Respuesta :
Answer:
[tex]Q_2=4.4293\times 10^{22}\ nC[/tex]
Explanation:
Given
charge on the fixed particle, [tex]Q_1=15\times 10^{-9}\ C[/tex]
mass of the second charge, [tex]m_2=9.5\times 10^{-3}\ kg[/tex]
distance between the fixed charge and the floating charge on the top, [tex]d=0.25\ m[/tex]
- According to the question the second charge is floating just above the fixed positive charge despite of gravity this means that the floating charge is also positive in nature and hence feels the repulsion from the fixed charge which is equal in magnitude to the gravitational force on the charge.
[tex]m_2.g=\frac{1}{4\pi\epsilon_0} \times \frac{Q_1.Q_2}{d^2}[/tex]
where:
[tex]\epsilon_0=[/tex] permittivity of free space
g = acceleration due to gravity
[tex]9.5\times 10^{-3}\times 9.8=8.85\times 10^{-9}\times \frac{15\times 10^{-9}\times Q_2}{0.25^2}[/tex]
[tex]Q_2=4.4293\times 10^{22}\ nC[/tex]
