What is the magnitude of the electric field, \(E\) such that an clectron placed in the field would experience an electric force equal to its weight? \(\left( M _{ c }=9.1 \times 10^{-31} kg , g=9.8 m / s ^{2}\right)\).
A. \(5.6 \times 10^{-11} N / C\) B. \(9.8 \times 10^{-11} N / C\) C. \(1.6 \times 10^{-19} N / C\) D. \(5.6 \times 10^{-19} N / C\)
Correct Answer: A
Explanation
given: Mass of electron \(M_{e}=9.1 \times 10^{-31} kg , g=9.8 m / s ^{2}\)orce experienced = weight of the electron \(=m g\)lso, \(E=\frac{\text { Force }}{\text { Charge }}\) \(E=\frac{m g}{q}\), where \(q=\) charge on electron \(=1.6 \times 10^{-19} C\) \(E=\frac{9.1 \times 10^{-31} \times 9.8}{1.6 \times 10^{-19}}\) \(E=\frac{8.918 \times 10^{-30}}{1.6 \times 10^{-19}}\) \(E=5.57 \times 10^{-11} N / C\) \(E \approx 5.6 \times 10^{-11} N / C\)
