Respuesta :
Answer:
The Magnitude of charge = q = 15.47 micro coulomb
Step-by-step explanation:
The attractive force between two points q and - 2 q = F = 2.2 Newton
The separation between two charges = 1.4 meters
Now from Coulomb's law = F = [tex](k \frac{q1 q2}{r^{2}})[/tex] where k= [tex]9 \times 10^{9}[/tex] [tex]\frac{N m^{2}}{c^{2}}[/tex]
So , 2.2 = [tex]9 \times 10^{9}[/tex] ( [tex](\frac{(q) ( - 2q)}{1.4^{2}})[/tex]
So , q ² = ([tex]\frac{0.479 \times 10^{-9}}{-2}[/tex])
For charge magnitude
I.e q ² = [tex]2.395 \times 10^{-10}[/tex]
so, q = [tex]1.547 \times 10^{-5}[/tex]
q = 15.47 \times 10^{-6} c
Or, q = 15.47 micro coulomb
Hence the Magnitude of charge = q = 15.47 micro coulomb Answer
By using the electromagnetic force equation, we will see that the magnitude of the charge is q = 1.55*10^5 Coulombs.
How to get the magnitude of the charge?
The force between two charges q₁ and q₂ is:
[tex]|F| = k*\frac{q_1*q_2}{r^2}[/tex]
Where r is the distance between the charges, r = 1.4 meters, and k is a constant.
k = 8.99*10^9 kg*m^3/s^2*C^2
In this case, we know that:
q₁ = q
q₂ = -2q
|F| = 2.2 N
Replacing all that we get:
[tex]2.2N = (8.99*10^9 kg*m^3/s^2*C^2)\frac{|q*-2q|}{(1.4m)^2} \\\\2.2N*\frac{(1.4m)^2}{ (8.99*10^9 kg*m^3/s^2*C^2)} = 2q^2\\\\\sqrt{2.2N*\frac{(1.4m)^2}{ 2*(8.99*10^9 kg*m^3/s^2*C^2)}} = q = 1.55*10^-5 C[/tex]
So we have that the magnitude of the charge q is 1.55*10^5 Coulombs.
If you want to learn more about electromagnetism,you can read:
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