Respuesta :
Answer:
[tex]41.4496148484\ m/s[/tex]
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
[tex]\sigma[/tex] = Surface charge density = [tex]5.9\times 10^{-8}\ C/m^2[/tex]
[tex]\Delta x[/tex] = 0.57-0.26
q = Charge = [tex]6.9\times 10^{-9}\ C[/tex]
m = Mass of object = [tex]8.3\times 10^{-9}\ kg[/tex]
Electric field due to a sheet is given by
[tex]E=\dfrac{\sigma}{2\epsilon_0}\\\Rightarrow E=\dfrac{5.9\times 10^{-8}}{2\times 8.85\times 10^{-12}}\\\Rightarrow E=3333.33\ V/m[/tex]
Electric field is given by
[tex]E=\dfrac{V}{d}[/tex]
Voltage is given by
[tex]V=E\Delta x[/tex]
Kinetic energy is given by
[tex]K=qV[/tex]
[tex]\dfrac{1}{2}mv^2=qE\Delta x\\\Rightarrow v=\sqrt{\dfrac{2qE\Delta x}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 6.9\times 10^{-9}\times 3333.33\times (0.57-0.26)}{8.3\times 10^{-9}}}\\\Rightarrow v=41.4496148484\ m/s[/tex]
The initial speed of the object is [tex]41.4496148484\ m/s[/tex]
The speed of an object depends on its position. The initial speed of the object is 41.43 m/s.
What is speed?
The speed of an object is defined as the magnitude of the rate of change of its position with time.
Given that the mass m of the object is 8.30×10-9 kg and positive charge q is 6.90×10-9 C. The uniform surface charge density [tex]\sigma[/tex] = [tex]5.90 \times 10^{ - 8} {\rm{ C/m}}^2[/tex]. The permittivity of free space [tex]\epsilon _0[/tex] = [tex]8.85\times 10^{-12} \;\rm F/m[/tex]. The initial distance d1 of the object is 0.570 m and the final distance d2 = 0.260 m.
The electric field due to the sheet is given below.
[tex]E = \dfrac {\sigma}{2\epsilon_0}[/tex]
[tex]E = \dfrac {5.90 \times 10^{-8}}{2\times 8.85\times 10^{-12}}[/tex]
[tex]E = 3333.33 \;\rm V/m[/tex]
The voltage across the sheet is given as below.
[tex]V = E (d_1-d_2)[/tex]
[tex]V = 3333.33 ( 0.57 -0.26)[/tex]
[tex]V = 1033.33 \;\rm V[/tex]
The object is in motion, hence it has some kinetic energy which is given as below.
[tex]KE = \dfrac {1}{2}mv^2[/tex]
The kinetic energy is also given in terms of voltage and charge.
[tex]KE = qV[/tex]
So, [tex]\dfrac {1}{2}mv^2 = qV[/tex]
[tex]\dfrac {1}{2} \times 8.30\times 10^{-9}\times v^2 = 6.90\times 10^{-9}\times 1033.33[/tex]
[tex]v^2 = \dfrac{1.425\times 10^{-5}}{8.30 \times 10^{-9}}[/tex]
[tex]v = \sqrt{1716.86}[/tex]
[tex]v = 41.43 \;\rm m/s[/tex]
Hence we can conclude that the initial speed of the object is 41.43 m/s.
To know more about the speed, follow the link given below.
https://brainly.com/question/706354.
