Respuesta :
Answer: v = [tex]1.19 * 10^{6} m/s[/tex]
Explanation: q = magnitude of electronic charge = [tex]1.609 * 10^{-19} c[/tex]
mass of an electronic charge = [tex]9.10 * 10^{-31} kg[/tex]
V= potential difference = 4V
v = velocity of electron
by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.
kinetic energy = [tex]\frac{mv^{2} }{2}[/tex], potential energy = qV
hence, [tex]\frac{mv^{2} }{2} = qV[/tex]
[tex]\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s[/tex]
The speed of the charge after accelerating through a potential difference of 4V is [tex]1.18\times 10^6[/tex] m/s.
How do you calculate the velocity of the charge?
Given that the charge q has a velocity of v and potential difference V. The charge is in motion, thus its potential energy will be equal to its kinetic energy.
[tex]\dfrac {1}{2}mv^2 = qV[/tex]
Let us consider that the charge q has a magnitude of 1.6*10^-19 C which is the charge of an electron and the mass is 9.1*10^-31 kg. For the potential difference of 4V, the velocity will be,
[tex]\dfrac {1}{2}\times 9.1\times 10^{-31}\times v^2 = 1.6\times 10^{-19}\times 4[/tex]
[tex]v^2= 1.41 \times 10^{12}[/tex]
[tex]v = 1.18 \times 10 ^6\;\rm m/s[/tex]
Hence we can conclude that the speed of the charge after accelerating through a potential difference of 4V is [tex]1.18\times 10^6[/tex] m/s.
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https://brainly.com/question/1329567.
