A proton is moving parallel to a uniform electric field. The electric field accelerates the proton and increases its linear momentum to 5.0 x 10^-23 kg x m/s from 1.5 x 10^-23 m/s in a time of 6.3 x 10^-6 s. What is the magnitude of the electric field?

Carol Gates 2021-02-09 Answered

A proton is moving parallel to a uniform electric field. The electric field accelerates the proton and increases its linear momentum to 5.0×1023kg x m/s from 1.5×1023m/s in a time of 6.3 x 10^-6 s. What is the magnitude of the electric field?

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Expert Answer

Leonard Stokes
Answered 2021-02-10 Author has 98 answers
Jeffrey Jordon
Answered 2021-09-30 Author has 2027 answers

Given that the initial momentum is p1=1.501023kg×m/s
the final momentum is p2=5.001023kg×m/s
the time taken is t=6.3×106 s.
-------------------------------------------------------------------------------
The charge of the proton is q=1.602×1019C

The force related with momentum is
F=(P2P1)t
Eq=(P2P1)t  (since F=Eq)
SO electric field ,
E=(P2P1)t×q
=(51.5)×1023[1.602×1019×6.3×106]
= 0.3468×102
= 34.68 N/C

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