Introductory Physics Volume Two
Introductory Physics Volume Two
Introductory Physics Volume Two
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180 Hints A<br />
2.11 First suppose that there is a charge Q on the length L of the<br />
central wire, and a charge −Q on the outer shield. Then use gauss’s<br />
law to show that the electric field strength between the wire and shield<br />
is E =<br />
Q 1<br />
2πLɛ 0 r<br />
. Next show that the potential difference between the<br />
wire and shield is ∆V =<br />
Q<br />
2πLɛ 0<br />
ln(b/a). Finally use the definition of<br />
capacitance.<br />
2.12 The field around a line charge has already been found, use this<br />
to find the electric potential around a single wire. Find the electric<br />
potential difference as you move from one to the other for each wire<br />
and then add. Once you have the electric potential difference between<br />
the wires you can find the capacitance.<br />
2.13 Since there is a maximum field strength there is also a maximum<br />
energy density, and the energy density is the energy stored divided by<br />
the volume of the capacitor.<br />
2.14 The change in the charge switches the direction of the electric<br />
field, which alters the dot product. Also q = −|q|.<br />
2.15 Use the work energy theorem.<br />
2.16 Use the work energy theorem.<br />
2.17 Use the work energy theorem.<br />
2.18 Use the work energy theorem.<br />
2.19 Use the work energy theorem.<br />
2.20 Assemble the particles one at a time. The work to bring the<br />
first particle in is zero since there is no field to do work against. The<br />
second particles feels the potential of the first as you bring it in. The<br />
third particle feels the potential of the first two, and so on.<br />
2.21 The field is the gradient of the potential.<br />
2.22 V = ∫ dq<br />
4πɛ . 0r<br />
2.23 Find the electric potential for each section first. You have actually<br />
done a problem like each of the sections before.<br />
2.24 Write out the field and electric potential of a point charge, then<br />
do some algebra.<br />
2.25 The potential is the sum of the three point potentials. The field<br />
can be found from the derivative of the potential.<br />
2.26 The answer will be in terms of the radius (r 0 ), field (E 0 ), charge<br />
density (σ 0 ), and potential (V 0 ) of the original drops. The charge is all<br />
on the surface of the drop.
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