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A proton is placed in an electric field of intensity 800 n/c. what are the magnitude and direction of the acceleration of the proton due to this field? (e = 1. 60 × 10-19 c, mproton = 1. 67 × 10-27 kg)

Sagot :

tanisharawat014

The acceleration of the proton is 7.66*10^10m/s^2 in the direction of the electric field.

To find the answer, we have to know more about the electric field.

How to find the acceleration of the proton?

  • We have the expression for electric field due to a accelerating particle  as,

             [tex]E=\frac{F}{q}[/tex] , where q is the charge of the proton and F is the force.

  • We have the expression for force as,

                [tex]F=ma[/tex]

  • Combining both and rearranging, we get,

                    [tex]ma=Eq\\\\a=\frac{Eq}{m} =\frac{800*1.60*10^{-19}}{1.67*10^{-27}} \\\\a=7.66*10^{10}m/s^2[/tex]

Thus, we can conclude that, the acceleration of the proton is 7.66*10^10m/s^2 in the direction of the electric field.

Learn more about the electric field here:

https://brainly.com/question/27752270

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