Question

what potential difference, acting over a distance of 4.4 cm, would be needed to balance the downward force of gravity so that an electron would remain stationary? assume that the electric field is uniform. express your answer to two significant figures and include the appropriate units.
δv = value units

Explanation:

Step1: Equate electric force and gravitational force

When the electron is stationary, \(F_{e}=F_{g}\). The electric force \(F_{e}=eE\), and the gravitational force \(F_{g} = mg\). So \(eE=mg\).

Step2: Relate electric field \(E\) and potential difference \(\Delta V\)

The electric field \(E=\frac{\Delta V}{d}\) (for a uniform electric field, where \(d\) is the distance). Substitute \(E\) into \(eE = mg\), we get \(e\frac{\Delta V}{d}=mg\).

Step3: Solve for \(\Delta V\)

Rearrange the equation \(\Delta V=\frac{mgd}{e}\).
We know \(m = 9.11\times10^{-31}\space kg\) (mass of electron), \(g = 9.8\space m/s^{2}\), \(d=4.4\times10^{-2}\space m\), \(e = 1.6\times10^{-19}\space C\).
Substitute the values:
\(\Delta V=\frac{9.11\times 10^{-31}\times9.8\times4.4\times 10^{-2}}{1.6\times10^{-19}}\)
\(\Delta V=\frac{9.11\times9.8\times4.4\times10^{-31 - 2}}{1.6\times10^{-19}}\)
\(\Delta V=\frac{9.11\times9.8\times4.4\times10^{-33}}{1.6\times10^{-19}}\)
Using the rule of exponents \(a^{m}\div a^{n}=a^{m - n}\), we have \(\Delta V=\frac{9.11\times9.8\times4.4}{1.6}\times10^{-33+ 19}\)
\(\Delta V=\frac{9.11\times9.8\times4.4}{1.6}\times10^{-14}\)
\(9.11\times9.8 = 89.278\), \(89.278\times4.4=392.8232\)
\(\Delta V=\frac{392.8232}{1.6}\times10^{-14}\)
\(\Delta V = 245.5145\times10^{-14}\)
\(\Delta V=2.455145\times10^{-12}\space V\approx2.5\times 10^{-12}\space V\)

Answer:

\(\Delta V = 2.5\times10^{-12}\space V\)