Approximations used in solving for the electric field

  • #1
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Homework Statement


Suppose we have a rod of length L oriented on the z-axis so that the bottom end is at z= - L/2 and the top end if at z = L/2. Furthermore, the rod is uniformly charged with charge density 'λ'. Now we are interested in finding the electric field at a point x away from the midpoint of the rod. If this distance is very far away so that x>>L. What is the electric field at x?

Homework Equations


E = (λ/4πεx)*(L/√(x_squared + (L_squared/4))), in the x hat direction.
(Sorry for the mess, I'm new here and am not familiar with Latex or any other methods of typing mathematical equations).

The Attempt at a Solution


Now my intuition tells me that if x>>L, then x/L>>1 or L/x must equal about 0. I see an L/x term in my equation and if I apply the latter approximation my electric field becomes zero. If I am very, very far away from the charged rod, I shouldn't feel its effects. However, another solution I see is to factor out the x_squared from the denominator to get E = (λ/4πεx_squared)*(L/√(1 + (L_squared/4x_squared))) THEN apply the approximation to get rid of the L_squared/4x_squared term and arrive at a final answer of E = λL/4πεx_squared, in the x hat direction. What is wrong with my initial attempt?
 

Answers and Replies

  • #2
BvU
Science Advisor
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Hi. Nothing wrong with your initial shot: the L/x factor stays in there (after all you are looking at te x-dependence) and you get $$|E| = {\lambda L\over 4 \pi\varepsilon_0 x^2} = {1\over 4 \pi\varepsilon_0} \, {Q\over x^2} $$ just like from a simple charge at the origin -- and just like your second approach.

By the way: Hello Karo, :welcome: (in case you weren't welcomed yet).

##\LaTeX## is fun. Check out the intros. Start with ##\#\# ## a^2 + b^2 = c^2 ##\#\# ## to get ## a^2 + b^2 = c^2 ##.
Proceed with right-mouse click over an equation and "Show math as ##\TeX## commands"
 

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