Calculate the initial (from rest) acceleration of a proton in a 5.00 x 10^6 N/C electric field (such as created by a research Van de Graaff). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.

Respuesta :

Answer:

Acceleration of the proton will be equal to [tex]4.79\times 10^{14}m/sec^2[/tex]

Explanation:

We have given electric field [tex]E=5\times 10^6N/C[/tex]

Mass of proton is equal to [tex]m=1.67\times 10^{-27}kg[/tex]

And charge on proton is equal to [tex]e=1.6\times 10^{-19}C[/tex]

Electrostatic force will be responsible for the motion of proton

Electrostatic force will be equal to [tex]F=qE=1.6\times 10^{-19}\times 5\times 10^6=8\times 10^{-13}N[/tex]

According to newton law force on the proton will be equal to F = ma, here m is mass of proton and a is acceleration

This newton force will be equal to electrostatic force

So [tex]1.67\times 10^{-27}\times a=8\times 10^{-13}[/tex]

[tex]a=4.79\times 10^{14}m/sec^2[/tex]

So acceleration of the proton will be equal to [tex]4.79\times 10^{14}m/sec^2[/tex]

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