Electric fields and acceleration

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SUMMARY

The discussion centers on calculating the initial acceleration of a proton placed above a long horizontal conducting wire with a charge density of 5.70 x 10^-12 C/m. The formula used for acceleration is a = q * E / m, where E is derived from the electric field equation for a uniform linear charge distribution. The correct calculation for the electric field E is 2 * k * charge density / distance to the proton, yielding an initial acceleration of 1.199 x 10^7 m/s² for the proton.

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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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