Speed of an Electron in an Electric Field help
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Speed of an Electron in an Electric Field help

[From: ] [author: ] [Date: 11-05-04] [Hit: ]
then you can figure its speed.Potential energy from each charge is -k*q1*q2/r. To calculate the PE for the electron in the initial position, do -k*(charge 1)*(-1.6 x 10^-19 C)/0.26 m,......
Two stationary positive point charges, charge 1 of magnitude 3.45nC and charge 2 of magnitude 1.95nC , are separated by a distance of 52.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.
What is the final speed of the electron when it is 10.0 cm from charge 1?

Thanks!

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Do it in terms of energy. Figure the change in PE. That turns into KE of the electron = (1/2)mv^2. If you know how much KE it has, then you can figure its speed.

Potential energy from each charge is -k*q1*q2/r. To calculate the PE for the electron in the initial position, do -k*(charge 1)*(-1.6 x 10^-19 C)/0.26 m, and -k*(charge 2)*(-1.6 x 10^-19 C)/0.26 m, then add them up. Pay attention to signs.

Note the distance r is 26 cm from each at the initial position, and it has to be expressed in meters. Charges have to be in Coulombs.

Now do the same calculation for the two PEs at the final position. Add up to get the total PE.

Calculate the change in PE. Set it equal to (1/2)mv^2.
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