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A proton with an initial horizontal velocity of 2.3*10^6 m/s enters a region of...

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undeadable | eNoter

Posted March 7, 2012 at 3:45 AM via web

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A proton with an initial horizontal velocity of 2.3*10^6 m/s enters a region of uniform electric field of strength 1.1*10^5 N/C between two oppositely charged parallel plates. If the plates are 1.3 cm apart and 6.7 cm long, how far below the top plate must the particle be shot so as to just miss hitting the right edge of the top plate?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted March 7, 2012 at 2:51 PM (Answer #1)

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A proton with an initial horizontal velocity of 2.3*10^6 m/s enters a region of uniform electric field of strength 1.1*10^5 N/C between two oppositely charged parallel plates.  The plates are 1.3 cm apart and 6.7 cm long.

It is assumed that the upper plate is negatively charged. Between the plates, the proton is accelerated upwards due to a force equal to 1.1*10^5*1.602*10^-19 N

This accelerates the proton by

1.1*10^5*1.602*10^-19/(1.672*10^-27) m/s^2

Let the distance below the top plate that the particle must be shot to miss the top plate be H. The proton travels 6.7 cm in 6.7/2.3*10^8 s.

In this time it rises upwards by 0.5*1.05*10^13*(2.91*10^-8)^2 = 4.44*10^-3 m

=> 0.44 cm

The proton must be shot at least 0.44 cm below the upper plate to pass between them without striking the upper plate.

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