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An electron acquires 7.45 x 10^-17 J of kinetic energy when it is accelerated by an...

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lkehoe | Valedictorian

Posted April 22, 2013 at 6:24 PM via web

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An electron acquires 7.45 x 10^-17 J of kinetic energy when it is accelerated by an electric field from plate A to plate B.  (a) What is the potential difference between the plates and (b) Which plate is at the higher potential?

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quirozd | High School Teacher | (Level 3) Adjunct Educator

Posted April 23, 2013 at 1:30 AM (Answer #1)

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Given

`\DeltaKE = 7.45\times10^(-17)\text(J)`

 

`q_e = -1.602\times10^(-19)\text(C) `

Due to the conservation of energy, the change in kinetic energy came from the change in potential energy due to the electric field working on the electron.

Thus,

`\DeltaPE = -\DeltaKE`

So, Since potential difference is defined as:

`\DeltaV_(ba) = (\DeltaPE)/q`

then,

`\DeltaV_(ba)=-(\DeltaKE)/q = -(7.45\times10^(-17) \text(J))/(-1.602\times10^(-19) \text(C))`

`\DeltaV_(ba) = +465 V`

Which means that plate B has a higher potential. The electric field lines point from B to A, if you were to sketch the plates.

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