To solve this problem we will start by calculating the electric field. This can be defined as the change in charge density over twice the permittivity constant in a vacuum. From this value we will proceed to calculate by the relations of energy and load, the relation with the speed and the position of the objective.
Our values,
[tex]\text{The mass of the object} = m = 7.5*10^{-9}Kg[/tex]
[tex]\text{The charge of the object} = q = 7.3*10^{-9}C[/tex]
[tex]\text{The charge density of the sheet} = \sigma = 5.9*10^{-8}C/m^2[/tex]
[tex]\text{The initial position of the object} = x_1 = 0.460m[/tex]
[tex]\text{The initial position of the another object} = x_2 = 0.100m[/tex]
The electric field due to very large insulating sheet can be calculated as
[tex]E = \frac{\sigma}{2\epsilon_0}[/tex]
[tex]E = \frac{5.9*10^{-8}C/m^2}{2(8.85*10^{-12}C^2/N\cdot m^2)}[/tex]
[tex]E = 3333.33V/m[/tex]
The relation between the electric field E and potential V is given by,
[tex]E = \frac{V}{d}[/tex]
Therefore, the potential in terms of electric field can be written as,
[tex]V = E\Delta x[/tex]
The kinetic energy of the object is given by
[tex]K = qV[/tex]
[tex]\frac{1}{2} mv^2 = q(E\Delta x)[/tex]
The speed of the object then is
[tex]v = \sqrt{\frac{2qE\Delta x}{m}}[/tex]
Replacing we have then,
[tex]v = \sqrt{\frac{2(7.3*10^{-9})(3333.33)(0.460-0.1)}{7.5*10^{-9}}}[/tex]
[tex]v = 48.3322m/s[/tex]
Therefore the initial speed is 48.3322m/s
