# 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?

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

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Leonard Stokes
F =qE
$F\mathrm{△}t=\mathrm{△}p$
$qE=\frac{\mathrm{△}p}{\mathrm{△}t}$
$E=\frac{\mathrm{△}p}{q\mathrm{△}t}$
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Jeffrey Jordon

Given that the initial momentum is ${p}_{1}=1.50\ast {10}^{-23}kg×m/s$
the final momentum is ${p}_{2}=5.00\ast {10}^{-23}kg×m/s$
the time taken is $t=6.3×{10}^{-6}$ s.
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The charge of the proton is $q=1.602×{10}^{-19}$C

The force related with momentum is
$F=\frac{\left({P}_{2}-{P}_{1}\right)}{t}$
${E}_{q}=\frac{\left({P}_{2}-{P}_{1}\right)}{t}$  (since $F={E}_{q}$)
SO electric field ,
$E=\frac{\left({P}_{2}-{P}_{1}\right)}{t×q}$
$=\frac{\left(5-1.5\right)×{10}^{-23}}{\left[1.602×{10}^{-19}×6.3×{10}^{-6}\right]}$
= $0.3468×{10}^{2}$
= 34.68 N/C