What is the lesser and larger charge?
Electric potential energy U = kq1q2/r = -270 μJ but q1+q2 = 25nC => q2 = 25e-9 - q1
-270e-6*r/k = q1*q2 = 9e-16 = q1*(25e-9 - q1) = 25e-9*q1- q1^2
q1^2 - 25e-9*q1 - 9e-16 = 0
Use the quadratic formula: -b/2*a +/- √b^2 - 4*a*c)/2*a
a =1, b = - 25e-9, c = - 9e-16
q1 = 25e-9/2 +/- √6.25e-16+36e-16)/2
q1 = 1.25e-8 +/- 3.25e-8 = 4.5e-8C or -2e-8C
q2 = 25e-9 -q1 = 25e-9 -4.5e-8 = -2e-8
q1 = 45nC, q2 = -20nC <-------------------- Answer
Check
q1*q2 = 45e-9 * -20e-9 => kq1q2/r = -9e9*9e-16/.03 = -2.7e-4 = -270e-6 CHECK
If this has helped you, please vote best answer
W=-270 μJ
r=3 cm=0.03 m
let 2 charges be q1 &q2
q1+q2=25 nC...1
W=kq1q2/r
-270*10^-6=9*10^9*q1*q2/0.03
q1*q2=-9*10^-16
q1=-9*10^-16/q2
putting the value of q1 in equation 1
-9*10^-16/q2+q2=25*10^-9
-9*10^-16+q2^2=25*10^-9*q2
q2^2-25*10^-9*q2-9*10^-16=0
q2=4.5*10^-8 C or -4*10^-8 C
when q2=4.5*10^-8 C
q1=-2*10^-8 C
when q2=-4.10^-8 C
q1=6.5*10^-8 C
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
Verified answer
Electric potential energy U = kq1q2/r = -270 μJ but q1+q2 = 25nC => q2 = 25e-9 - q1
-270e-6*r/k = q1*q2 = 9e-16 = q1*(25e-9 - q1) = 25e-9*q1- q1^2
q1^2 - 25e-9*q1 - 9e-16 = 0
Use the quadratic formula: -b/2*a +/- √b^2 - 4*a*c)/2*a
a =1, b = - 25e-9, c = - 9e-16
q1 = 25e-9/2 +/- √6.25e-16+36e-16)/2
q1 = 1.25e-8 +/- 3.25e-8 = 4.5e-8C or -2e-8C
q2 = 25e-9 -q1 = 25e-9 -4.5e-8 = -2e-8
q1 = 45nC, q2 = -20nC <-------------------- Answer
Check
q1*q2 = 45e-9 * -20e-9 => kq1q2/r = -9e9*9e-16/.03 = -2.7e-4 = -270e-6 CHECK
If this has helped you, please vote best answer
W=-270 μJ
r=3 cm=0.03 m
let 2 charges be q1 &q2
q1+q2=25 nC...1
W=kq1q2/r
-270*10^-6=9*10^9*q1*q2/0.03
q1*q2=-9*10^-16
q1=-9*10^-16/q2
putting the value of q1 in equation 1
-9*10^-16/q2+q2=25*10^-9
-9*10^-16+q2^2=25*10^-9*q2
q2^2-25*10^-9*q2-9*10^-16=0
q2=4.5*10^-8 C or -4*10^-8 C
when q2=4.5*10^-8 C
q1=-2*10^-8 C
when q2=-4.10^-8 C
q1=6.5*10^-8 C