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So i suppose we can call it and write x y the same

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Unformatted text preview: tial close to the wire, or by doing a brute force integration using the electrostatic force law and the charge on the wire. By superposition, if we now have two wires located at z = c and z = -c, and oppositely charged (positive at c and negative at -c), we obtain F(z) = A[Ln(z-c) - Ln(z+c)] This gives the real part (actual potential) as Ôz-cÔ ∞(z) = A lnÔ Ô Ôz+cÔ The equipotentials are therefore given by Ôz-cÔ Ô Ô = constant Ôz+cÔ If we use a little coordinate geometry, and find all (x, y) whose distance from one point = constant time its distance from another, we get the equation of a circle (or in one special case when the constant is 1, a line). This gives the equipotential lines as shown: 30 - + Question What is the significance of the conjugate potential? Answer. Since ∞ and § are conjugate, ∂∞ ∂§ ∂∞ ∂§ = and =∂x ∂y ∂y ∂x In other words, Ô ∞ and Ô § are orthogonal. However, since these gradients are themselves orthogonal to the lines ∞ = const and §...
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