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

Question

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

Leonard Stokes · Accepted Answer

F =qE    F△t=△p    qE=△p△t    E=△pq△t

Jeffrey Jordon · Accepted Answer

Given that the initial momentum is p1=1.50∗10−23kg×m/s the final momentum is p2=5.00∗10−23kg×m/s the time taken is t=6.3×10−6 s. ------------------------------------------------------------------------------- The charge of the proton is q=1.602×10−19C  The force related with momentum is F=(P2−P1)t Eq=(P2−P1)t  (since F=Eq) SO electric field , E=(P2−P1)t×q =(5−1.5)×10−23[1.602×10−19×6.3×10−6] = 0.3468×102 = 34.68 N/C

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?

Answered question

Answer & Explanation

New Questions in Other