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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`\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.
`\DeltaPE = -\DeltaKE`
So, Since potential difference is defined as:
`\DeltaV_(ba) = (\DeltaPE)/q`
`\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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