Question

A very small object with mass 7.50×10−9 kg and positive charge 7.30×10−9 C is projected directly toward a very large insulating sheet of positive charge that has uniform surface charge density 5.90×10−8C/m2. The object is initially 0.460 m from the sheet.What initial speed must the object have in order for its closest distance of approach to the sheet to be 0.100 m?

Answer 1

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,

$\text{The mass of the object} = m = 7.5*10^{-9}Kg$

$\text{The charge of the object} = q = 7.3*10^{-9}C$

$\text{The charge density of the sheet} = \sigma = 5.9*10^{-8}C/m^2$

$\text{The initial position of the object} = x_1 = 0.460m$

$\text{The initial position of the another object} = x_2 = 0.100m$

The electric field due to very large insulating sheet can be calculated as

$E = \frac{\sigma}{2\epsilon_0}$

$E = \frac{5.9*10^{-8}C/m^2}{2(8.85*10^{-12}C^2/N\cdot m^2)}$

$E = 3333.33V/m$

The relation between the electric field E and potential V is given by,

$E = \frac{V}{d}$

Therefore, the potential in terms of electric field can be written as,

$V = E\Delta x$

The kinetic energy of the object is given by

$K = qV$

$\frac{1}{2} mv^2 = q(E\Delta x)$

The speed of the object then is

$v = \sqrt{\frac{2qE\Delta x}{m}}$

Replacing we have then,

$v = \sqrt{\frac{2(7.3*10^{-9})(3333.33)(0.460-0.1)}{7.5*10^{-9}}}$

$v = 48.3322m/s$

Therefore the initial speed is 48.3322m/s