Respuesta :
Electrostatic force between two sphere is given by
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we know that
r = 0.60 m
F = 10.8 N
[tex]q_1 + q_2 = -24 \mu C[/tex]
now we will have
[tex]10.8 = \frac{9 \times 10^9 q_1q_2}{0.60^2}[/tex]
since the two charges are attracting each other so here the two charges must be opposite in nature
[tex]q_1q_2 = -4.32 \times 10^{-10}[/tex]
now we will use
[tex]q_1 - \frac{4.32 \times 10^{-10}}{q_1}= -24 \times 10^{-6}[/tex]
[tex]q_1^2 + 24 \times 10^{-6} q_1 - 4.32 \times 10^{-10} = 0[/tex]
by solving above equation we will have
[tex]q_1 = \pm 36\mu C[/tex]
[tex]q_2 = \mp 12\mu C[/tex]
now the force between two sphere when connected by wire is to be determine
so here we have net charge on each sphere is half of total charge
so we will have
[tex]q_1 = q_2 = -12\mu C[/tex]
so force now is given as
[tex]F = \frac{9\times 10^9 (-12\times 10^{-6})^2}{0.60^2}[/tex]
[tex]F = 3.6 N[/tex]
this is repulsive force between two charges
Electric force on each small conducting spheres when, two spheres are connected by a slender conducting wire, which is then removed is 3.6 N.
What is electric force?
Electric force is the force of attraction of repulsion between two bodies.
According to the Coulombs law, the electric force of attraction of repulsion between charged two bodies is directly proportional to the product of charges of them and inversely proportional to the square of distance between them. It can be given as,
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
Here, (k) is the coulombs constant.
Given information-
The distance between the spheres is 0.6 m.
The electric force experience by each sphere is 10.8 N.
The total charge on the two spheres is -24
Let Q₁ is the charge of the first sphere and Q₂ is the charge of sphere 2. As the total charge on the two spheres is -24. Thus,
Q₁+Q₂=-24 X 10⁻⁶
Let the above equation as equation 1.
As the electric force experience by each sphere is 10.8 N and the distance between the spheres is 0.6 m. Thus by using the formula of electric force,
[tex]10.8=9\times10^9\times\dfrac{Q_1Q_2}{0.6^2}[\tex]
Let the above equation as equation 2.
On solving equation 1 and 2, we get,
Q₁=±36 μC
Q₂=±12 μC
Now the two spheres are now connected by a slender conducting wire. Thus the net charge on each square should be half of the total charge. Thus,
Q₁=Q₂=-12 μC.
Put the values in the formula of electric force, we get,
F=3.6 N.
Hence, the electric force on each sphere when, two spheres are connected by a slender conducting wire, which is then removed is 3.6 N.
learn more about the electric force here;
https://brainly.com/question/14372859
