Answer:
The value is [tex]n = 1.527 *10^{14} \ electrons [/tex]
Explanation:
From the question we are told that
The mass of the sphere is m = 18.5 kg
The diameter is [tex]d= 25.0 \ cm = 0.25 \ m[/tex]
The number of electrons is [tex]N = 8.10 *10^{15} \ electrons[/tex]
The mass of the plastic ball is [tex]m_b = 0.120 g[/tex]
The initial acceleration of the ball is [tex]u = 1525\ m/s^2[/tex]
Generally the radius of the sphere is mathematically evaluated as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{0.25}{2}[/tex]
=> [tex]r = 0.125 \ m [/tex]
Generally the force between the ball and the sphere is
[tex]F = \frac{k * q_1 * q_2 }{r^2}[/tex]
Generally this force can also be mathematically represented as
[tex]F = m_b * a[/tex]
So
[tex] m_b * a = \frac{k * q_1 * q_2 }{r^2}[/tex]
Here [tex]q_1[/tex] is the charge removed from the sphere whicn is evaluated as
[tex]q_1 = N * e[/tex]
substituting [tex]1.60*10^{-19} \ C[/tex] for e
[tex]q_1 = 8.10 *10^{15} * 1.60*10^{-19}[/tex]
[tex]q_1 = 0.0013 \ C [/tex]
and [tex]q_2[/tex] is the charge to be removed from the ball
So
[tex] 0.120 * 1525 = \frac{9*10^9 * 0.0013* q_2 }{0.125^2}[/tex]
[tex]q_2 = 2.44 *10^{-7} \ C [/tex]
Generally the number of electron to be removed from the ball is mathematically represented as
[tex]n = \frac{q_2}{e}[/tex]
=> [tex]n = \frac{ 2.44 *10^{-7}}{1.60*10^{-19}}[/tex]
=> [tex]n = 1.527 *10^{14} \ electrons [/tex]
