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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.

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