Question

A free electron and a free proton at exactly 1.0 CM apart. Find the magnitude and direction of the acceleration of the proton and the acceleration of the electron

Answer 1

$F=k\frac{q\cdot q}{r^2}=9\cdot10^9\cdot \frac{1.6\cdot10^{-19}\cdot1.6\cdot10^{-19}}{0.01^2}=23.04\cdot10^{-25}(N)$

$a_e=F/m_e=23.04\cdot10^{-25}/9.1\cdot10^{-31}=2.5\cdot10^6(m/s^2)$ to the proton

$a_p=F/m_p=23.04\cdot10^{-25}/1.67 \cdot10^{-27}=1380(m/s^2)$ to the electron