Answer:
The magnitude of the electric field is [tex]5.1 \times 10^{11} \frac{N}{C}[/tex]
Explanation:
Given:
Charge of electron [tex]q = 1.6 \times 10^{-19}[/tex]C
Separation between two charges [tex]r = 5.3 \times 10^{-11}[/tex] m
For finding the magnitude of the electric field,
[tex]E= \frac{kq}{r^{2} }[/tex]
Where [tex]k = 9 \times 10^{9}[/tex]
[tex]E = \frac{9 \times 10^{9} \times 1.6 \times 10^{-19} }{(5.3 \times 10^{-11} )^{2} }[/tex]
[tex]E = 5.1 \times 10^{11}[/tex] [tex]\frac{N}{C}[/tex]
Therefore, the magnitude of the electric field is [tex]5.1 \times 10^{11} \frac{N}{C}[/tex]
