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How Is the Electric Potential Difference Calculated Near an Infinite Wire?
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[QUOTE="sandy.bridge, post: 4520544, member: 356724"] [h2]Homework Statement [/h2] [itex]\rho_{wire}=a[/itex], surface charge density [itex]\rho_S[/itex] What is potential difference of a point a distance of [itex]b[/itex] measured from the centre of the infinitely long wire, and the surface of the wire? Since all of the charge is dispersed on the surface of the conductor, there will exist no charge within the surface, and henceforth the electric field inside the conductor is zero. In addition, symmetry of the problem tells us that the electric field radiates outwards from the conductor. Let the origin be centred in the wire, such that the z-axis runs parallel to the infinitely long wire. We can apply Gauss's Law utilizing a gaussian surface such that [itex]\rho_{gaussian}=g > a[/itex], and the height from the xy-plane will be [itex]l[/itex]. At this point, I just want to know if I am solving for the electric field properly. Thanks in advance. We know that [tex]\frac{1}{ε_o}\int _V \rho dV=\oint_S\vec{E}\cdot\vec{dS}[/tex] [tex]\frac{\rho_S(2\pi al)}{ε_o}=\oint_S\vec{E}\cdot\vec{dS}=E\int_SdS=E(2\pi gl)\rightarrow E = \frac{\rho_Sa}{b}[/tex], [/QUOTE]
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How Is the Electric Potential Difference Calculated Near an Infinite Wire?
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