Electric fields and acceleration

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To find the initial acceleration of a proton above a charged wire, the electric field (E) generated by the wire is calculated using the formula E = 2*k*charge density/distance. Substituting the given values, E is determined to be approximately 1.199*10^7 N/C. The acceleration (a) of the proton is then found using the equation a = q*E/m, where q is the charge of the proton and m is its mass. The calculations indicate that the initial acceleration of the proton should indeed yield the expected result of 1.199*10^7 m/s². Clarification on the electric field equation for a linear charge distribution is suggested for accurate results.
eltel2910
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Consider a long horizontal conducting wire with charge density
= 5.70*10^-12 C/m. A proton (mass = 1.67*10^-27 kg) is placed and
released 0.820 m above the wire. What is the magnitude of initial
acceleration of the proton?

My thought is: a=q*E/m

and E is: 2*k*charge density/distance to the proton or 2*8990000000*5.7*10^-12/.820m

but I can't get the right answer, which I know to be: 1.199*10^7.

Any thoughts??
 
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Sounds like you are on the right track, but I don't have my E&M books at home (where I am now). What is the equation that you are using for the electric field from a uniform linear charge distribution?
 

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