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)

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ANSWER:

5.58*10^-11 N/C, downward

STEP-BY-STEP EXPLANATION:

Given:

m = 9.11 x 10^-31 kg

The electric force can be calculated as follows:

[tex]\begin{gathered} F=qE \\ \text{ we solve for E} \\ E=\frac{F}{q} \\ F=W=m\cdot g \\ \text{ we replacing} \\ E=\frac{mg}{q} \end{gathered}[/tex]

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:

[tex]\begin{gathered} E=\frac{9.11\cdot10^{-31}\cdot9.8}{1.6\cdot10^{-19}} \\ E=\: 5.58\cdot10^{-11}\text{ N/C} \end{gathered}[/tex]

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

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