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 removedwith
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
pipethat 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 antiderivative 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 zeroit is negative. Consequence: If you let a 118 ...
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 Spring '11
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