Electric Field along the x axis of a dipole

  • Thread starter Libohove90
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Homework Statement


Show that the field on the x axis for the dipole (negative on the right, positive on the left separated by 2a) is given by the equation E = kq / x^3 for x >>a.

Homework Equations


Coulomb's Law F = kq1q2 / r^2
E = F/q = kQ / r^2


The Attempt at a Solution



Since its a vector sum, I just add the two equations to get E = kq / r^2 + k(-q) / r^2

The solutions manual gives me E = kq / (x-a)^2 + k(-q) / (x+a)^2

I feel stupid but I seem to be hitting a wall. Why does the positive charge have (x-a)^2, while the negative charge has a (x+a)^2? What does 'a' represent? I keep confusing myself, can someone clarify this for me? I understand everything except the nature of the denominators. Thanks
 

Answers and Replies

  • #2
tiny-tim
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Hi Libohove90! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)
Why does the positive charge have (x-a)^2, while the negative charge has a (x+a)^2? What does 'a' represent?

If you're at position x, you're x-a from the nearer charge, and x+a from the further charge. :wink:
 

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