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23.12.2022 05:58
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A very small object with mass 7.50Γ—10βˆ’9 kg and positive charge 7.30Γ—10βˆ’9 C is projected

A very small object with mass 7.50Γ—10βˆ’9 kg and positive charge 7.30Γ—10βˆ’9 C is projected directly toward a very large insulating sheet of positive charge that has uniform surface charge density 5.90Γ—10βˆ’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?
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23ricorvan
23ricorvan
4,4(2 marks)

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

mprjug6
mprjug6
4,8(21 marks)

C. Procedural

Explanation:

Because he step by step investigated it

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