Problem 23.47 A point charge q1 = 4.10 nC is placed at the origin, and a second

Question

Problem 23.47 A point charge q1 = 4.10 nC is placed at the origin, and a second point charge q2 = -3.00 nC is placed on the x-axis at x=+ 21.0 cm . A third point charge q3 = 1.90 nC is to be placed on the x-axis between q1 and q2. (Take as zero the potential energy of the three charges when they are infinitely far apart.)

Part A What is the potential energy of the system of the three charges if q3 is placed at x=+ 10.5 cm ? U = J SubmitMy AnswersGive Up

Part B Where should q3 be placed between q1 and q2 to make the potential energy of the system equal to zero? x = cm SubmitMy AnswersGive Up

Explanation / Answer

First put q1. So work needed to be done to put q1= 0. Then put q2;

work needed (V1)= k*q1*q2/r1 ;r1-distance betweenq1 and q2=21 cm

V1= -9*109 *4.1*3/0.21 = -5.27 x 10-7

then place q3.

work needed(V2)= k*q1*q3/r2 + k*q2*q3/r3 ; distance between q1 and q3(r2) =10.5 ; r3= 21-10.5= 10.5

V2 = 1.79 x 10-7 J

potential energy= V1+V2= -3.48 x 10-7 J

B) total potential energy as before

V= k*q1*q2/r1+ k*q1*q3/r2 + k*q2*q3/r3 ; where r2=x ; r3= 21- x ; for V=0

-k*3*4.1/21+ k*q1*q3/x + k*q2*q3/(21-x) =0 ;

4.1/x - 3/(21-x) = 4.1*3/1.9 =123/19

19[4.1(21-x) -3x] = 123x(21-x)

123x2 - 2717.9 x +1635.9 =0

x=21.48 or 0.62 cm

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