162 Quiz Help
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Question 2
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=VbVa 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.
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=VbVa 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 Guest
Question #3
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=1.6E19*(135.6V/.0728m)
F=2.84428E16
Answer is positive since the electric force points toward the positive force.
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=1.6E19*(135.6V/.0728m)
F=2.84428E16
Answer is positive since the electric force points toward the positive force.
Malcolm Guest
Question #1
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=VbVa
In this case Vab=128.8V128.8V=257.6V
Then we can solve for deltaPE:
deltaPE=2{Since its 2 positive elementary electron charges we use 2 here}*(1.6E19)*(257.6)
deltaPE=8.234e17
Answer is negative, the done is by the electric force, not someone against the electric force pushing the particle over.Not so bad. Good luck!
Since everyone should have finished the hw, everyone come out to IFSC night tonight in the SAC Ballroom A annnnnd**************The OFFICIAL After Party to FnS Night**************
The Craziest 4Way Party Of HUMANKIND
Why Settle for PARTYING At A BORING CLUB?! COME THRU and PARTY With The Best!!
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Code Red Entertainment
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and
Purple Haze Productions
Causes a Night of Mayhem and Destruction,
So Be Prepared For...
 A R M A G E D D O N 
 February 5th 2009 
*****Ground Control*****
18 to Board
21 to Vaporize
*****Casualty*****
Ladies.... $ 5
Gentlemen.... $8
Inflation after 12:30 am
Greeks, Fraternities & Sororities NO inflation
*****DJ Evolution*****
DJ will be spinning the hottest and your favorite Reggae, Reggaeton, Salsa, Bachata, Merengue, and HipHop ALL NIGHT LONG!!!
*****Evacuation Method*****
Free Transportation at the SAC Loop all night
OR
We will pick you up at your door step
  There will be special treats if you take the VAN!!!  
*****Coordinates*****
Club Pollo Rico
2435 Middle Country Road
Centereach, NY
*****Path of Havoc*****
From Stony Brook
Take Nichols Road South to Route 25
Make a right onto Route 25
Continue on Route 25
Pollo Rico will be on your right side
From New York City
Take I495 to exit 62 North (Stony Brook)
Continue North on Nicholls Road
Exit off of Route 25
Make a left onto Route 25
Continue on Route 25
Pollo Rico will be on your right side
*****Last Dance*****
Drinks Special All NIGHT
that means de todo everything...
even sangria for those who love that lol
State ID or SBU ID Required!
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=VbVa
In this case Vab=128.8V128.8V=257.6V
Then we can solve for deltaPE:
deltaPE=2{Since its 2 positive elementary electron charges we use 2 here}*(1.6E19)*(257.6)
deltaPE=8.234e17
Answer is negative, the done is by the electric force, not someone against the electric force pushing the particle over.Not so bad. Good luck!
Since everyone should have finished the hw, everyone come out to IFSC night tonight in the SAC Ballroom A annnnnd**************The OFFICIAL After Party to FnS Night**************
The Craziest 4Way Party Of HUMANKIND
Why Settle for PARTYING At A BORING CLUB?! COME THRU and PARTY With The Best!!
Angry Afrikan Entertainment
Code Red Entertainment
Green Wave Productions
and
Purple Haze Productions
Causes a Night of Mayhem and Destruction,
So Be Prepared For...
 A R M A G E D D O N 
 February 5th 2009 
*****Ground Control*****
18 to Board
21 to Vaporize
*****Casualty*****
Ladies.... $ 5
Gentlemen.... $8
Inflation after 12:30 am
Greeks, Fraternities & Sororities NO inflation
*****DJ Evolution*****
DJ will be spinning the hottest and your favorite Reggae, Reggaeton, Salsa, Bachata, Merengue, and HipHop ALL NIGHT LONG!!!
*****Evacuation Method*****
Free Transportation at the SAC Loop all night
OR
We will pick you up at your door step
  There will be special treats if you take the VAN!!!  
*****Coordinates*****
Club Pollo Rico
2435 Middle Country Road
Centereach, NY
*****Path of Havoc*****
From Stony Brook
Take Nichols Road South to Route 25
Make a right onto Route 25
Continue on Route 25
Pollo Rico will be on your right side
From New York City
Take I495 to exit 62 North (Stony Brook)
Continue North on Nicholls Road
Exit off of Route 25
Make a left onto Route 25
Continue on Route 25
Pollo Rico will be on your right side
*****Last Dance*****
Drinks Special All NIGHT
that means de todo everything...
even sangria for those who love that lol
State ID or SBU ID Required!
Malcolm Guest
16_2 Question 4
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?
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?
Guest1 Guest
Re: 162 Quiz Help
for number 4
use the equation E= deltaV/delta x and rearrange to solve for deltaV
the answer is negative.
use the equation E= deltaV/delta x and rearrange to solve for deltaV
the answer is negative.
guest 00 Guest
Question 4
Dude question 4 is the easiest one of the assignment. Open your cd and just go to 27' man
wow Guest
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