If you put a total of 4.65×106 electrons on an intially electrically neutral wir

Question

If you put a total of 4.65×106 electrons on an intially electrically neutral wire of length 1.43 m, what is the magnitude of the electric field a perpendicular distance of 0.401 m away from the center of the wire?

-What is the magnitude of the acceleration that a proton experiences at this point in space?

I believe this is a Line of charge problem but i get lost after obtaining (Linear Charge Denisty),

This is what i have so far,

since 4.65e6 electrons are applied to the wire each with a charge of -1.603e-19 results in a net charge application of q = 7.45e-13 C.

Taking q/dx, where dx is the length of the wire (1.43m) i obtain a value of 5.209e-13 C/m

I get stuck here. Some guidance would be appreciated

_

Thanks,

Ryan

Explanation / Answer

The number of charge carries n = 4.65*10^6

The charge of the electron q = 1.6*10^-19C

thne the net charge Q = nq = 7.45*10^-13C

The length of the wire l = 1.43m

the perpendicular distance a = 0.401m

Then the electric field at a point perpendicular to the wire is

E = Q/20a [ 1/(l^2 + 4a^2)]

= (7.45*10^-13C)/20(0.401) [ 1/(1.43^2 + 4(0.401)^2)]

= (0.6099)

= 0.0204 N/C

The relation between electric filed and force

  F = Eq

ma = Eq

Therefore the accleration of the proton

a = EQ/m

     = (0.0204)(7.45*10^-13)/1.67*10^-27)

      = 0.091*10^14

       = 9.1*10^12 m/s^2

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