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Page 1: Quiz 6 - agni.phys.iit.eduagni.phys.iit.edu/~hanlet/phy221/Exams/Quiz6-s.pdf · Quiz 6 Name: Phys221 Fall 2007 Dr. P. Hanlet Show your work!!! If I can read it, I will give you partial

Quiz 6

Name:Phys221 Fall 2007 Dr. P. Hanlet

Show your work!!! If I can read it, I will give you partialcredit!!! Correct answers without work will NOT get full credit.

Concept (3 points)

1. In your own words, state Ohm’s Law. Is it always valid? If it is, give an example of when it is valid,

if it is not, also give an example of when it is not valid.

Ohm’s law relates the voltage with the current and resistance in circuit:

V = iR

The equation is only valid for components which are ohmic in nature; i.e.those whose resistivity is independent of the applied electric field:

ρ 6= ρ( ~E)

An example of Ohm’s law is circuit with a resistor. An example of whenthe law does not apply is for a transistor.

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Page 2: Quiz 6 - agni.phys.iit.eduagni.phys.iit.edu/~hanlet/phy221/Exams/Quiz6-s.pdf · Quiz 6 Name: Phys221 Fall 2007 Dr. P. Hanlet Show your work!!! If I can read it, I will give you partial

Problem (5 points)

2. An isolated conducting sphere has a radius r = 10cm. One wire carries a current of iin = 1.000002A

to the sphere, and another wire carries iout = 1.000000A off of the sphere. How long will it take for

the sphere to increase in potential by V = 1kV ?

(k=8.99 × 109 N ·m2/C and ε0 =8.85 × 10−12C/(N ·m2))

The potential on the sphere is:

V = kq

r

and the change in potential is:

∆V = k∆q

r

such that:

∆q =r∆V

kAlso, since:

∆i = iin − iout =∆q

∆tOne can solve this for ∆t:

∆t =∆q

iin − iout

Substituting in for ∆q:

∆t =∆q

iin − iout

=r∆V

k(iin − iout)=

0.1m · 1000N · m

(8.99× 109N · m2/C)(1.000002A− 1.000000A)

or∆t = 5.56 × 10−ss = 5.56ms

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