# 16-2 Quiz Help

## 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=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.

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**- 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=1.6E-19*(135.6V/.0728m)

F=2.84428E-16

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=1.6E-19*(135.6V/.0728m)

F=2.84428E-16

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=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. 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 F-n-S Night**************

The Craziest 4-Way 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 Hip-Hop 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 I-495 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=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. 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 F-n-S Night**************

The Craziest 4-Way 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 Hip-Hop 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 I-495 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: 16-2 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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