Question

An electron is released at rest from point A. It moves in response to the electric field of a fixed charge distribution along a path that takes it through point B. If the potential has a value of -11 V at point A and 25 V at point B, what is the electron's speed as it passes point B?

Answer 1

`v=8.89* 10^(15)\ m/s`

Step-by-step explanation:

An electron is released at rest from point A. It moves in response to the electric field of a fixed charge distribution along a path that takes it through point B.

Potential at point A,
`V_A=-11\ V`

Potential at point B,
`V_B=25\ V`

Let v is the speed of the electron as it passes point B. It can be calculated using the conservation of energy theorem as :

`(1)/(2)mv^2-(1)/(2)mu^2=V_B-V_A`

Here, u = 0

`(1)/(2)mv^2=25-(-11)`

`mv^2=72`

`v=\sqrt{(72)/(m)}`

m is the mass of electron

`v=\sqrt{(72)/(9.1* 10^(-31))}`

`v=8.89* 10^(15)\ m/s`

So, the electron's speed as it passes point B is
`8.89* 10^(15)\ m/s`. Hence, this is the required solution.