Answered

Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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

#SPJ4

Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.