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 = (\sigma)/(2\epsilon_0)`

`E = (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 = (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`

`(1)/(2) mv^2 = q(E\Delta x)`

The speed of the object then is

`v = \sqrt{(2qE\Delta x)/(m)}`

Replacing we have then,

`v = \sqrt{(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