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Q: College Physics Help ( No Answer,   0 Comments )
Question  
Subject: College Physics Help
Category: Science > Physics
Asked by: ph2005-ga
List Price: $35.00
Posted: 19 Jun 2004 13:10 PDT
Expires: 19 Jun 2004 14:19 PDT
Question ID: 363427
I need the answers for the following three questions with work shown
(this way I can learn what is going on).

1.	A wire with cross-sectional area A carries a current I. Show that
the electric field strength E in the wire is proportional to the
current per unit area (I/A) and identify the constant of
proportionality. [Hint: Assume a length L of wire. How is the
potential difference across the wire related to the electric field in
the wire? (Which is uniform?) Use V = IR and the connection between
resistance and resistivity.]
2.	Near Earth?s surface the air contains both negative and positive
ions, due to radioactivity in the soil and cosmic rays from space. As
a simplified model, assume there are 600 singly charged positive lions
per cm^3 and 500 singly charged negative ions per cm^3; ignore the
presence of multiply-charged ions. The electric field is 100 V/m,
directed downward.
	(a.) In which direction do the positive ions move? In which direction
do the negative ions move?
	(b.) What is the direction of the current due to these ions?
	(c.) The measured resistivity of the air in the region is 4.0 x 10^13
?*m. Calculate the drift speed of the ions, assuming it to be the same
for positive and negative ions. [Hint: Consider a vertical tube of air
of length L and cross-sectional area A. How is the potential
difference across the tube related to the electric field strength?]
	(d.) If these conditions existed over the entire surface of the
Earth, what is the total current due to the movement of ions in the
air?

3.	(a.) Given two identical, ideal batteries (emf = E) and identical
light bulbs (resistance = R), design a circuit to make both bulbs glow
as brightly as possible. (b.) What is the power dissipated by each
bulb? (c.) Design a circuit to make both bulbs glow, but one more
brightly than the other. Identify the brighter bulb.
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