Introductory Physics Volume Two
Introductory Physics Volume Two
Introductory Physics Volume Two
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A Hints 179<br />
1.14 Since this is only an estimate, assume that you are composed<br />
entirely of water.<br />
1.16 Be sure to add the parts as vectors.<br />
1.17 Let the charge elements dq be little sections of arc that subtend<br />
and angle dθ. Notice that the full charge is spread over a half a circle<br />
so that the charge density (charge per angle) is −7.5µC<br />
π<br />
, so that dq =<br />
−7.5µC<br />
π<br />
dθ.<br />
1.18 Use the work energy theorem. Recall that work is force time<br />
distance.<br />
1.19 This is like a projectile motion problem, but this time we have<br />
a constant electric force rather than an constant gravitational force.<br />
1.20 Start by making a free body diagram and be sure to include the<br />
force of gravity and the force of the string.<br />
1.21 A flux is negative if it is into the volume of the box and positive<br />
if the flux is out of the volume of the box.<br />
1.22 There is a very easy way to do this problem, by using the fact<br />
that there is no charge inside the volume of the nose cone.<br />
1.25 ⃗r = −xî<br />
2.1 Look at the definition of electric potential, how is the electric<br />
potential related to potential energy?<br />
2.2 Use the work energy theorem.<br />
2.3 Follow the example in the text for a nonuniform field.<br />
2.4 Follow the example in the text.<br />
2.6 Consider a closed surface that is totally within the outer conductor<br />
and surrounds the inside surface of the outer conductor, as indicated<br />
by the dotted line in the figure. Using the fact that there is no field<br />
inside a conductor argue that the electric flux through this surface is<br />
zero. From this use Gauss’s law to find the charge on the inside surface<br />
of the outer conductor. Next use the fact that the total charge on the<br />
outer conductor is equal to the sum of the charge on its two surfaces,<br />
to find the charge on the outside surface of the outer conductor.<br />
2.7 Use Gauss’s law. Remember that the electric field is zero inside<br />
the body of a conductor.<br />
2.8 Use the definition of capacitance.<br />
2.9 Use the fact that the electric field strength between the plates is<br />
both σ/ɛ 0 and ∆V/d, where σ is the charge density on the plates.<br />
2.10 Assume that the capacitor is charged so that the inside sphere<br />
has a charge Q and the outside sphere has a charge −Q. Use Gauss’s<br />
law to get the field strength between the shells.