Unformatted text preview: FIGURE 1116 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 iii2m)2 {so x iii2m = 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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 Spring '11
 Getahneh

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