Question

If we are viewing the atom in such a way that the electron's orbit is in the plane of the paper with the electron moving clockwise, find the magnitude of the electric field that the electron produces at the location of the nucleus (treated as a point).

Answer 1

Answer:

The magnitude  of the electric field is $5.1 \times 10^{11} \frac{N}{C}$

Explanation:

Given:

Charge of electron $q = 1.6 \times 10^{-19}$C

Separation between two charges $r = 5.3 \times 10^{-11}$ m

For finding the magnitude of the electric field,

   $E= \frac{kq}{r^{2} }$

Where $k = 9 \times 10^{9}$

   $E = \frac{9 \times 10^{9} \times 1.6 \times 10^{-19} }{(5.3 \times 10^{-11} )^{2} }$

   $E = 5.1 \times 10^{11}$ $\frac{N}{C}$

Therefore, the magnitude  of the electric field is $5.1 \times 10^{11} \frac{N}{C}$