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# Page 6 - insuring zero electric ﬁeld inside the conductor...

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Unformatted text preview: insuring zero electric ﬁeld inside the conductor) leaving negative charge in the amount — Z. on the outside surface of the pipe. Using a Gaussian surface on the outside of the conﬁguration would yield a net charge - 2. which would necessitate electric field lines in the sketch. There are no electric ﬁeld lines outside the pipe, so there must be extra free charge in the system. In fact, the additional free charge is in the amount of +2. , and it is located on the inside surface of the pipe. Why inside surface? Some positive charge must reside there so that the charge enclosed inside a Gaussian surface inside the conductor is zero. If there is free charge available to do that job, it will do that job (actually, what really happens when the positive free charge is put on the pipe is that electrons are removed--with that, the electrons that are left rearrange themselves so as to cover over all the positive charge in the structure except positive charge on the inside surface of the pipe--that free positive charge will be in the amount +2). In short, we have negative charge down the wire and an equal amount of positive charge on the pipe at r = b. b.) What do you know about the electrical potential outside the pipe? Splutim: The temptation is to assume that the negatively charged wire is grounded and, hence, should be assumed to have zero electrical potential. The problem with doing this is that with no electric ﬁeld outside the pipe, the electrical potential there must be a constant throughout the region all the way to inﬁnity. Because the electrical potential in that region all the way to inﬁnity must also be the same as the electrical potential at the outside surface of the pipe (remember, electrical potential functions must be continuous), which in turn must be equal to the electrical potential on the inside surface of the pipe (there is no electric ﬁeld between r = b and r = c, so the electrical potential difference between any two points in the pipe from r = b outward must be zero, and the electrical potential must be equal throughout), it leaves us with the distasteful prospect of having the electrical potential of the inside surface of the pipe equaling the electrical potential at inﬁnity. WE COULD DO THIS (again, electrical potential diﬂ'erences are all that are important), but let's not. Instead, let's assume that the electrical potential at inﬁnity is zero and see where that takes us. c.) What do you know about the electrical potential inside the dotted area? m: Inside area B, the electric ﬁeld is zero (no ﬁeld lines), so the electrical potential is a constant. From Parts 0. and b, that constant is zero. d.) What do you know about the electrical potential inside the hollow? Solution: We can use Gauss's Law to derive the electric field expression for the region inside the hollow. Doing so yields -il(21zeor), where - 2. is the linear density function for the charge on the wire, the negative sign signiﬁes that the electric ﬁeld is not outward but, instead, inward, and r is the perpendicular distance between the central axis and a point of interest inside the hollow. As this is a Mr function, the electrical potential will be a 111 r function (remember, the electrical potential is the anti-derivative of the electric ﬁeld). In any case, the electrical potential turns out to be -[- l /(21:so)][ln(r) - ln(b)]. Notice that at r = b, the electrical potential is zero, and inside the hollow at r < b, the electrical potential is negative. This all makes sense, given our choice for zero electrical potential. That is, the electrical potential at r = b is zero, but the electrical potential for r < b is less than zero--it is negative. Consequence: If you let a 118 ...
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