Question

What is the magnitude and direction of an electric field that would just support the weight of an electron? (m = 9.11 x 10^-31 kg)

Answer 1

5.58*10^-11 N/C, downward

Explanation:

Given:

m = 9.11 x 10^-31 kg

The electric force can be calculated as follows:

`\begin{gathered} F=qE \\ \text{ we solve for E} \\ E=(F)/(q) \\ F=W=m\cdot g \\ \text{ we replacing} \\ E=(mg)/(q) \end{gathered}`

We substitute the values, knowing that the acceleration is 9.8 m/s^2 and q is equal to 1.6*10^-19 C, therefore:

`\begin{gathered} E=(9.11\cdot10^(-31)\cdot9.8)/(1.6\cdot10^(-19)) \\ E=\: 5.58\cdot10^(-11)\text{ N/C} \end{gathered}`

Which means that the magnitude is equal to 5.58*10^-11 N/C and with downward direction