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= 0.48 mC = 4.8 x 10 1 . C q = 2 0.48mC—4.8X10C = 12.3km1.23X iO 1 r m 4 N• I C kc = 8.99 x i0 2 m 1 q 12.3km = 1 q 0.48mG 12.3 km Unknown: = E=? 0=? = 1.00, or 0 = = = 2 1 r 1 q 8.99xl0 8.99x10 1.43 x 10—2 N/C 2 q 2.85 x lO_2 N/C E k 1 c 9 N .m 2 C2 2 N 9 .m = (4.8X10 —4 C) (l.74xl0 m 4 2 ) X 10 C 4 ) (1.23 x 10 4 m) 2 (4,8 45.0° 1. Calculate the magnitude of the electric field strength produced by each charge: Because we are finding the magnitude of the electric field strength, we can dis regard the signs of each charge. tan 0 The angle that r 2 makes with the coordinate system equals the inverse tangent of the ratio of the vertical to the horizonta l components. Because these compone nts are equal, The distance r 2 must be calculated from the information in the diagram. Because r 2 forms the hypotenuse of a right triangle who se sides equal r , it follows that 1 = 2 J(rj) + (rj) 2 = .Jir 1 1.74 X i0 m Given: = 0.48 mC 1 r The Seto-Ohashi bridge, linking the two Japa nese islands of Honshu and Shikoku, is the longest “rail and road” bridg e, with an overall length of 12.3 km. Suppose two equal charges of 0.48 mC each are placed at the op posite ends of the bridge. Find the electric field strength, E, due to these charges at the point exactly 12.3 km abov e one of the bridge’s ends. .ECTRIC FIELD STRENGTH Ey,2 1 E i_2 N/C 2 (cos 45.0°) E 2.85 x = (1.43 x iO N/C)(cos 45.0°) 1.01 X iO_2 N/C = E 2 (sin 45.0°) = (1.43 X l0 N/C)(sin 45.0°) =i.01 N 2 /C x10 = = = 1.01 X 102 N/C 3.86 x l0_2 N/C X 2 1 N /C 0 N/C 75.3° 7. Evaluate your answer: The electric field strergth has a magnitude of nearly 0.04 N/C at the point above the bridge. The vector’s net direction is to the right and up from the pOSiL)( axka a.fl øf7S.3° = (3.86 x i02 N/C) =3.82 (1.01 x 2 io N /C 6 = tan(3.82) E()( tan6— 10 E, 6. Use a suitable trigonometric function to find the direc tion of the resultant electric field strength vector: ln this case, you can use the inverse tangent function. X /(L0l X l0 N/C 2) + (3.86 x 10-2 2 N/C) = = f()2+ (E 2 ) 0 = 5. Use the Pythagorean theorem to find the magnitude of the resul tant electric field strength vector: = 0 tE =Ey,i+Ey,2=t 2 1 /C 0 2.85X N + 1.01 = Extot= E 1+E = 0 + 1.01 X iO_2 N/C 2 4. Calculate the magnitude of the total electric field stren gth in both the x and y directions: For E : 2 1 For : E unt. q is positive. slyLing 3, Find the xand y components of each electric field strength vecto r: At this point, the direction of each component must be taken into acco The vector E 2 is directed at an angle upward and to the right because The vector E 1 is directed vertically upward because q 1 is positive. 2. Determine the direction of each electric field strength vector I the signs of the charges: - 218 m ‘ 2 1 q 1 5 0.Om is the electric field strength at the center of the rectang le? The value of q is 6.4 nC. 5 The world’s largest windows, which are in the Palace of Industry and Technology in Paris, France, have a maximum width of 218 m and a maximum height of 50.0 m. Consider a rectangle with these dimensions. If charges are placed at its corners, as shown in the figure below, what the ground 120 m from the first charge, the horizontal component of the resultant electric field strength is found to be E 1.60 x 102 N/C. Using this information, calculate the unknown quantity of charge. 3. Suppose three charges of 3.6 pC each are placed at three corners of the Imperial Palace in Beijing, China, which has a length of 960 m and a width of 750 m,What is the strength of the electric field at the fourth corner? 4. The largest fountain is found at Fountain Hills, Arizon a. Under ideal conditions, the 8000 kg column of water can reach as high as 190 m. Suppose a 12 nC charge is placed on the ground and anot her charge of unknown quantity is located 190 m above the first charge . At a point on The wheel held 10 first-class and 30 second-class cabins , and each cabin was capable of carrying 30 people. Consider two cabins positioned exactly opposite each other. Suppose one cabin has an unbalanced charge of 4.8 nC and the other cabin has a charge of 16 nC. At what distance from the 4.8 nC charge along the diameter of the wheel would the strength of the resultant electric field be zero? 2. In 1897, a Ferris wheel with a diameter of 86.5 m was built in London. 1. Pontiac Silverdome Stadium, in Detroit, Michigan, is the largest airsupported building in the world, with a length of 2.20 x 102 m and a capacity to hold more than 80 000 people. If a charge of 18.0 pG is placed at one end of the stadium and a charge of —12.0 pC is placed at the other end of the stadium, what will be the strength of the electric field halfway between the charges? Fet4 — ‘.‘t •a. r’ _____________ ) ‘ tie ‘LI • sic ‘t lh., S 61 1 L. I, ) -— e bAA a (l:ca.) L ML—’, sic’ -4- C’ I •;;,; Fe in. £ .. • J •_c--. *- i•,•• ‘C. &Xe.s )4 i 1 L i’t’ e j 2 at 42 — 4 -a-’ . 54q ti-Ic 1ç;; ç ( ft. F iv, I J ‘(L%’c Sr C-. 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