Ra a2 232 1 the contribution from the lower pair of

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Unformatted text preview: nt A each positive charge contribution is compensated by a negative charge contribution in the current geometry. There is no contradiction with the finite value of the field (3) because the field is determined by how the potential changes in space (E = - ), not by the value of the potential at a point. Moreover, it's clear that all points of type A (along that vertical bisecting line on the figure) have the same zero potential, therefore the field there cannot have non-zero y component, and indeed, we found the field (3) to be directed along x axis. Extra: Expanding the result (3) in powers of a/R 1 and leaving only the main term, one easily obtains 3ke qa2 E= . (4) R4 The same result one would obtain by using the expression for the field of a dipole d: E = ke 3R(R d) - |R|2 d , |R|5 that we derived in the classroom. Contributions from upper (|R| = R - a/2) and ke qa lower (|R| = R + a/2) dipoles are then, respectively...
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This note was uploaded on 09/18/2008 for the course PHYS 2326 taught by Professor Ingor during the Spring '08 term at University of Texas at Dallas, Richardson.

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