Respuesta :
Answer:
2.24×10⁻⁸ C
Explanation:
From coulomb's law,
F = kq1q2/r² ............................ Equation 1
Where F = Repulsive force between the two charges, q1 = charge on the first sphere, q2 = charge on the second sphere, r = distance between the sphere, k = proportionality constant.
Note: q1 = q2 = q,
Then, we can rewrite equation 1 as,
F = kq²/r²
making q the subject of the equation
q = √(Fr²/k)................................. Equation 2
Given: F = 21 mN = 0.021 N, r = 15 mm = 0.015 m
Constant: k = 9.0×10⁹ Nm²/C²
Substituting these values into equation 2
q = √(0.021×0.015²/9.0×10⁹)
q = √(5×10⁻¹⁶)
q = 2.24×10⁻⁸ C.
Hence the charge on each sphere = 2.24×10⁻⁸ C
The magnitude of the charge on each sphere is [tex]2.29*10^{-8}C[/tex]
The force between two charges is given by coulombs law,
[tex]F=k\frac{Q_{1}*Q_{2}}{r^{2} }[/tex]
Where r is distance between charges.
And k is constant, [tex]k=9*10^{9} Nm^{2}/C^{2}[/tex]
Given that, [tex]Q_{1}=Q_{2},r=15mm=15*10^{-3}m,F=21mN=21*10^{-3}N[/tex]
Substitute above values in above formula.
[tex]21*10^{-3}=9*10^{9}*\frac{Q^{2} }{(15*10^{-3} )^{2} } \\\\Q^{2}=\frac{21*10^{-3}*225*10^{-6} }{9*10^{9} } \\\\Q^{2}=5.25*10^{-16} \\\\Q=\sqrt{5.25*10^{-16}} =2.29*10^{-8}C[/tex]
Hence, the magnitude of the charge on each sphere is [tex]2.29*10^{-8}C[/tex]
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