Answer:
The charge on the sphere is = 309 nC
Explanation:
The electric potential energy is given by the equation:
[tex]U = \frac{1}{4 \pi \epsilon_o} \frac{q_1q_2}{r}[/tex]
where;
Permittivity constant [tex]\epsilon_o[/tex] [tex]= 8.85 *10^{-12} \ C^2/Nm^2[/tex]
the distance between the two charges = r
The kinetic energy that is said to be given by the spring to the bead is converted to electric potential energy at the closest approach.
So; the kinetic energy is equal to the the elastic potential energy;
The elastic potential energy is represented by the equation:
[tex]U_s = \frac{1}{2} kx^2[/tex]
where
k = spring constant
x = displacement from the equilibrium position
Converting the charge of the bead from nano coulombs to coulombs ; we have:
[tex]q_b = 2.5 \ nC \\ \\ q_b = 2.5 \ nC ( \frac{10^{-9} \ C}{1 \ nC } )[/tex]
[tex]q_b = 2.5*10^{-9} \ C[/tex]
Thus, the bead is approaching the metal sphere of Van de Graff generator. Then the kinetic energy is being transferred to electric potential energy. The process ends when it reaches to the sphere with the Closest approach.
∴ The charge of the sphere can be determined by relating the electric potential energy to elastic potential energy.
[tex]\frac{1}{4 \pi \epsilon_o} \frac{q_bQ_s}{r} = \frac{1}{2} kx^2[/tex]
Making [tex]Q_s[/tex] the subject pf the formula; we have:
[tex]Q_s = \frac{2 \pi \epsilon_o kx^2 r}{q_b}[/tex]
[tex]Q_s = \frac{2 (8.85*10^{-12} \ C^2 /Nm^2 )(0.65 \ N/m) (2.141*10^{-5 \ m^3)}}{2.5*10^{-9} \ C }[/tex]
[tex]Q_s = 3.09 *10^{-9} \ C[/tex]
[tex]Q_s = 3.09 *10^{-9} \ C (\frac{1 \ nC}{10^{-9} \ C})[/tex]
[tex]Q_s = 309 \ nC[/tex]
Thus the charge on the sphere is = 309 nC
