What is the electric field strength of a dipole rod on the x axis?

AI Thread Summary
The discussion revolves around calculating the electric field strength of a charged dipole rod along the x-axis. The rod, with a line charge density defined as T = T(x/L), is analyzed for points where x > L. The initial attempt yields an expression for the electric field, but the user mistakenly finds a 1/x^2 dependence instead of the expected 1/x^3 for dipole fields at large distances. Participants clarify that the electric field behavior should indeed reflect the 1/x^3 dependence characteristic of dipoles, regardless of whether the points of interest are parallel or perpendicular to the dipole. The conversation emphasizes the importance of correctly applying the dipole field equations in this context.
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Homework Statement



A rod of length 2L lies on the x axis, centered at the origen, and carries a line charge

density given by t=T(x/L), where T is a constant. (a) Find an expression for the electric

field strength at points on the x axis, for x>L (b) Show that for x>>L, your result has the

1/x^3 dependence of a dipole field, and by comparison, determine the dipole moment of the

rod.

Homework Equations



dE = (k dq)/r^2

The Attempt at a Solution



dq=Tda, dE= (k dq)/(x-a)^2 = (kT/L) * (a da) / (x-a)^2

integral dE form -L to +L

E = (kT/L) * ( - x/(x+L) + x/(x-L) + ln( (x-L)/(x+L)) )

when x>>L , by using binomial expanding, the result what I have

1/x^2 dependence. Help !
 
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Could I get confirmation on this question? Normally when electric dipoles are talked of, we would look to find fields for distances perpendicular to the dipole rather than parallel.
 
The situation fof that problem, To find the fields for distances parallel to the dipole.

I think it doesn't matter whether perpendicular or parallel.

The result of both situation has 1/^3 dependence of a dipole field, I think

If you know the answer about perpendicular case, I really appreciate your sincerity

although that is not exact answer about this problem. Thank you for your reply.
 
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