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Since . all eight charges are the same all of the forces will be repulsive. We need to sketch a diagram to show how the charges are labeled. 2 6 4 1 7 3 8 The magnitude of the force of charge 2 on charge 1 is F12 = 1 q2 2 , 4 0 r12 5 where r12 = a, the length of a side. Since both charges are the same we wrote q 2 . By symmetry we expect that the magnitudes of F12 , F13 , and F14 will all be the same and they will all be at right angles to each other directed along the edges of the cube. Written in terms of vectors the forces 5 would be F12 F13 F14 The force from charge 5 is 1 q2 2 , 4 0 r15 and is directed along the side diagonal away from charge 5. The distance r15 is also the side diagonal distance, and can be found from 2 r15 = a2 + a2 = 2a2 , F15 = then 1 q2 . 4 0 2a2 By symmetry we expect that the magnitudes of F15 , F16 , and F17 will all be the same and they will all be directed along the diagonals of the faces of the cube. In terms of components we would have F15 = F15 F16 F17 = = = 1 q2 4 0 2a2 1 q2 4 0 2a2 1 q2 4 0 2a2 ^ ^ 2 + k/ 2 , j/ ^ 2 + k/ 2 , ^ i/ ^ 2 +^ 2 . i/ j/ = = = 1 q2 ^ i, 4 0 a2 1 q2 ^ j, 4 0 a2 1 q2 ^ k. 4 0 a2 The last force is the force from charge 8 on charge 1, and is given by F18 = 1 q2 2 , 4 0 r18 and is directed along the cube diagonal away from charge 8. The distance r18 is also the cube diagonal distance, and can be found from 2 r18 = a2 + a2 + a2 = 3a2 , then in term of components F18 = 1 q2 ^ ^ i/ 3 + ^ 3 + k/ 3 . j/ 2 4 0 3a We can add the components together. By symmetry we expect the same answer for each components, so we'll just do one. How about ^ This component has contributions from charge 2, 6, 7, i. and 8: 1 1 q2 1 2 , + + 2 4 0 a 1 2 2 3 3 or 1 q2 (1.90) 4 0 a2 The three components add according to Pythagoras to pick up a final factor of 3, so F net = (0.262) q2 . 2 0a 6 E25-14 (a) Yes. Changing the sign of y will change the sign of Fy ; since this is equivalent to putting the charge q0 on the "other" side, we would expect the force to also push in the "other" direction. (b) The equation should look Eq. 25-15, except all y's should be replaced by x's. Then Fx = 1 q0 q . 4 0 x x2 + L2 /4 (c) Setting the particle a distance d away should give a force with the same magnitude as F = 1 4 0 d q0 q d2 + L2 /4 . This force is directed along the 45 line, so Fx = F cos 45 and Fy = F sin 45 . (d) Let the distance be d = x2 + y 2 , and then use the fact that Fx /F = cos = x/d. Then Fx = F and Fy = F 1 x q0 q x = . 2 + y 2 + L2 /4)3/2 d 4 0 (x 1 y q0 q y = . 2 + y 2 + L2 /4)3/2 d 4 0 (x E25-15 (a) The equation is valid for both positive and negative z, so in vector form it would read ^ F = Fz k = 1 q0 q z ^ k. 4 0 (z 2 + R2 )3/2 (b) The equation is not valid for both positive and negative Reversing the sign of z should z. reverse the sign of Fz , and one way to fix this is to write 1 = z/ z 2 . Then ^ F = Fz k = 1 2q0 qz 4 0 R2 1 1 - 2 z z2 ^ k. E25-16 Divide the rod into small differential lengths dr, each with charge dQ = (Q/L)dr. Each differential length contributes a differential force dF = Integrate: F = = 1 qQ dr, 4 0 r2 L x 1 qQ 1 1 - 4 0 L x x + L dF = x+L 1 q dQ 1 qQ = dr. 2 4 0 r 4 0 r2 L E25-17 You must solve Ex. 16 before solving this problem! q0 refers to the charge that had been called q in that problem. In either case the distance from q0 will be the same regardless of the sign of q; if q = Q then q will be on the right, while if q = -Q then q will be on the left. Setting the forces equal to each other one gets 1 qQ 4 0 L or r= x(x + L). 7 1 1 - x x+L = 1 qQ , 4 0 r2 E25-18 You must solve Ex. 16 and Ex. 17 before solving this problem. If all charges are positive then moving q0 off axis will result in a net force away from the axis. That's unstable. If q = -Q then both q and Q are on the same side of q0 . Moving q0 closer to q will result in the attractive force growing faster than the repulsive force, so q0 will move away from equilibrium. E25-19 We can start with the work that was do of incidence will increase by an angle , and so will the angle of reflection. But that means that the angle between the incident angle and the reflected angle has increased by twice. E40-4 Sketch a line from Sarah through the right edge of the mirror and then beyond. Sarah can see any image which is located between that line and the mirror. By similar triangles, the image of Bernie will be d/2 = (3.0 m)/2 = 1/5 m from the mirror when it becomes visible. Since i = -o, Bernie will also be 1.5 m from the mirror. E40-5 The images are fainter than the object. Several sample rays are shown. E40-6 The image is displaced. The eye would need to look up to see it. E40-7 The apparent depth of the swimming pool is given by the work done for Exercise 3925, dapp = d/n The water then "appears" to be only 186 cm/1.33 = 140 cm deep. The apparent distance between the light and the mirror is then 250 cm + 140 cm = 390 cm; consequently the image of the light is 390 cm beneath the surface of the mirror. E40-8 Three. There is a single direct image in each mirror and one more image of an image in one of the mirrors. 188 E40-9 We want to know over what surface area of the mirror are rays of light reflected from the object into the eye. By similar triangles the diameter of the pupil and the diameter of the part of the mirror (d) which reflects light into the eye are related by (5.0 mm) d = , (10 cm) (24 cm) + (10 cm) which has solution d = 1.47 mm The area of the circle on the mirror is A = (1.47 mm)2 /4 = 1.7 mm2 . E40-10 (a) Seven; (b) Five; and (c) Two. This is a problem of symmetry. E40-11 Seven. Three images are the ones from Exercise 8. But each image has an image in the ceiling mirror. That would make a total of six, except that you also have an image in the ceiling mirror (look up, eh?). So the total is seven! E40-12 A point focus is not formed. The envelope of rays is called the caustic. You can see a similar effect when you allow light to reflect off of a spoon onto a table. E40-13 The image is magnified by factor a of 2.7, so the image distance is 2.7 times farther from the mirror than the object. An important question to ask is whether or not the image is real or virtual. If it is a virtual image it is behind the mirror and someone looking at the mirror could see it. If it were a real image it would be in front of the mirror, and the man, who serves as the object and is therefore closer to the mirror than the image, would not be able to see it. So we shall assume that the image is virtual. The image distance is then a negative number. The focal length is half of the radius of curvature, so we want to solve Eq. 40-6, with f = 17.5 cm and i = -2.7o 1 1 1 0.63 = + = , (17.5 cm) o -2.7o o which has solution o = 11 cm. E40-14 The image will be located at a point given by 1 1 1 1 1 1 = - = - = . i f o (10 cm) (15 cm) (30 cm) The vertical scale is three times the horizontal scale in the figure below. 189 E40-15 This problem requires repeated application of 1/f = 1/o + 1/i, r = 2f , m = -i/o, or the properties of plane, convex, or concave mirrors. All dimensioned variables below (f, r, i, o) are measured in centimeters. (a) Concave mirrors have positive focal lengths, so f = +20; r = 2f = +40; 1/i = 1/f - 1/o = 1/(20) - 1/(10) = 1/(-20); m = -i/o = -(-20)/(10) = 2; the image is virtual and upright. (b) m = +1 for plane mirrors only; r = for flat surface; f = /2 = ; i = -o = -10; the image is virtual and upright. (c) If f is positive the mirror is concave; r = 2f = +40; 1/i = 1/f - 1/o = 1/(20) - 1/(30) = 1/(60); m = -i/o = -(60)/(30) = -2; the image is real and inverted. (d) If m is negative then the image is real and inverted; only Concave mirrors produce real images (from real objects); i = -mo = -(-0.5)(60) = 30; 1/f = 1/o + 1/i = 1/(30) + 1/(60) = 1/(20); r = 2f = +40. (e) If r is negative the mirror is convex; f = r/2 = (-40)/2 = -20; 1/o = 1/f - 1/i = 1/(-20) - 1/(-10) = 1/(20); m = -(-10)/(20) = 0.5; the image is virtual and upright. (f) If m is positive the image is virtual and upright; if m is less than one thfrom the converging lens is a virtual object in the diverging lens, so that the object distance for the diverging lens is o = -30 cm. The image formed by the diverging lens is located by solving 1 1 1 1 1 1 = - = - = , i f o (-15 cm) (-30 cm) -30 cm or i = -30 cm. This would mean the image formed by the diverging lens would be a virtual image, and would be located to the left of the diverging lens. The image is virtual, so it is upright. The magnification from the first lens is m1 = -i/o = -(40 cm)/(40 cm)) = -1; 195 the magnification from the second lens is m2 = -i/o = -(-30 cm)/(-30 cm)) = -1; which implies an overall magnification of m1 m2 = 1. E40-32 (a) The parallel rays of light which strike the lens of focal length f will converge on the focal point. This point will act like an object for the second lens. If the second lens is located a distance L from the first then the object distance for the second lens will be L - f . Note that this will be a negative value for L < f , which means the object is virtual. The image will form at a point 1/i = 1/(-f ) - 1/(L - f ) = L/f (f - L). Note that i will be positive if L < f , so the rays really do converge on a point. (b) The same equation applies, except switch the sign of f . Then 1/i = 1/(f ) - 1/(L - f ) = L/f (L - f ). This is negative for L < f , so there is no real image, and no converging of the light rays. (c) If L = 0 then i = , which means the rays coming from the second lens are parallel. E40-33 The image from the converging lens is found from 1 1 1 1 = - = i1 (0.58 m) (1.12 m) 1.20 m so i1 = 1.20 m, and the image is real and inverted. This real image is 1.97 m - 1.20 m = 0.77 m in front of the plane mirror. It acts as an object for the mirror. The mirror produces a virtual image 0.77 m behind the plane mirror. This image is upright relative to the object which formed it, which was inverted relative to the original object. This second image is 1.97 m + 0.77 m = 2.74 m away from the lens. This second image acts as an object for the lens, the image of which is found from 1 1 1 1 = - = i3 (0.58 m) (2.74 m) 0.736 m so i3 = 0.736 m, and the image is real and inverted relative to the object which formed it, which was inverted relative to the original object. So this image is actually upright. E40-34 (a) The first lens forms a real image at a location given by 1/i = 1/f - 1/o = 1/(0.1 m) - 1/(0.2 m) = 1/(0.2 m). The image and object distance are the same, so the image has a magnification of 1. This image is 0.3 m - 0.2 m = 0.1 m from the second lens. The second lens forms an image at a location given by 1/i = 1/f - 1/o = 1/(0.125 m) - 1/(0.1 m) = 1/(-0.5 m). Note that this puts the final image at the location of the original object! The image is magnified by a factor of (0.5 m)/(0.1 m) = 5. (c) The image is virtual, but inverted. 196 E40-35 If the two lenses "pass" the same amount of light then the solid angle subtended by each lens as seen from the respective focal points must be the same. If we assume the lenses have the same round shape then we can write this as do /f o = de /f e . Then de fo = = m , do fe or de = (72 mm)/36 = 2 mm. E40-36 (a) f = (0.25 m)/(200) 1.3 mm. Then 1/f = (n - 1)(2/r) can be used to find r; r = 2(n - 1)f = 2(1.5 - 1)(1.3 mm) = 1.3 mm. (b) The diameter would be twice the radius. In effect, these were tiny glass balls. E40-37 (a) In Fig. 40-46(a) the image is at the focal point. This means that in Fig. 40-46(b) i = f = 2.5 cm, even though f = f . Solving, 1 1 1 1 = + = . f (36 cm) (2.5 cm) 2.34 cm (b) The effective radii of curvature must have decreased. E40-38 (a) s = (25 cm) - (4.2 cm) - (7.7 cm) = 13.1 cm. (b) i = (25 cm) - (7.7 cm) = 17.3 cm. Then 1 1 1 1 = - = o (4.2 cm) (17.3 cm) 5.54 cm. The object should be placed 5.5 - 4.2 = 1.34 cm beyond F1 . (c) m = -(17.3)/(5.5) = -3.1. (d) m = (25 cm)/(7.7 cm) = 3.2. (e) M = mm = -10. E40-39 Microscope magnification is given by Eq. 40-33. We need to first find the focal length of the objective lens before we can use this formula. We are told in the text, however, that the microscop... View Full Document