Question

If the electric intensity between two parallel plates, placed 1cm apart is 104 NC-1and the direction of the field of this intensity is vertically upward. Find the force on an electron in this field and compare the force on the electron with the electron’s weight

Answer 1

The force of a charge in an electric field is:

`\vec{F}_e=q\vec{E}`

In this case we know the electric field is:

`\vec{E}=104\hat{j}`

and that the charge is that of the electron, then we have:

`\begin{gathered} \vec{F}_e=-1.6*10^(-19)(104\hat{j}) \\ \vec{F}_e=-1.664*10^(-17)\text{ N} \end{gathered}`

Therefore, the magnitude of the force is

`1.664*10^(-17)\text{ N}`

and in points down.

The weight of the electron is:

`\begin{gathered} W=1.67*10^(-27)(9.98) \\ W=1.6366*10^(-26) \end{gathered}`

Making the quotient between the force we have:

`(1.664*10^(-17))/(1.6366*10^(-26))=1.02*10^9`

Therefore, the electric force is approximately 1e9 times the weight.