An electron initially at rest is accelerated between two parallel plate 2.000 cm

Question

An electron initially at rest is accelerated between two parallel plate 2.000 cm apart. The pace between the plates is a vacuum. The potential difference across the plates is 300.0 V.


What kinetic energy does the electron have when it hits the positive plate?


I have a feeling we would solve for potential energy when it is still on the negative side and just convert that to kinetic energy as we know that the potential energy is transformed into kinetic once the electron moves to the positive.

But, any explanation and steps toward solving this problem is appreciated.

Explanation / Answer

change in potential energy = change in kinetic energy

q v = 0.5 mv^2

1.6 x 10-19 *300 = 0.5 * 9.11 x 10-31 * v2

where, mass of electron = 9.11 x 10-31kg

charge on electron = 1.6 x 10-19C

v= 10.27 * 106 m/s so kinetic energy at +ve plate = 480eV ( electron-volts)

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