One of the smallest revolvers ever created fires a tiny bullet, with a mass of o

Question

One of the smallest revolvers ever created fires a tiny bullet, with a mass of only 0.7 grams, but with a velocity of 134 m/s. The projectile has a final kinetic energy of 6.28 J. Prof. Askew has long had a fascination with attempting to recreate traditional firearms with electromagnetism. In his empty, pocket universe, he creates a device similar to a parallel plate capacitor: a constant electric field is produced between two plates (with small holes through which a projecile can travel), separated by 20 cm. A projectile with the same mass as the tiny bullet is produced, and a charge of 125.6 micro-coulombs is placed on the projectile. What potential difference between the plates is required in order to accelerate this projectile to the same kinetic energy as the tiny bullet? Ignoring any fringe field effects, how does the kinetic energy of this projectile change after the projectile leaves the plates?

Explanation / Answer

let potential difference required is V volts.

then work done by electric field=kinetic energy of the bullet

==>charge*potential=kinetic energy

==>125.6*10^(-6)*V=6.28

==>V=50000 volts

so potential difference between the plates should be 50000 Volts.

part b:

after the projectile leaves the pate, if there is any friction force (air friction/drag force) present in the pocket universe, then the kinetic energy will reduce.

other wise the kinetic energy will remain same if there is no energy loss due to work done against friction .

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