Question

An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitude and direction of its velocity after it moves 0.10 cm form rest. Does the electron gain or lose potential energy?

Answer 1

Step-by-step explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence,
`qE * S + mg * S = 0.5 * mv^(2)`

`1.6 * 10^(-19) * 1000 + 9.1 * 10^(-31) * 9.8 * ((0.10)/(100)) = 0.5 * 9.1 * 10^(-31) * v^(2)`

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight and the above equation will be as follows.

`(1.6 * 10^(-19) * 1000) * ((0.10)/(100)) = 0.5 * 9.1 * 10^(-31) * v^(2 )`

v =
`sqrt{(1.6 * 10^(-19))/((0.5 * 9.1 * 10^(-31)))`

= 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.