Question about electrial potential

  • Thread starter leolaw
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  • #1
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The work done by an external force to move a [tex]-8.50{\mu}C[/tex] charge from a point a to point b is [tex]15.0*10^{-4}J[/tex]. If the charge was started from rest and had [tex] 4.82 * 10^{-4}J[/tex] of kinetic energy when it reached point b, what must be the potential difference between a and b.

I am not sure if this is suppose to be a trick question or not, but if the final kinetic energy is 4.82 * 10^ -4, then the initial potential difference has to be 4.82 * 10^-4 right?
 

Answers and Replies

  • #2
vsage
Consider the units of electric potention (V or J/C) and work (J)
 
  • #3
dextercioby
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What is the unit of potential difference...?

Daniel.
 
  • #4
85
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so
[tex]{\Delta} KE = -{\Delta}PE}[/tex]
[tex] \frac{1}{2}mv^2 = -qV_{ba}[/tex]
[tex]4.82*10^{-4}J=8.5*10^{-6} V_{ba}[/tex]
So V_ba = 5.6 * 10 V

But when I use
[tex] W = -qV_{ba}[/tex]
[tex] 15 * 10^{-4} = 8.5 * 10^{-6} V_{ba}[/tex]
[tex] V_{ba} = 1.76 * 10 ^2 [/tex]

I am kind of lost here, can somebody help
 
  • #5
xanthym
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You're not properly using the energy information provided. What's the relationship between the 2 energy values given in the problem?

~~
 
Last edited:
  • #6
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[tex] 4.82 * 10^{-4}J[/tex], which is the kinetic energy
 
  • #7
xanthym
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What about:

[tex] Work_{ext} = \Delta K.E. + \Delta P.E. [/tex]


~~
 
  • #8
xanthym
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HINT:

[tex] :(1): \ \ \ \ Work_{ext} = \Delta K.E. + \Delta P.E. [/tex]

[tex] :(2): \ \ \ \ (15x10^{-4} Joules) \ \ = \ \ (4.82x10^{-4} Joules) \ \ + \ \ (Charge)*(PotentialDifference) [/tex]

Do you see the technique?


~~
 
Last edited:
  • #9
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I have the answer -1.19 x 10 ^ -2 V
 
  • #10
dextercioby
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I've said that b4.It's the OP-s job to do the "dirty" work.Namely the simple calculations...We assume that the person coming in for advice/help has minimum knowledge of arithmetics...

Daniel.

P.S.Your result is 10.000 times smaler...
P.P.S.xanthym,the latex code for multiplication is either "\times" or "\cdot" or leaving a space...
 
  • #11
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xanthym said:
What about:

[tex] Work_{ext} = \Delta K.E. + \Delta P.E. [/tex]


~~
I didnt know this forumla before because I started from the middle of the book (since that i am taking second half of the course), so i skipped all the Mechnical energy stuff

But once i have this, it makes everything easier
 
  • #12
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dextercioby said:
I've said that b4.It's the OP-s job to do the "dirty" work.Namely the simple calculations...We assume that the person coming in for advice/help has minimum knowledge of arithmetics...

Daniel.

P.S.Your result is 10.000 times smaler...
P.P.S.xanthym,the latex code for multiplication is either "\times" or "\cdot" or leaving a space...
you mean it is -1.2 x 10 ^3?
 
  • #13
xanthym
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LEOLAW --
Don't guess. Repeat the calculation. (10^3 is still wrong.)
 
  • #14
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xanthym said:
HINT:

[tex] :(1): \ \ \ \ Work_{ext} = \Delta K.E. + \Delta P.E. [/tex]

[tex] :(2): \ \ \ \ (15x10^{-4} Joules) \ \ = \ \ (4.82x10^{-4} Joules) \ \ + \ \ (Charge)*(PotentialDifference) [/tex]

Do you see the technique?


~~
[tex] (15*10^{-4} Joules) = (4.82x10^{-4} Joules) + (Charge)*(PotentialDifference) [/tex]

[tex]1.018 * 10 ^{-3} = 8.5*10^{-6} * (PotentialDifference)[/tex]

and i still get the result -1.19 x 10 ^2

now I get the idea.... but not the answer?/
 
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  • #15
xanthym
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CORRECT!! :smile:
 
  • #16
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xanthym said:
CORRECT!! :smile:
finally..wholala
 

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