F13 16.1 Coulomb’s Law Magnitude , F1 Fx Fy 2 2 F1 q3 - r13 - r12 + q1 q2 Direction, tan F y F x 18 16.1 Coulomb’s Law Example 16.6 Figure below shows the three point charges are placed in the shape of triangular. - q3 r13 q1 - r12 + q2 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. 19 16.1 Coulomb’s Law Solution 16.6 q1= 1.2x10-6 C, q2= 3.7x10-6 C, q3= 2.3x10-6 C, r12= 15x10-2 m, r13= 10x10-2 m kq q - F12 12 2 q3 r12 r13 F12 q1 - 58 r12 + q2 F13 kq1q3 F13 2 r13 16.1 Coulomb’s Law Solution 16.6 q1= 1.2x10-6 C, q2= 3.7x10-6 C, q3= 2.3x10-6 C, r12= 15x10-2 m, r13= 10x10-2 m Fx F12 F13 cos 58 q3 - 34.2 r13 F x q1 - r12 + q2 F y 0 F13 sin 58 F y F1 3.74 N Magnitude ; F1 F F x 2 y 2 F1 3.74 N Direction ; tan F y F x 21 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 q 5 C q2 2 C q3 - 3 C 1 (18.75 N , to the left , + + - 191.25 N to the right , 4.0 cm 2.0 cm 172.51 N to the left) 3. Calculate the net electrostatics force on charge q2 . q1 5C q2 2C + 4m + 3m (4.11 x10-3 N , 18.4 o below the x-axis) - q3 3C 22 16.1 Coulomb’s Law 4. What is the electrostatics force on charge q3 ? q1 2.5C (0,+0.30 m) + (0.43 N in the +x direction) + q3 3C (0.40 m,0) (0,-0.30 m) + q2 2.5C 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 q + + q are l = 60.0 cm and w = 15.0 cm. Calculate the magnitude and direction of the resultant w electric force exerted on the charge at the lower left corner l by the other three charges. q+ +q 23 (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 : a) Define electric field. F b) Define and use electric field strength, E . qo c) 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 24 charges. 16.2 Electric field Definition of electric field : A region of space around isolated charge where an electric force is experienced if a positive test charge 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. 25 16.2 Electric field • Figures below show the electric field patterns around the charge. Field direction -q +q a. Single positive charge b. Single negative charge the lines point radially the lines point radially outward from the inward toward the charge charge 26 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 ) Field direction Is defined as a point +q X +q (region) where the resultant electric force is zero. It lies along the vertical dash line. 27 e. Two opposite unequal charges, +2q and -q 16.2 Electric field Field direction • note that twice as many lines leave +2q as there are lines entering –q. -q +2q • 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. 28 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 tangent 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 stronger the field. •Electric field lines start on positive charges and end on negative charges, and the number starting or ending is proportional to the magnitude of the charge. •The field lines never cross because the electric field does not have two values at the same point. 29 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 F E : magnitude of the electric field strength E q0 F : magnitude of the electric force q0 : magnitude of test charge • 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. 30 16.2 Electric field F F From E E q0 q0 (ve) kqq0 q and F 2 r r kqq0 A positive isolated point charge. 2 thus E r q0 F kq q E q0 (ve) E 2 r r In the calculation of r : distance between the magnitude E, substitute the MAGNITUDE of the point and isolated point charge charge only. A negative isolated point charge. 31 16.2 Electric field Example 16.7 2.0 cm + A q1 5.0C 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 32 16.2 Electric field Example 16.8 2.0 cm - A q1 5.0C Calculate the electric field strength at point A, 2.0 cm from a point charge q1. Solution 33 16.2 Electric field Example 16.9 q1 + A - q2 2.0 cm 3.0 cm 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) 34 16.2 Electric field Solution 16.9 q1 + E A2 q A E A1 - 2 2.0 cm 3.0 cm a) kq1 E A1 2 r1 kq2 E A2 2 r2 The electric field strength at point A due to the charges is given by E E E A A1 A2 35 16.2 Electric field Solution 16.9 q1 + - q2 A 2.0 cm 3.0 cm b) FA EA q0 FA q0 E A 36 16.2 Electric field Example 16.10 q1 - A - q2 2.0 cm 3.0 cm 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) 37 16.2 Electric field Solution 16.10 E A1 q1 - E A 2 - q2 A 2.0 cm 3.0 cm a) kq1 E A1 2 r1 kq2 E A2 2 r2 The electric field strength at point A due to the charges is given by E E E A A1 A2 38 16.2 Electric field Solution 16.10 q1 - - q2 A 2.0 cm 3.0 cm b) FA EA q0 FA q0 E A 39 16.2 Electric field Example 16.11 q1 - 4.0 cm - q2 3.0 cm A Two point charges, q1= -1.0 C and q2= -4.0 C, are separated by 4.0 cm from each other as shown in figure above. Find the magnitude and direction of the electric field intensity at point A 3.0 cm from charge q1. (Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2) 40 16.2 Electric field Solution 16.11 q1 - 4.0 cm - q2 E A1 3.0 cm A E A2 4 tan 53o 5 kq1 E A1 2 r1 kq2 E A2 2 r2 16.2 Electric field Solution 16.11 q1 - 4.0 cm - q2 E A1 3.0 cm 53o A E A2 42 16.2 Electric field Solution 16.11 q1 - - q2 EA Ey 58.7 oE A x The electric field strength at point A due to the charges is given by Direction ; 43 16.2 Electric field Example 16.12 q1 8.00C + - q2 5.00C q3 4.00C - A 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? 44 16.2 Electric field Solution16.12 30.0 cm q1 8.00C + - q2 5.00C 30.0 cm EA2 30.0 cm 42.4 cm q3 4.00C - EA3 A EA1 kq1 (9.0 109 )(8.00 106 ) E A1 2 r1 ( 42.4 102 )2 kq2 (9.0 109 )(5.00 106 ) E A2 2 r2 (30.0 102 )2 kq3 (9.0 109 )( 4.00 106 ) E A3 2 (30.0 102 )2 45 r3 16.2 Electric field Solution 16.12 30.0 cm q1 8.00C + - q2 5.00C E A1 4.00 105 N C-1 30.0 cm EA2 E A2 5.00 105 N C-1 30.0 cm 42.4 cm E A3 4.00 105 N C-1 q3 4.00C - EA3 A 45o EA1 46 16.2 Electric field Solution 16.12 E E AY =61.7o Magnitude ; E AX Direction ; 47 16.2 Electric field Exercise 1. 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) q1 + X 2. - q2 20 cm 20 cm 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 a) The resultant electric field strength at point X. (1.13 x 103 kN C-1 towards q2) b) The electric force that acts on a point charge q = 0.50 µC placed at X. (0.57 N) 48 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. (Given 0=8.85 x 10-12 C2 N-1 m-2) (HRW. pg. 540.11) Ans. : zero. 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. 540.13) Ans. : 1.02x105 N C-1, upwards. 49 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 : a) 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 50 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) Fe vo Fe vo , constant, v =0, a=0 v 0, a 0 a 0 E W E W Figure a : Case 1 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 51 qE = mg 16.3 Charge in a uniform electric field Case 2 : Charge moving perpendicularly to the field x v0 q0 - y E FE vx θ vy v • When the electron enters the electric field , E the only force that acts on the electron is electrostatic force , FE=qE. 52 16.3 Charge in a uniform electric field • This causes the electron of mass m to accelerate downwards with an acceleration a. q0 E ay m • 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 x t From R S x vo t vo 53 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 v y uy a y t qo E x vy 0 m vo q Ex vy o mvo • After emerging from the electric field, the electron travels with constant velocity v, where v vx2 v y2 54 16.3 Charge in a uniform electric field • The direction of the velocity v is at an angle vy 1 tan to the horizontal. vx • The position of the electron at time t is s x v0 t or v y 2 u y 2 2a y s y where u y 0 1 2 v y 2 s y uy t a y t sy 2 2a y 1 2 sy ayt 2 55 16.3 Charge in a uniform electric field Example 16.13 + uo = 1.5 x 107m/s 20 mm - 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. a) Sketch the path of the electron in between plates, and after emerging from the space between the plates. b) Find the magnitude and direction of the acceleration of the electron in between the plates. c) Calculate the vertical and horizontal components of the electron velocity when it emerges from the space between the plates. d) Find the angle of deflection of the electron beam. 56 16.3 Charge in a uniform electric field Solution 16.13 + uo = 1.5 x 107m/s 20 mm - 60 mm a) DIY b) 57 16.3 Charge in a uniform electric field Solution 16.13 c) Time taken for electron to travel through the space between l 60 103 the plates is t 4.0 109 s v x 1.5 107 Vertical component of velocity, v y u y at v y 0 7.03 1014 4.0 09 2.81106 m s-1 Horizontal component of velocity, v x= 1.5 x 107 m s-1 v2.81106 1 y d) Angle of deflection, tan 10.60 v x 1.5 107 58 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 Fe a v 0 a v0 E Fe E Figure 1 Figure 2 59 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 qE a m • 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 . 60 16.3 Charge in a uniform electric field • The velocity v of the particle after time t is given by v vo at , qE v vo t m • The displacement s after time t is given by 1 2 s vot at , 2 1 qE 2 s vot t 2 m • The displacement s in terms of velocity v is given by v 2 vo 2as, 2 v 2 vo v 2 vo 2 2 s 2a qE 2 61 m 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 direction opposite the electric field is given by qE a m • 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 62 16.3 Charge in a uniform electric field • The velocity v of the particle after time t is given by v vo at , qE v vo t m • The displacement s after time t is given by 1 2 s vot at , 2 1 qE 2 s vot t 2 m • The displacement s in terms of velocity v is given by v 2 vo 2as, 2 vo v 2 vo v 2 2 2 s 2a qE 2 63 m 16.3 Charge in a uniform electric field Example 16.14 A pair of flat parallel metal plates A and B, are separated by 1.0 cm and the electric field strength that exists between the plates is 10 kV m-1. An electron emerges from plate A at a speed 5.0 x 106 m s-1 and moves towards B in a linear path which is perpendicular to B. Determine a) the acceleration of the electron b) the speed of the electron when it reaches B c) the time taken by the electron to travel to B. 64 16.3 Charge in a uniform electric field Solution 16.14 s= 1.0 x 10 -2 m , vo = 5.0 x 106 m s-1, q = 1.6 x 10 -19 C, me = 9.11 x 10 -31 kg qE 1.6 10 1.0 10 19 4 a) a 1.8 10 m s-2 15 m 9.11 1031 a is in opposite direction to E b) c) 65 16.3 Charge in a uniform electric field Example 16.15 An electron is released from rest and allowed to accelerate in a straight line in a uniform electric field of strength 3.0 kV m-1. Determine a) its acceleration b) its speed after 3.0 s. Solution q = 1.6 x 10 -19 C, me = 9.11 x 10 -31 kg a) a qE 1.6 1019 3.0 103 5.3 10 m s-2 14 m 9.11 1031 a is in opposite direction to E b) 66 SUBTOPIC : 16.4 Electric Potential (1 hour) LEARNING OUTCOMES : At the end of this lesson, the students should be able to : a) Define electric potential. b) Define and sketch equipotential lines and surfaces of i) an isolated charge , ii) a uniform electric field Q c) Use V k for a point charge and a system of charges. r d) Calculate potential difference between two points. V V V AB A B WBA V VAB e) Use E for uniform E. q d f) Deduce the change in potential energy,U between two points in electric field. U qV g) Calculate potential energy of a system of point charges. q1q2 q1q3 q2q3 U k 67 r12 r13 r23 16.4 Electric Potential • The electric potential V at a point in an electric field is the work done to bring a unit positive charge from infinity to that point. (energy required to bring 1 C of positive charge from infinity to that point) • The electric potential at infinity is considered zero. V 0 • Scalar quantity. • Its unit is volt (V) or J/C. Q • Its formula is V k r 68 16.4 Electric Potential Case A Case B Q Q + • A - • A r r The electric potential at The electric potential at point A at distance r from point A at distance r from a positive point charge Q a negative point charge Q is is VA k Q VA k Q r r Q Q VA k VA k r r 69 16.4 Electric Potential Case A Q Fext FE + • A r +qo r The electric potential V at a point in an electric field is the work done to bring a unit positive charge from infinity to that point. dW Fext dr kQq0 Wr dW FE dr but FE 2 VA r q0 r r r r r dW r FE dr kQqo r VA W kQqo r 2dr qo r r Q W kQqo r 1 VA k r Qqo W k 70 r 16.4 Electric Potential Case B Q FE - • A r +qo r The electric potential V at a point in an electric field is the work done to bring a unit positive charge from infinity to that point. kQq0 Wr dW FE dr but FE 2 VA r q0 r r r r r dW r FE dr kQqo r VA W kQqo r 2dr qo r r Q W kQqo r 1 VA k r Qqo W k 71 r 16.4 Electric Potential Example 16.16 q+ A 10 m Figure above shows a point A at distance 10 m from the positive point charge, q = 5C. Calculate the electric potential at point A and describe the meaning of the answer. (Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2) kq (9.0 109 )(5) VA r (10) 72 16.4 Electric Potential Example 16.17 Q1 =2µC Q2 =2µC + + 3 cm 5 cm A• Calculate the electric potential at point A . 73 16.4 Electric Potential Example 16.18 Q1 =2µC Q2 =2µC + - 3 cm 5 cm A• Calculate the electric potential at point A . 74 16.4 Electric Potential Example 16.19 +0.01 µC +0.02 µC + + 10 cm o 10 cm + - +0.01 µC -0.02 µC Calculate the electric potential at point O . VA = 2546 V 75 16.4 Electric Potential Potential Difference (a) • The potential difference between points A and B, VAB is given by WBA VAB VA VB q WBA : work done in bringing a charge from point B to point A VA : electric potential at point A VB : electric potential at point B q : a charge WBA VAB In calculation if q i) W positive : work done by electric field or (work done by electric force) i) W negative : work done on electric field WBA qVAB (work done by external force) 76 16.4 Electric Potential Potential Difference (b) • The potential difference between points A and B, VAB is given by WBA VAB VA VB q WBA : work done in bringing a charge from point B to point A VA : electric potential at point A VB : electric potential at point B q : a charge WBA VAB In calculation if q i) W positive : work done on electric field or (work done by external force) i) W negative : work done by electric field WBA qVAB (work done by electric force) 77 16.4 Electric Potential Example 16.20 A point charge of q = +50 µC moves from point A to point B in an electric field. The work done by the field is 10 µJ. Determine the electric p.d crossed by the charge. 78 16.4 Electric Potential Example 16.21 A point charge of q =-2.0 µC travels from point X to point Y in an electric field, crossing a potential rise of 200 V. a) Determine the work done in transferring the charge. b) Is the work done on or by the electric field ? c) An electric potential of -20 V exists at point Y. Determine the potential at point X. Solution a) 79 16.4 Electric Potential Example 16.22 Two point charges q1=+2.40 nC and q2=-6.50 nC are 0.100 m apart. Point A is midway between them, point B is 0.080 m from q1 and 0.060 m from q2 as shown in figure below. B 0.080 m 0.060 m A + - q1 q2 0.050 m 0.050 m Find a) the electric potential at point A, b) the electric potential at point B, c) the work done by the electric field on a charge of 2.5 nC 80 that travels from point B to point A. 16.4 Electric Potential Solution 16.22 B q1=+2.40 nC , q2=-6.50 nC 0.080 m 0.060 m A + - q1 q2 0.050 m 0.050 m a) VA VA1 VA2 c) b) VB VB1 VB 2 81 16.4 Electric Potential Example 16.23 a) What is the electric potential at the point i) 10 cm and ii) 50 cm from a point charge of 2 µC ? b) Find the work done to move a charge of 0.05 µC from a point 50 cm from the point charge 2 µC to a point 10 cm from the point charge. State whether the work done by or on the electric field. a) b) 82 Exercise 16.4 Electric Potential 1. What is the electric potential 11.0 cm from a 4.25 µC point charge ? (3.47 x 10 5 V) 2. What is the electric potential 2.0 x 10 -10 m from a proton (charge +e) ? (7.2 V) 3. The electric potential and the electric field strength at a point in an electric field produced by a point charge Q are + 500 V and 150 V/m respectively. Determine a) the distance of the point from the point charge (3.33 m) b) the charge Q. (1.85 x 10 -7 C) 4. A negative point charge of 0.75 mC travelling from point X to point Y in an electric field, experiences a pd drop of 200 V. a) Determine the potential at point Y if a potential of +20 V exists at point X. (-180 V) b) Determine the work done in this charge transfer. Is the work done on or by the field ? (-0.15 J, work done on the field) 5. A rectangle ABCD has length AB = 4.0 m and breadth AD = 3.0 m. Point charges Q1 = + 0.08µC and Q2 = - 0.03 µC are placed at A and C respectively. Determine a) the electric potential VB-VD across B and D (-72.8 V) b) the work required in moving a point charge q = =0.50 µC from D to B. 83 (+36.4 J) 16.4 Electric Potential Relation Between V and E • The relationship between electric field intensity E at a point in a uniform electric field and electric potential V is given by V E d d E (uniform) V • Unit ; Vm-1 84 • Vector quantity : directed towards negative plate 16.4 Electric Potential Example 16.24 Two parallel plates are charged to a voltage of 50 V. If the separation between the plates is 0.050 m, calculate the electric field strength between them. Exercise 1. How strong is the electric field between twp parallel plates 5.0 mm apart if the potential difference between them is 110 V ? (22 kV m-1) 2. An electric field of 800 V/m is desired between two parallel plates 6.0 cm apart. How large a voltage should be applied ? (48 V) 3. The electric field between two parallel plates connected to a 85 45-V battery is 600 V/m. How far apart are the plates ? (75 mm) 16.4 Electric Potential Equipotential Lines and Surfaces • The electric potential can be represented graphically by drawing equipotential lines or in three dimensions, equipotential surfaces. • An equipotential surface is a surface on which all points are at the same potential. • An equipotential line is a line on which all points are at the same potential. • No work is done when a charge moves from one point on an equipotential surface to another point on the same surface (because the potential difference is zero) • An equipotential surface must be perpendicularly to the electric field at any point. 86 16.4 Electric Potential i) An isolated charge A E VA VB VC B WBA qVAB q(VA VB ) WBA 0 C The dashed lines represent the equipotential surface (line). 87 16.4 Electric Potential ii) A uniform electric field + - VA VB VC A + - C WBA qVAB q(VA VB ) + - + B - WBA 0 + - E The dashed lines represent the equipotential surface (line). 88 16.4 Electric Potential Electric Potential Energy, U • The electric potential energy U of a charge q at a point in an electric field is given by U qV Q Q + • A - • A q q r r U A qVA U A q( VA ) • Scalar quantity. • Its unit is joule (J). • The sign of charges must be substituted in the calculation. Exercise What is the potential energy of an electron that is 0.53 x 10 -10 m from a proton ? (-4.35 x 10-18 J) 89 16.4 Electric Potential The Change in Electric Potential Energy, ∆U U U AB U A UB qVA qVB The difference between two electric potential energy The difference between electric potential energy at point A and electric potential energy at point B 90 16.4 Electric Potential Example 16.25 a) What is the electric potential energy of a charge 0.05µC at the point i) 10 cm and ii) 50 cm from a point charge of 2 µC ? b) Find the change in electric potential energy between i) and ii). a) b) 91 16.4 Electric Potential Electrical Potential Energy of A System of Point Charges • Consider a system with three point charges, q1 , q2 and q3. • The potential energy of the system is given by U U12 U13 U 23 q1q2 q1q3 q2 q3 The electric potential U k energy of the system of r12 r13 r23 charges is the work done q2 + to move all the charges from infinity to the r23 points where the charges q3 + r12 are placed. r12 + q1 92 • The sign for the charge (+ or -) must be substituted in the calculation. 16.4 Electric Potential Example 16.26 + q2 = 10µC r23 = 5.0 cm + r12= 4.0 cm q3 = 10µC r13 = 5.0 cm - q1 = - 10µC Calculate the potential energy for the system of charges shown above. 93 16.4 Electric Potential Exercise 1. A point charge Q = +9.10 µC is held fixed at the origin. A second point charge with a charge of q = -0.420 µC is placed on the x -axis, 0.960 m from the origin. Calculate the electric potential energy of the pair of charges. (-3.58 x 10-2 J) 2. +200 pC -100 pC • • • • A D C B 20 cm 60 cm 20 cm In figure above, the charge at A is +200 pC, while the charge at B is -100 pC. a) Calculate the electric potentials at points C and D. (-2.25 V, +7.88 V) b) If a charge of +500 µC is placed at points C and D, calculate the electric potential energy at points C and D. (- 1.13 x 10 -3 J, 3.94 x 10-3 J) c) The change in potential energy between points C and D. (-5.07 mJ) 94 16.4 Electric Potential Exercise 3. A point charge q1 = +2.00 nC is placed at the origin and a second point charge q2 = -3.00 nC is placed on the x -axis, at x = +20.0 cm. A third point charge q3 = 5.00 nC is to be placed on the x -axis between q1 and q2. What is the electric potential energy of the system of the three charges if q3 is placed at x = +10.0 cm ? (-7.20 x 10 -7 J) 4. Three equal point charges q = 8.40 x 10-7 C are placed at the corner of an equilateral triangle whose side is 1.00 m. What is the potential energy of the system? ( 19.1 mJ) 95