Steve Smith Tuition: Physics Notes

Transcription

1 Steve Smith Tuition: Physics Notes E = mc 2 F = GMm r 2 sin θ m = mλ d hν = φ mv2 Static Fields V: Electric Field Examples Contents 1 Electric Field Equations Radial Electric Field Equations Uniform Electric Field Equations The Questions Example Example Example Example Example Example Example Example Example Example Example The nswers Example Example Example Example Example Example Example Example Example Example Example ppendices 21 Useful Data 21 1

2 Prerequisites You should really have read my other document on electric fields: Smith (2014). Notes None. Document History Date Version Comments 22nd December Initial creation of the document. 28th March Simplifying uniform electric fields. 2

3 1 Electric Field Equations s a quick reminder, here are the pictures that you should learn to master electric fields. You should know what all the symbols mean by now! 1.1 Radial Electric Field Equations Figure 1: Obtaining Radial Electric Field Equations IV: The Full Monty Force F = kqq r 2 integrate differentiate Potential Energy E p = kqq r vectors q q q q scalars Field Strength E = kq r 2 integrate differentiate Potential V = kq r 1.2 Uniform Electric Field Equations Figure 2: Obtaining Uniform Electric Field Equations IV: The Full Monty Force integrate Potential Energy F = qe c differentiate E p = (E c r + V c )q vectors q q q q scalars Field Strength integrate Potential E = E c differentiate V = E c r + V c Don t forget that with uniform electric fields, r is measured f rom the plate the field lines point towards. 3

4 2 The Questions 2.1 Example 1 Figure 3: Equipotentials 10V 20V 40V 80V Q B Figure 3 shows the equipotential lines around a point charge Q. (a) What are equipotential lines? (b) Is the charge Q positive, or negative? (c) How much energy would it take to move an electron from to B in this field? 2.2 Example 2 Define potential at a point in an electric field. 2.3 Example 3 (a) What is meant by the terms (i) potential and (ii) field strength in electrostatics? State whether each quantity is a scalar or a vector. (b) Write down the law which gives the force between two point charges Q and q at a distance r apart and use it to derive the electric field strength and the potential due to a point charge Q at a distance r from it. 2.4 Example 4 charged oil-drop of radius m is prevented from falling under gravity by the vertical field between two horizontal plates charged to a difference of potential of 8340 volts. The distance between the plates is 16 mm, and the density of oil is 920 kg m 3. Calculate the magnitude of the charge on the drop. [ g = 9.81 ms 2 ] 4

5 2.5 Example 5 Define the electric potential V and the electric field strength E at a point in an electrostatic field. 2.6 Example 6 (a) Two plane parallel conducting plates 15 mm apart are held horizontal, one above the other, in air. The upper plate is maintained at a positive potential of 1500 volts while the lower plate is earthed. Calculate the number of electrons which must be attached to a small oil drop of mass kg, if it remains stationary in the air between the plates. [ ssume that the density of air is negligible in comparison with that of oil. ] (b) If the potential of the upper plate is suddenly changed to 1500 volts what is the initial acceleration of the charged drop? 2.7 Example 7 (a) What is an electric field? With reference to such a field define electric potential. (b) Two plane parallel conducting plates are held horizontal, one above the other, in a vacuum. Electrons having a speed of m s 1 and moving normally to the plates enter the region between them through a hole in the lower plate which is earthed. What potential must be applied to the other plate so that the electrons just fail to reach it? What is the subsequent motion of these electrons? [ ssume that the electrons do not interact with one another. ] [ Ratio of charge to mass of electron is C kg 1. ] 2.8 Example 8 Figure 4 shows an arrangement of two point charges in air, Q being 0.3 µc. Figure 4: Charges in a Line : Q 3Q +Q 0.8 m P B 1.2 m (a) Find the electric field strength and the electric potential at P. (b) Find the point on B where the electric potential is zero. (c) Explain why the electric potential to the left of the 3Q charge is always negative. 5

6 2.9 Example 9 (a) Define the terms potential and field strength at a point in an electric field. How are they related? Figure 5: Parallel Plates +350 V 10 cm +0 V (b) Figure 5 shows two horizontal parallel conducting plates in a vacuum. small particle of mass kg, carrying a positive charge of C, is released at near the upper plate. What total force acts on the particle? (c) Calculate the kinetic energy of the particle when it reaches the lower plate Example 10 Figure 6: More Charges in a Line : Q X C C 6.0 cm B 4.0 cm F C + P F 1 Figure 6 shows three small charges,, B and P in a line. The charge at is positive, that at B is negative, and that at P is positive. The values are those shown. (a) Calculate the force on the charge at P due to and B. (b) t what point X on the line B should there be no force on the charge at P due to and B if P were placed there? 6

7 2.11 Example 11 In Figure 7, two small equal charges C are placed at and B, one positive and the other negative. B is 6 cm. X is the mid-point of B. Figure 7: Even More Charges in a Line : Q C P 4 cm C X B C 3 cm 6 cm Find the force on a charge C placed at P, where P is 4 cm from the line B along the perpendicular bisector XP. 7

8 3 The nswers 3.1 Example 1 Figure 8: Equipotentials 10V 20V 40V 80V Q B Figure 8 shows the equipotential lines around a point charge Q. (a) What are equipotential lines? (b) Is the charge Q positive, or negative? (c) How much energy would it take to move an electron from to B in this field? (a) n equipotential line represents points in the field where the potential is the same. They re like the contour lines on a map. (b) Since this is a radial electric field, from Figure 1, the potential at a point a distance r from Q is given by V = kq r Since k and r are both positive, then the signs of the potentials in the field will be the same as the sign on the charge Q. So this charge Q must be negative. (c) From Figure 1, the relationship between potential V and potential energy E p is E p = Vq So to find the potential energy difference between and B we find the potential difference, and multiply by q, the charge on the electron: E p = (potential at B potential at ) q so E p = ( 20 ( 10)) = J 8

9 3.2 Example 2 Define potential at a point in an electric field. Potential is defined to be the work done in taking a unit positive charge from infinity to the point in the field. nd potential difference between points and B will be the work done in taking a unit positive charge from B to. 3.3 Example 3 (a) What is meant by the terms (i) potential and (ii) field strength in electrostatics? State whether each quantity is a scalar or a vector. (b) Write down the law which gives the force between two point charges Q and q at a distance r apart and use it to derive the electric field strength and the potential due to a point charge Q at a distance r from it. (a) (i) Potential is defined to be the work done in taking a unit positive charge from infinity to the point in the field. nd potential difference between points and B will be the work done in taking a unit positive charge from B to. Field strength is the force per unit charge that the field exerts at a point in the field. Field strength is a vector, while potential is a scalar. (a) (ii) Since we are dealing with a radial field here, from Figure 1, the force between two charges Q and q a distance r apart is given by F = kqq r 2 = Qq 4πɛr 2 depending on which constant you use in your formulas. To get from the force to the field strength, we divide the force by q: E = F q = kqq r 2 q = Qq 4πɛr 2 q = kq r 2 = Q 4πɛr 2 nd to get from the field strength to the potential, you have to integrate: V = = = kq r E dr kq r 2 = Q 4πɛr dr = Q 4πɛr 2 dr 9

10 3.4 Example 4 charged oil-drop of radius m is prevented from falling under gravity by the vertical field between two horizontal plates charged to a difference of potential of 8340 volts. The distance between the plates is 16 mm, and the density of oil is 920 kg m 3. Calculate the magnitude of the charge on the drop. [ g = 9.81ms 2 ] The first thing that we have to realise here is that there are two forces on the drop. They are: the force due to the uniform electric field. This will act upwards because of the way the question is worded; the force due to gravity. This will act downwards because gravity always does. I guess we would have to assume that these two forces were balanced, so that the drop was held stationary in space. The size of the downward force F g due to gravity is going to be given by F g = mg To work out this force we will need to know the mass of the drop. We don t know that, but we do know the radius and the density of the drop. ssuming the drop is a sphere, then its volume will be given by V = 4 3 πr3 and density ρ is defined in terms of mass m and volume V by So the mass of the drop will be given by ρ = m V m = ρv = 4 3 πρr3 and so Phew! Plugging the numbers in, F g = 4 3 πρgr3 F g = 4 3 π ( ) 3 = N (to 3 sf) OK, let s turn our attention to the electric field. The question is interestingly worded, in that it does not say whether the charge on the oil drop is positive or negative; neither does it say which way the electric field is oriented. The only thing we know is that the force on the oil drop must be directed upwards. So, let s use the uniform field shown in Figure 9. I ve chosen that the field direction is down because it is Figure 9: Uniform Field V 16 mm +0 V easier to make an oil drop negative (by attaching electrons to it) than it would be to make it positive. That 10

11 being the case, the field shown in Figure 9 would impart an upward force on the negatively charged drop. We would use Figure 2 to give us the equations for this problem. Now from Figure 2 the size of the upward force F e due to the electric field (of strength E) will be given by F e = Eq where q is the charge on the drop. We can work out the size of the field strength using Figure 2: it is E = V r where V is the potential difference across the plates, which are a distance r apart. Notice here that as the bottom plate (where r = 0) has a potential of zero, the constant of integration, V c, in the equations of Figure 2 will be zero. So, F e = Vq r = 8340 q = q V m 1 (to 3 sf) So, if the two forces balance, then F g = F e, and so and so q = which is about the charge on an electron = q = C (to 3 sf) 3.5 Example 5 Define the electric potential V and the electric field strength E at a point in an electrostatic field. n electric field is a volume of space where charged particles would experience a force. Electric potential V is a property of the space in an electric field. Every point in the field has a value of potential. The potential at a point is defined to be the work done in moving a unit positive charge from infinity to that point. The electric field strength E is also a property of the space in an electric field. The electric field strength at a point in the field is defined to be the force per unit mass on an object placed at that point in the field. 11

12 3.6 Example 6 (a) Two plane parallel conducting plates 15 mm apart are held horizontal, one above the other, in air. The upper plate is maintained at a positive potential of 1500 volts while the lower plate is earthed. Calculate the number of electrons which must be attached to a small oil drop of mass kg, if it remains stationary in the air between the plates. [ ssume that the density of air is negligible in comparison with that of oil. ] (b) If the potential of the upper plate is suddenly changed to 1500 volts what is the initial acceleration of the charged drop? (a) See Figure 10 for a picture of this situation. The equations from Figure 2 will apply, so long as we Figure 10: Uniform Field V 15 mm +0 V measure r away from the bottom plate. Notice that since the bottom plate has a potential of zero, then V c in the equations of Figure 2 will be zero. We have a situation where there will be two forces on this oil drop: the force F g of gravity acting down, and the force F e, due to the uniform electric field, acting up. If the two forces are to balance, so that F g = F e, then from Figure 2, mg = Eq where m is the mass of the drop, q is the size of the charge on the drop, E is the size of the electric field strength between the plates, and g is the acceleration due to gravity. Now, using Figure 2, we can get the electric field strength E between the plates from the potential difference between the plates V and the distance r between them: E = V r Putting these two equations together, and plugging the numbers in we get q = mg E = mgr V q = = C (to 3 sf) nd since the charge on an electron is C, this charge represents the charge on three electrons. (b) If the potential on the upper plate was changed to 1500 V, the size of the electric force will stay the same (because the size of the field strength would stay the same), but would just be in the opposite direction (that is, down). Since the size of electric force was the same as the size of the gravitational force on the drop in part (a), this will still be true. But both forces are acting down now. So the net force down on the drop will be F = F g + F e = 2mg and so the initial acceleration will be 2g (Newton s Second Law). 12

13 3.7 Example 7 (a) What is an electric field? With reference to such a field define electric potential. (b) Two plane parallel conducting plates are held horizontal, one above the other, in a vacuum. Electrons having a speed of m s 1 and moving normally to the plates enter the region between them through a hole in the lower plate which is earthed. What potential must be applied to the other plate so that the electrons just fail to reach it? What is the subsequent motion of these electrons? [ ssume that the electrons do not interact with one another. ] [ Ratio of charge to mass of electron is C kg 1. ] (a) n electric field is a volume of space where charged particles would experience a force. Electric potential is a property of the space in an electric field. Every point in the field has a value of potential. The potential at a point is defined to be the work done in moving a unit positive charge from infinity to that point. (b) See Figure 11 for a picture of this situation. The lower of the two plates is Earthed, which means Figure 11: Uniform Field V V electrons +0 V that the potential on it is zero. So the top plate will have a negative potential to repel the electrons. The equations shown in Figure 2 will apply, so long as we measure r down from the top plate. In this case, the constant of integration will be V. The electrons will have an initial kinetic energy of 1 2 mv2 as they enter the space between the plates. If they are to lose all this kinetic energy when just reaching the top plate, then they must gain an equivalent amount of potential energy. From Figure 2, the potential energy of the electrons at the top plate would be E p = (E c 0 + V)q = Vq since r = 0 at the top plate. So, putting these two equations together, 1 2 mv2 = Vq and so V = 1 mv 2 2 q = 1 2 ( m q ) v 2 = ( v2 ) 2 mq nd plugging the numbers in So the potential on the upper plate must be 100 V. V = (6 106 ) 2 = 100 V The subsequent motion of these electron will be to accelerate back down toward the bottom plate. 13

14 3.8 Example 8 Figure 12: Charges in a Line : 3Q +Q 0.8 m P B 1.2 m Figure 12 shows an arrangement of two point charges in air, Q being 0.3 µc. (a) Find the electric field strength and the electric potential at P. (b) Find the point on B where the electric potential is zero. (c) Explain why the electric potential to the left of the 3Q charge is always negative. (a) From Figure 1, the electric field strength at a point in a radial electric field is given by E = kq r 2 So the electric field strength E 3Q at P due to the charge 3Q will be given by E 3Q = = 5620 N C 1 (to 3 sf) The sign indicating that the direction of the electric field strength will be toward the 3Q charge. (Remember that forces are attractive? Well, electric field strength is a vector too, right?) nd the electric field strength E +Q at P due to the charge +Q will be given by E +Q = = N C 1 (to 3 sf) The + sign indicating that the direction of the electric field strength will be away from the +Q charge. Consequently, the net electric field strength at P will be the (vector) sum of 5620 N C 1 to the left and 4210 N C 1 to the right. This will be 1410 N C 1 to the left. We can work out the potential at P in a similar way. From Figure 1, the potential V at a point in a radial electric field is given by V = kq r So the potential at P, V 3Q, due to the 3Q charge will be V 3Q = nd the potential at P, V +Q, due to the +Q charge will be = 6740 V (to 3 sf) V +Q = = V (to 3 sf) Since potentials are just scalars, we can simply add them to get the total potential V P at P: V P = = 3370 V (to 3 sf) (b) I m guessing that the point on B where the potential will be zero is to the right of the +Q charge. This might not be right, but we should discover whether we are right or not as the calculation rolls on. OK, let s say that it is x m to the right of the +Q charge. In that case, the potential at the neutral point due to the 3Q charge will be V 3Q = x = x 14

15 since the distance between the 3Q and +Q charges is 0.4 m. nd the potential at the neutral point due to the +Q charge will be V +Q = x = x But when we add these values together we should get zero, so (keeping 4sf through to the end) x = 0 x Or nd so Multiplying out the brackets nd so So that x = 2696 x 8089x = 2696(0.4 + x) 8089x = x 5393x = 1078 x = 1078 = 0.20 m (to 3 sf) 5393 Now it could be that the point on B where the potential will be zero is to the left of the +Q charge. Let s try that. OK, let s say that it is y m to the left of the +Q charge. In that case, the potential at the neutral point due to the 3Q charge will be V 3Q = y = y since the distance between the 3Q and +Q charges is 0.4 m. nd the potential at the neutral point due to the +Q charge will be V +Q = y = y But when we add these values together we should get zero, so (keeping 4sf through to the end) Or nd so Multiplying out the brackets nd so y = 0 y y = 2696 y 8089y = 2696(0.4 y) 8089y = y 10780y = 1078 So that y = 1078 = 0.10 m (to 3 sf) So there are two solutions to this problem!! Interesting! (c) t any point to the left of the 3Q charge, the potential ue to the 3Q charge will be negative. nd it s size will be larger than the potential due to the +Q charge, because the charge is three times bigger, and we are closer to it. So anywhere to the left of the 3Q charge, the potential will be negative. 15

16 3.9 Example 9 (a) Define the terms potential and field strength at a point in an electric field. How are they related? Figure 13: Parallel Plates +350 V 10 cm +0 V (b) Figure 13 shows two horizontal parallel conducting plates in a vacuum. small particle of mass kg, carrying a positive charge of C, is released at near the upper plate. What total force acts on the particle? (c) Calculate the kinetic energy of the particle when it reaches the lower plate. (a) n electric field is a volume of space where charged particles would experience a force. Electric potential V is a property of the space in an electric field. Every point in the field has a value of potential. The potential at a point is defined to be the work done in moving a unit positive charge from infinity to that point. The electric field strength E is also a property of the space in an electric field. The electric field strength at a point in the field is defined to be the force per unit mass on an object placed at that point in the field. Using Figure 2 they are related by: and V = E = dv dr (b) In Figure 14 I ve added the electric field lines to the diagram of this question. Using our standard idea E dr Figure 14: Parallel Plates +350 V 10 cm +0 V of measuring r up from the bottom plate, then equations from Figure 2 will apply. lso, since the potential is zero when r is zero, then the constant of integration will be zero. There will be two forces acting on the particle: the gravitational force F g acting down and the force F e, due to the uniform electric field, which also acts down because the particle has a positive charge and so will be repelled from the top plate. The sizes of these forces will be and F g = mg = = N (to 3 sf) F e = Eq = V r q = = N using the connection between electric field strength and potential from Figure 2. So the total downward force F on the particle will be F = = N (to 3 sf) 16

17 (c) You could solve this problem using SUVT: the distance s is 10 cm; the initial velocity u is 0 ms 1 ; the final velocity v is what we want because that will give us the kinetic energy; the acceleration a is F = m = 36.1 ms 2 ; we neither know nor want the time for the journey. See Table 1. Table 1: SUVT Problem Quantity Wot We Know Wot We Want s u v a t m 0 ms ms 2 Since we can t use a formula with t in it, we will have to use v 2 = u 2 + 2as. So: v 2 = = 7.21 m 2 s 2 So to find the kinetic energy, K E = 1 2 mv2 = = J (to 3 sf) Now we didn t have to use SUVT to solve this problem. We could have used energy considerations. We could have said that the kinetic energy gained in crossing the gap between the plates is equal to the potential energy lost. nd using Figure 2, potential energy lost will be the gravitational potential energy lost plus the electric potential energy lost: so the kinetic energy gained will be equal to E p = mgh + Vq K E = = J (to 3 sf) 17

18 3.10 Example 10 Figure 15: More Charges in a Line : X C C 6.0 cm B 4.0 cm F C + P F 1 Figure 15 shows three small charges,, B and P in a line. The charge at is positive, that at B is negative, and that at P is positive. The values are those shown. (a) Calculate the force on the charge at P due to and B. (b) t what point X on the line B should there be no force on the charge at P due to and B if P were placed there? (a) Since we are dealing with radial electric fields here, from Figure 1, the electric force F e between two charges Q and q a distance r apart is F e = kqq r 2 So the force F 1 on P due to the charge is F 1 = ( ) 2 = N (to 3 sf) This force is positive, showing that it is a repulsion. That s why it s the force F 1 in Figure 15. nd the force F 2 on P due to the charge B is F 2 = ( ) 2 = N (to 3 sf) This force is negative, showing that it is an attraction. That s why it s the force F 2 in Figure 15. So the total force on P will be = N to the left (to 3 sf). (b) The point X where there is no force on P would have to be to the left of (why??). Let the distance that X is to the left of be x. So the force F on P due to the charge this time is F = (x 2 = x 2 This force is positive, showing that it is a repulsion. So it will act to the left. nd the force F B on P due to the charge B this time is F B = ( x) 2 = ( x) 2 This force is negative, showing that it is an attraction. So it will act to the right. If there is no net force on P at X then these two forces must add to zero: dding the second term to both sides Multiplying out to get rid of the fractions, x 2 ( x) 2 = x 2 = ( x) ( x) 2 = x 2 18

19 Divide both sides by ( x) 2 = 2.50 x 2 Square root both sides Subtract x from both sides So or x = 10.3 cm from (to 3 sf) x = 1.58 x = 1.58 x x = 0.58 x x = = m 3.11 Example 11 Figure 16: Even More Charges in a Line : C P 4 cm C X B C 3 cm 6 cm In Figure 16, two small equal charges C are placed at and B, one positive and the other negative. B is 6 cm. X is the mid-point of B. Find the force on a charge C placed at P, where P is 4 cm from the line B along the perpendicular bisector XP. (a) Since we are dealing with radial electric fields here, from Figure 1, the electric force F e between two charges Q and q a distance r apart is F e = kqq r 2 First problem then is to work out the distance from to P. Triangle XP is a right-angled triangle whose base is 3 cm and height 4 cm. The hypotenuse will be 5 cm then (Pythagoras). So the force F 1 on P due to the charge is F 1 = ( ) 2 = N (to 3 sf) This force is positive, showing that it is a repulsion. That s why the force F 1 is pointing the way it does in Figure 17. Similarly, the force F 2 on P due to the charge B is F 2 = ( ) 2 = N (to 3 sf) 19

20 Figure 17: Even More Charges in a Line : B F C P θ F F 2 4 cm θ C X B C 3 cm 6 cm This force is negative, showing that it is an attraction. That s why the force F 2 is pointing the way it does in Figure 17. To find the resultant F for of F 1 and F 2 added together, we need to know the value of the cosine of the angle θ. Now since P = 5 cm, and X = 3 cm, then cos(θ) = 3 5 The horizontal component of F 1 will consequently be = N (to 3 sf) From symmetry, this will also be the horizontal component of F 2. So the total horizontal component of F is = N (to 3 sf) lso by symmetry, the vertical components of F 1 and F 2 cancel out. So the resultant force on the charge at P will be N to the right (to 3 sf). 20

21 4 ppendices Useful Data Quantity Earth - Moon Distance Gravitational Constant Mass of the Earth Radius of the Earth Mass of the Moon Gravitational Field Strength at the surface of the Earth k Value m m 3 kg 1 s kg m kg 9.81 ms m 3 kg s 4 2 ɛ m 3 kg 1 s

22 References Smith, S. (2014). Static Fields III: Electric Field Essentials. How to obtain all the electric field equations. 22