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21-7 - FIGURE 11-16 Example 21—1 111(1 we Qt = —25 MC...

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Unformatted text preview: FIGURE 11-16 Example 21—1 111(1)), we Qt = —25 MC P {22 = +£01.11? .—d—. F—rl = 2.0 cm«+-—r2 = 3.1.] cot—d don’t know the relative lengths of E L and E: (a) until we do the calculation. Q E"1 Q .1 4:. .2 ii, (In) E at a point between tum charges. Two point charges are separated by a distance of 10:0 cm. IDne has a charge of —25 aC and the other +50 put]. (a) Determine the direction and magnitude of the electric field at a point P between the two charges that is 2.0en1 from the negative charge (Fig. 21—2tia}. {b} If an electron [mass = 9.11 x I'll—31kg} is placed at rest at P and then released, what will he its initial acceleration {direction and magnitude}? APPRMCH The electric field at P will he the vector sum of the fields m‘eated sepa— rately by 91 and 9;. The field due to the negative charge Q, points toward 91 , and the field due to the positive charge Q1 points away from Q1. Thus both fields point to the left as shown in Fig. Zl—Zfih and we can add the magnitudes of the two fields together algebraically, ignoring the signs of the charges In (it) we use Newton's second law {F = me) to determine the acceleration, 1where F = :35 (Eel, 21—5]. SOLUHDN [HJEaeh field is due to a point charge as given by Eq. 21—4, is = We. The total field is E = tend; = liege) "1 "z "1 J’2 '5 -6 (an x tflgN.m1yC2}(M + L00?) (an x iii-2m)2 {so x iii-2m = as x 10mm. (in) The electric field points to the left, so the electron will feel a force to the right since it is negativelyr charged. Therefore the acceleration a = Ffm (Newton’s second law} will he to the right. The force on a charge q in an electric field E is F = {(15 (Eq. 21—5]. Hence the magnitude of the acceleration is 1.6!;I X Ill—“C 63 X lflsN C a = E = E = —{ I": _31 I} =1.1><10me52. m m 9.11 X 10 kg More By carefully considering the directions or and: field (E, and fig before doing any calculations, we made sure our calculation could be done simply and correctly. ...
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