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1 month ago

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Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after

accelerating through potential difference 4V?

1 answer:

Ostrovityanka [3.2K]1 month ago

7 0

Answer: v = $1.19 * 10^{6} m/s$

Explanation: q signifies the amount of electric charge = $1.609 * 10^{-19} c$

mass of an electron = $9.10 * 10^{-31} kg$

V = potential difference = 4V

v = speed of the electron

Using the work-energy theorem, the kinetic energy acquired by the electron must equal the work done in propelling it.

The kinetic energy = $\frac{mv^{2} }{2}$ , and potential energy = qV

Thus, $\frac{mv^{2} }{2} = qV$

$\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s$

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