- #1

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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?

- Thread starter leolaw
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- #1

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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?

- #2

vsage

Consider the units of electric potention (V or J/C) and work (J)

- #3

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What is the unit of potential difference...?

Daniel.

Daniel.

- #4

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[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

Science Advisor

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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?

~~

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Last edited:

- #6

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[tex] 4.82 * 10^{-4}J[/tex], which is the kinetic energy

- #7

xanthym

Science Advisor

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What about:

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

~~

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

~~

- #8

xanthym

Science Advisor

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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?

~~

[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

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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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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 stuffxanthym said:What about:

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

~~

But once i have this, it makes everything easier

- #12

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you mean it is -1.2 x 10 ^3?dextercioby said:

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...

- #13

xanthym

Science Advisor

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LEOLAW --

Don't guess. Repeat the calculation. (10^3 is still wrong.)

Don't guess. Repeat the calculation. (10^3 is still wrong.)

- #14

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

- #15

xanthym

Science Advisor

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- #16

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finally..wholalaxanthym said:CORRECT!!

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