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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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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?

Posted by lkehoe on April 22, 2013 at 6:24 PM via web and tagged with accelerated, electric field, electric potential, kinetic energy, physics, plate, science

High School Teacher

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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.

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

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