Question

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.

Answer 1

Answer:

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

Explanation:

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

Mass of proton is equal to $m=1.67\times 10^{-27}kg$

And charge on proton is equal to $e=1.6\times 10^{-19}C$

Electrostatic force will be responsible for the motion of proton

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

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 $1.67\times 10^{-27}\times a=8\times 10^{-13}$

$a=4.79\times 10^{14}m/sec^2$

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