Willing to pay fg to the person that get it right.








the answers i got are:

Submitted Answers
ANSWER 1:
q =
1.46⋅10−6
C
ANSWER 2:
q =
2.995⋅10−7
C
ANSWER 3:
q =
9.471⋅10−8
C



The half angle is θ = arcsin(20/22) = 65.38 degrees
T*cos(65.38) = m*g => T = m*g/cos(65.38) = .0018 kg*9.8/cos(27) = .04234 N
FE = k*q^2/d^2 so k*q^2/d^2 - T*sin(65.38) = 0
q = sqrt(T*sin(65.38)*d^2/k) = sqrt(.043234*sin(65.38)*.4meters^2/9.0x10^9) =1.46 x 10^-6 C


This post was edited by Stolem on Nov 27 2013 01:23am

(0.0485288337*(.22^2)/(8.987*10^9))^.5= 5.11228712e-7

Still not it

my bad, i misread the .18 grams; i had 18 grams
my angle was also a bit off

(0.0005050682991*(.22^2)/(8.987*10^9))^.5=5.21543294e-8

This post was edited by saber_x3 on Nov 27 2013 05:21pm

calculate angle:
ArcSin(Ѳ) = 11/40
Ѳ = 15.9620141628

Calculate tension:
T*Cos(Ѳ)=m*g
T=0.001834739813

Calculate the coulomb force:
F = k * q1 * q2 * (1/r^2)
Since they have equal charges can alter the equation:
F = k * q1^2 * (1/r^2)

Set the coulomb force equal to the tension in i:
k * q1^2 * (1/r^2) = T*Sin(Ѳ)
q1^2 = T*Sin(Ѳ)*r^2 * (1/k)
q1 = r*sqrt[T*Sin(Ѳ)*(1/k)]

q1 = 5.2126*10^(-8)