Question

What is the magnitude of electric field intensity that balances the weight of an electron.It is given that $e=1.6\times {10}^{-19}C$ and $m_e=9.1\times {10}^{-31}kg$.

Answer 1

For the equilibrium of electron

$mg=eE\Rightarrow E=\frac{mg}{e}$

$=\frac{9.1\times {10}^{-31}\times 9.8}{1.6\times {10}^{-19}}=5.57\times {10}^{-11}N/C$