Question

You have two conducting plates with nothing in between them. Using a voltmeter, you measure the voltage difference between the plates to be 465 V. You release an alpha particle with mass of 6.5×10-27 kg and a charge of 3.2×10-19 C from near the positive plate. What is the kinetic energy of the alpha particle when it reaches the other plate? The distance between the plates is 44.0 cm. b.) You observe an electron move through a uniform electric field. The electron/field-system loses 2480 pJ of electrostatic potential energy as the electron picks up speed. What was the voltage difference between the two points

Answer 1

Given that
The potential between the plates is V = 465 V

the alpha particle with mass m = 6.5*10^-27 kg

charge q = 3.2*10^-19 C

which is initially at rest vi = 0 m/s

when it is released due to potential difference the alpha particle can move so that

the potential energy between the plates shoudl equal to the kinetic energy

and we know that the relation between potential, potential energy and charged particle in electric field is

   V = U/q
   U = V*q
her U = k.e

   k.e = V*q

   k.e = 465*3.2*10^-19 J

   k.e = 1.488*10^-16 J

the kinetic energy of the alpha particle when it reaches the other plate i s k.e = 1.488*10^-16 J

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when electron is moving electrostatic potential energy lost is dU = 2480*10^-12 J

the voltage difference between the two points is V = U/q

       V = dU/q

       V = (2480*10^-12)/(1.6*10^-19) v = 155*10^8 V