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obj-4 - 23.13. IDENTIFY: E points from high potential to...

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Unformatted text preview: 23.13. IDENTIFY: E points from high potential to low potential. W“"" = V —V,,. qo SET UP: The force on a positive test charge is in the direction of E. EXECUTE: V decreases in the eastward direction. A is east of B, so VB > VA. C is east of A, so VC < VA. The force on a positive test charge is east, so no work is done on it by the electric force when it moves due south (the force and displacement are perpendicular), and VD : VA. EVALUATE: The electric potential is constant in a direction perpendicular to the electric field. 23.17. IDENTIFY: Apply the equation that precedes Eq.(23.17): WHb : q’fE : di. SET UP: Use coordinates where +y is upward and +x is to the right. Then E = with E = 4.00X104 N/C. (a) The path is sketched in Figure 23.17a. y r d4: dxi‘ X a b Figure 23.17a EXECUTE: E-di=(Ej‘)-(dxi)=0 so Wm:q’fE-dl"=0. EVALUATE: The electric force on the positive charge is upward (in the direction of the electric field) and does no work for a horizontal displacement of the charge. (b) SET UP: The path is sketched in Figure 23.17b. 3’ di = dyj' x a Figure 23.17b EXECUTE: E -di =(1Z})-(dy}) = Edy WM = q’fE ‘di = (I’Efdy = q’Em — ya) yb — ya = «10.670 m, positive since the displacement is upward and we have taken +32 to be upward. WM = q'E(yb — ya) = (+28.0><10‘9 C)(4.00><104 N/C)(+0.670 m) = +7.50x104 J. EVALUATE: The electric force on the positive charge is upward so it does positive work for an upward displacement of the charge. (c) SET UP: The path is sketched in Figure 23.17c. K Figure 23.17c y=0 yb =—rsin6: —(2.60 m)sin45°=—1.838 m The vertical component of the 2.60 m displacement is 1.838 m downward. EXECUTE: di = + dj (The displacement has both horizontal and vertical components.) E -di A: (E?) - (dxi + dyj) = Edy (Only the vertical component of the displacement contributes to the work.) WM = q'fE -d7 = q'E fdy = q’E(y,, — y.) W a—>b = q’E( y — y )= (+28.0><10'9C)(4.00x104 N/C)(—1.838 m) = —2.O6><10‘3 J. b a EVALUATE: The electric force on the positive charge is upward so it does negative work for a displacement of the charge that has a downward component. 23.29. 23.40. 23.41. (a) IDENTIFY and SET UP: The direction of 13:” is always from high potential to low potential so point :5 is at higher potential. (1)) Apply Eq.(23.17) to relate Vb —Va to E. EXECUTE: Vb ~Va =—f1§ -di =fde= E(xb —xa). Vb -—Va _ +240 V xb — xa 0.90 In — 0.60 In E = = 800 V/rn (c) WM : qtvb —Va) : (—0.200x10‘6 C)(+240 V) : e4.80><l0'5 r. EVALUATE: The electric force does negative work on a negative charge when the negative charge moves from high potential (point b) to low potential (point a). 7 w TDENTIFY and SET UP: For oppositely charged parallel plates. E = 6/ 60 between the plates and the potential difference between the plates is V = Ed . -9 2 EXECUTE: (a) E z 2 ~ M 50 60 (b) V 2 Ed 2 (5310 N/C)(0.0220 m) = l 17 V. (c) The electric field stays the same if the separation of the plates doubles. The potential difference between the plates doubles. EVALUATE: The electric field of an infinite sheet of charge is uniform, independent of distance from the sheet. The force on a test charge between the two plates is constant because the electric field is constant. The potential difference is the work per unit charge on a test charge when it moves from one plate to the other. When the distance doubles the work, which is force times distance, doubles and the potential difference doubles. IDENTIFY and SET UP: Use the result of Example 23.9 to relate the electric field between the plates to the potential difference between them and their separation. The force this field exerts on the particle is given by Eq.(21.3). Use the equation that precedes Eq.(23.17) to calculate the work. I =5310 N/C. EXECUTE: (a) From Example 23.9, E = Vab = fl]— = 8000 V/m d 0.0450 m (b) F = |q|E = (2.40x10‘9 C)(8000 V/m) = +1.92x10“5 N (c) The electric field between the plates is shown in Figure 23.41. ‘ + + + + + aunt. Figure 23.41 The plate with positive charge (plate a) is at higher potential. The electric field is directed from high potential toward low potential (or, E is from + charge toward -— charge), so E points from a to [2. Hence the force that E exerts on the positive charge is from a to b, so it does positive work. W = ff?“ . d? = Fd, where d is the separation between the plates. W = Fd = (1.92X10"5 N)(0.0450 m) = +8.64x10'7 I (d) Va —~Vb = +360 V (plate a is at higher potential) AU = Ub —Ua = q(Vb —Va) = (2.40x10‘9 C)(—360 V) = —8.64><10'7 J. EVALUATE: We see that WHb = ~(Ub —Ua) =Ua —Ub. 23.63. IDENTIFY and SET UP: Use Eq.(21.3) to calculate 17‘ and then F‘ = mii gives ii. EXECUTE: (3) FE = qE‘. Since q = —e is negative FE and E are in opposite directions; E is upward so FE is downward. The magnitude of F E is ; FE = |q|E = eE = (1.602><10‘19 C)(1.10><103 N/C) =1.76><10“6 N. (b) Calculate the acceleration of the electron produced by the electric force: —16 a :5 zw = 1.93x10” m/s2 m 9.109X10” kg EVALUATE: This is much larger than g = 9.80 m/sz, so the gravity force on the electron can be neglected. [7 E is downward, so [i is downward. . . t . . (c) IDENTIFY and SET UP: The acceleration is constant and downward, so the motion is like that of a pI‘OJCCtlle. Use the horizontal motion to find the time and then use the time to find the vertical displacement. EXECUTE: x—component vex = 6.50><106 m/s; ax : 0; )6 ~ x0 = 0.060 m; 1:? x — x0 = VOXI+%(1XZ2 and the ax term is zero, so I: x_x° =—0‘96£6m—=9.231><10‘9 s Vox 6.50>< 10 m/s y-component voy =0; ay =1.93><10‘4 m/52; t=9.231x10‘9 m/s; y— yo = ? y—y0 =v0yt+%ayt2 y — y0 = §(l.93>><10l4 m/s2)(9.231><10'9 s)2 = 0.00822 m = 0.822 cm (d) The velocity and its components as the electron leaves the plates are sketched in Figire 23.63. V VI = vex = 6.50><106 m/s (since ax =0) x , vy = voy + a yr E vy : vy 2 0+ (1.93x1014 m/s2)(9.231><10'9 s) ' ‘ “ ' " ‘ " “ ‘ 'V vy =1.782><106 m/s Figure 23.63 6 tana = 1y— = = 0.2742 so 0! =15.3°. vx 6.50><10 m/s EVALUATE: The greater the electric field or the smaller the initial speed the greater the downward deflection. (e) IDENTIFY and SET UP: Consider the motion of the electron after it leaves the region between the plates. Outside the plates there is no electric field, so a = 0. (Gravity can still be neglected since the electron is traveling at such high speed and the times are small.) Use the horizontal motion to find the time it takes the electron to travel 0.120 m horizontally to the screen. From this time find the distance downward that the electron travels. EXECUTE: x-component v0x =6.50><106 m/s; ax =0; x—xO =0.120 In; t: ? x — x0 = vext + #1th and the ax term is term is zero, so t = x"x° 52% =1.846><10‘8 s Vox 6.50X10 m/s y-component v0y =l.782><106 m/s (from part (b)); ay =0; t =1.846X10‘8 m/s; y — y0 = ? y— y0 = voyt+%ayt2 = (1.782><106 m/s)(l.846><10"8 s) = 0.0329 m = 3.29 cm EVALUATE: The electron travels downward a distance 0.822 cm while it is between the plates and a distance 3.29 cm while traveling from the edge of the plates to the screen. The total downward deflection is 0.822 cm + 3.29 cm = 4.11 cm. The horizontal distance between the plates is half the horizontal distance the electron travels after it leaves the plates. And the vertical velocity of the electron increases as it travels between the plates, so it makes sense for it to have greater downward displacement during the motion after it leaves the plates. ...
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This homework help was uploaded on 04/09/2008 for the course PH 1120 taught by Professor Keil during the Spring '08 term at WPI.

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obj-4 - 23.13. IDENTIFY: E points from high potential to...

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