16-2 Quiz Help

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16-2 Quiz Help

Post  Kathleen on Thu Feb 05, 2009 1:41 pm



Last edited by Kathleen on Thu Feb 05, 2009 10:03 pm; edited 1 time in total

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q2 and q4

Post  hwilson on Thu Feb 05, 2009 4:33 pm

I cant figure out #2 and 3 can someone help plz?

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Question 2

Post  Malcolm on Thu Feb 05, 2009 5:11 pm

An electron starts with a kinetic energy of 383.8 eV (electronvolts) from a potential of181 V and moves
to a positive potential with the same magnitude. What is the final kinetic energy in eV (see sheet
19,20) ? Indicate with a negative (positive) sign whether the electron loses (gains) potential energy.

Its good to start out by making a drawing of the difference between the positive and negative potentials.

O181<---------<-------<------<------(-181)O All you really have to do to find the difference in potentials is to multiply the potential given by two and reverse the sign. Or just use the eqn they give you.

Vab=Vb-Va Vab=181-(-181)=362V.

Since you have 1 electron, use the Equation DeltaPE=q*Vab
q=# of electrons. 1e-(One electron)*Vab(362V)=362eV(electron volts)

Finally, just add the initial electron volts given, 383.3eV with the eV change in PE, 362eV
383.3eV+362eV=745.8eV

The answer is negative because the electron gains KE, and therefore loses potential energy.

Malcolm
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number 3

Post  sck105 on Thu Feb 05, 2009 5:33 pm

I need help with number 3 has anyone figured it out yet plz help me?

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Question #3

Post  Malcolm on Thu Feb 05, 2009 6:41 pm

An electron enters a region with uniform electric field between two parallel plates which have a potential
difference of 135.6 V and a distance of 7.628 cm between them. What is the magnitude of the electric force on
the electron (see sheet 26,27) ? Indicate with a positive (negative) sign whether the electric force points
toward the positive (negative) plate.

Use two equations on sheet 26 and 27
F=qE F=Electric force q=Charge on an electron E=electric field
E=V/d V=Potential difference d=distance between two plates in meters
So, put the two equations together and you get
F=q*(V/d)
Convert the cm to meters by dividing the cm by 100 so that d=7.628cm/(100cm/m)=.07628m

Then solve, F=1.6E-19*(135.6V/.0728m)
F=2.84428E-16
Answer is positive since the electric force points toward the positive force.

Malcolm
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Question #1

Post  Malcolm on Thu Feb 05, 2009 6:56 pm

An alpha- particle (a Helium ion carrying 2 positive elementary electron charges) starts from at rest at a
potential of 128.8 V and moves to a negative potential with the same magnitude. What is the magnitude of
the change in potential energy (see sheet 19,20) ? Indicate with a negative (positive) sign whether work
is done on the particle by the electric force (by somebody against the electric force pushing the particle
over).

In this one we use the equations: Wab=deltaPE=q*Vab
Vab=Vb-Va

In this case Vab=-128.8V-128.8V=-257.6V

Then we can solve for deltaPE:
deltaPE=2{Since its 2 positive elementary electron charges we use 2 here}*(1.6E-19)*(-257.6)
deltaPE=-8.234e-17

Answer is negative, the done is by the electric force, not someone against the electric force pushing the particle over.Not so bad.Smile Good luck!

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yo

Post  lupe on Thu Feb 05, 2009 8:57 pm

wus good jeezy

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16_2 Question 4

Post  Guest1 on Thu Feb 05, 2009 10:46 pm

In a region with a uniform electric field of 45.37 N/C you make a step of 5.149 cm in a direction parallel to the
field lines. What is the magnitude of the potential change (see sheet 27') ? Indicate with a positive
(negative) sign whether you step parallel (perpendicular) to the equipotential lines.


how do you do this?

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Re: 16-2 Quiz Help

Post  guest 00 on Fri Feb 06, 2009 1:28 am

for number 4
use the equation E= deltaV/delta x and rearrange to solve for deltaV
the answer is negative.

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Question 4

Post  wow on Fri Feb 06, 2009 1:35 am

Dude question 4 is the easiest one of the assignment. Open your cd and just go to 27' man

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