Answer:
[tex]12.4 \mu C, 40.2 \mu C[/tex]
Explanation:
The electric force between the two spheres is given by
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where
k is the Coulomb's constant
q1, q2 are the two charges
r is the distance between the spheres
Here we know
F = 1.19 N
r = 1.94 m
So we can solve to find [tex]q_1 \cdot q_2[/tex]:
[tex]q_1 \cdot q_2 = \frac{Fr^2}{k}=\frac{(1.19)(1.94)^2}{9\cdot 10^9}=4.98\cdot 10^{-10} C^2 =[/tex] (1)
We also know that the sum of the two charges is
[tex]q_1 + q_2 = 52.6 \mu C = 52.6\cdot 10^{-6}C[/tex] (2)
So we have a system of 2 equations; from (1) we get
[tex]q_1 = \frac{4.98\cdot 10^{-10}}{q_2}[/tex]
and substituting into (2)
[tex]\frac{4.98\cdot 10^{-10}}{q_2}+ q_2 = 52.6\cdot 10^{-6}C\\\rightarrow q_2^2 -52.6\cdot 10^{-6} q_2 + 4.98\cdot 10^{-10} =0[/tex]
which has two solutions:
[tex]q_1 = 40.2\cdot 10^{-6} = 40.2 \mu C\\q_2 = 12.4\cdot 10^{-6} = 12.4 \mu C[/tex]
Answer:
q1= 4.26×10^-6C
q2= 1.0×10^-6C
Explanation:
q1+q2= 5.26×10^-6C
F=1.19N
F= kq1q2/r^2 =1.19
q1q2= (1.19)(r^2/k)
q1q2= 1.19×(1.94^2)×8.99×10^9=4.026×10^-10C^2
q2= 5.26×10^-6 -q2 ....eq1
q1q2=4.026×10^-10 .....eq2
Put eq1 into eq2
q1(5.26×10^-6)- q1=4.026×10^-10
5.26×10^-6q1-q1^2-4.026×10^-10
q1^2-(5.26×10^-6)q1 +4.026×10^-10
Using almighty formular to solve the quadratic equation
= 5.26×10^-6 +-sqrt(5.26×10^-6)^2 -4(4.026×10^-10) /2
q1=5.26×10^-6 +-sqrt(1.065x10^-11)/2
q1= (5.26×10^-6 )+-3.26×10^-6/2
q1= (2.26×10^-6) + 3.26×10^-6/2
q1= 4.26×10^-6C
q2= (5.26×10^-6 ) -(3.26×10^-6)/2
q2= 1.0×10^-6C
