Solutions to Homework Questions 1
Solutions to Homework Questions 1
Solutions to Homework Questions 1
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Physics 112 <strong>Homework</strong> 1 (solutions) (2004 Fall)<br />
Chapt15, Problem-28: Figure P15.28 shows<br />
the electric field lines for two point charges separated<br />
by a small distance. (a) Determine the ratio q 1 /q 2 . (b)<br />
What are the signs of q 1 and q 2 ?<br />
Solution:<br />
The magnitude of q 2 is three times the magnitude of q 1<br />
because 3 times as many lines emerge from q 2<br />
as enter q 1 . q 2 = 3 q 1<br />
(a) Then, q 1 q 2 =−1 3<br />
(b)<br />
q 2 > 0 because lines emerge from it,<br />
and q 1 < 0 because lines terminate on it<br />
Chapt15, Problem-35: If the electric field strength in air exceeds 3.0 x 10 6 N/C, the air becomes a<br />
conduc<strong>to</strong>r. Using this fact, determine the maximum amount of charge that can be carried by a metal sphere<br />
2.0 m in radius.<br />
Solution:<br />
For a uniformly charged sphere, the field is strongest at the surface.<br />
Thus, E max = k e q max<br />
R 2 ,<br />
or<br />
q max = R2 E max<br />
k e<br />
( )<br />
( = 2.0 m)2<br />
3.0 ×10 6 N C<br />
8.99 ×10 9 N⋅ m 2 C 2 = 1.3 ×10 −3 C<br />
Chapt15, Problem-41: A 40-cm-diameter loop is rotated in a uniform electric field until the<br />
position of maximum electric flux is found. The flux in this position is measured <strong>to</strong> be 5.2 x 10 5 N · m 2 /C.<br />
Calculate the electric field strength in this region.<br />
Solution:<br />
From the definition of Electric Flux Φ E = EAcosθ and Φ E = Φ E, max when θ = 0°<br />
Thus, E = Φ E, max<br />
A<br />
( )<br />
= Φ E, max<br />
πd 2 4 = 4 5.2 ×105 N⋅ m 2 C<br />
π 0.40 m<br />
( ) 2 = 4.1× 106 N C<br />
5