Answer:
The charge of each sphere is [tex]q = - 1.84 *10^{-4}C[/tex]
Explanation:
The free body diagram of this question is shown on the first uploaded image
From the question we are told that
The mass of the two sphere is [tex]m = 3 \ kg[/tex]
The length of the connection is [tex]l = 20\ m[/tex]
The angle between the suspended side is [tex]\theta = 17 ^o[/tex]
At Equilibrium the force acting at the horizontal is = 0N and the net force acting at the vertical is zero
This can be represented mathematically as
[tex]\sum F_y = 0N , \ \ and \ \ \sum F_x = 0N[/tex]
For [tex]\sum F_y = 0N[/tex]
[tex]T cos \theta = mg[/tex]
=> [tex]cos \theta = \frac{mg}{T}[/tex]
For [tex]\sum F_x = 0N[/tex]
[tex]T sin \theta = F[/tex]
=> [tex]sin \theta = \frac{F}{T}[/tex]
Now [tex]tan \theta = \frac{sin \theta }{cos \theta }[/tex]
[tex]= \frac{\frac{F}{T} }{\frac{mg}{T} }[/tex]
[tex]= \frac{F}{mg}[/tex]
Where F is the electrical force which is mathematically represented as
[tex]F = \frac{kq^2}{r^2}[/tex]
Therefore
[tex]tan \theta = \frac{kq^2}{mgr^2}[/tex]
Where [tex]r[/tex] is the distance between the two masses and from the diagram it is
[tex]r = 2l sin \theta[/tex]
So
[tex]tan \theta = \frac{kq^2 }{(2l sin \theta )^2 * mg}[/tex]
making q the subject of the formula
[tex]q = \sqrt{\frac{mg}{k} * tan \theta (2l sin \theta )^2 }[/tex]
Where k is the Coulomb's constant with a value of [tex]k = 9*10^{9} kg \ cdot m^3 s^{-4} A^{-2}[/tex]
Substituting values
[tex]q = \sqrt{\frac{3 * 9.8 }{9*10^9} * tan (17) (20 sin 17)^2 }[/tex]
[tex]q = - 1.84 *10^{-4}C[/tex]
The negative sign is because we are told from the question that they are negatively charged
