electric fields and acceleration?

Question

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??

Answer 1

I calculate the answer to be:
1.199 E7, just like you say it should be.

Are you remembering to multiply by the charge on the Proton?
The charge on a proton is +1.602177 E-19 C.

E = 2 * k_e * lambda / r
E = 2 * 8.99 E9 * 5.7E-12 / .820
E = .125 N/C

F = q*E = ma
a = q*E / m
a = 1.602177 E-19 C * .125 N/C / 1.67*10^-27 kg
a = 1.199 E7 m/s^2