The Electric Field Strength of a Proton Accelerated at 1 Million g

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Homework Help Overview

The discussion revolves around calculating the electric field strength experienced by a proton that is accelerated at a rate of 1 million "g". Participants are exploring the relationship between force, mass, charge, and electric field strength in the context of this acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between force and electric field strength, questioning how to apply the known values of mass and acceleration to find the electric field. There is uncertainty regarding the initial setup and the relevance of gravitational fields in this context.

Discussion Status

Some participants have provided calculations and attempted to derive the electric field strength, while others have pointed out potential errors in the order of magnitude. The conversation reflects a mix of interpretations and calculations without reaching a definitive conclusion.

Contextual Notes

There is mention of the acceleration value being converted from "g" to meters per second squared, and participants are navigating through the implications of using mass and charge of the proton in their calculations. The discussion highlights the complexity of relating gravitational concepts to electric fields.

joej
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What is the electric field strength at a point in space where a proton experiences an acceleration of 1 million "g"?

I'm completely lost on this one I know it's bound to be something simple but I just can't figure out where to start with it, well other than that 1 million "g" should be 9.8x10^6 m/s^2

I'm also not quite sure, but I'm guessing this will have something to do with gravitational fields.
 
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The question seems to infer that the proton is accelerated by the electric field.

Think about the relationship of Force to the electric field strength and as a function of mass and acceleration.

[tex]\vec{F} = m\vec{a} = q\vec{E}[/tex]
 
okay, so I get...

ma = (1.64x10^-21)/(1.6x10^-19) = 0.01N/C

would that be right?
 
joej said:
okay, so I get...

ma = (1.64x10^-21)/(1.6x10^-19) = 0.01N/C

would that be right?

U missed one order of size:
[tex]|\vec{E}|=\frac{m_{proton}|\vec{a}|}{q_{proton}}[/tex]
[tex]|\vec{E}|=\frac{1.67\cdot 10^{-27}Kg\cdot 9.8\cdot 10^{6} ms^{-2}}{1.6\cdot 10^{-19}C}\sim 0,1N/C[/tex]

Daniel.
 

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