A proton and an electron in a hydrogen atom are separated by a distance of
5.35 x 10-12m. What is the magnitude and direction of the electric field set up by the proton at the position of the electron? Remember that the charge of the electron is
1.6 x 10-19 C and E=( kc q) / r2

Question

A proton and an electron in a hydrogen atom are separated by a distance of
 5.35 x 10-12m.  What is the magnitude and direction of the electric field set up by the proton at the position of the electron?  Remember that the charge of the electron is 
1.6 x 10-19 C and E=( kc q) / r2

anonymous · Answer

$$E = \frac{ kq _{1} }{ R ^{2} }$$

anonymous · Answer

$$\frac{ (8.99*10^{9})(1.6*10^{-19}) }{ (5.35*10^{-12})}=E$$

anonymous · Answer

You know that the direction direction of the electric field will be outward because the proton is positive and the electric field vectors for a positive charge points outward.

anonymous · Answer

ok