Electric fields and acceleration

[eltel2910]

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??

 

[berkeman]

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?

 

[kyle8921]

Your thought process is correct. The formula for acceleration in an electric field is a=qE/m, where q is the charge of the particle, E is the electric field, and m is the mass of the particle. In this case, the charge of the proton is 1.6*10^-19 C and the mass is 1.67*10^-27 kg.

The electric field can be calculated using the formula E=2*k*lambda/d, where k is the Coulomb constant (8.99*10^9 N*m^2/C^2), lambda is the charge density (5.7*10^-12 C/m), and d is the distance between the proton and the wire (0.820 m).

Plugging in the values, we get E=2*8.99*10^9*5.7*10^-12/0.820=1.18*10^7 N/C.

Substituting this into the acceleration formula, we get a=(1.6*10^-19)(1.18*10^7)/(1.67*10^-27)=1.199*10^7 m/s^2. This is the correct answer of 1.199*10^7 m/s^2.

Make sure to check your units and use the correct value for the Coulomb constant. Also, double check your calculations to ensure accuracy.