Answer:
1.611008 N/C
[tex]8.96\times 10^{-12}\ C[/tex]
away
Explanation:
n = Number of charge = [tex]5.6\times 10^{11}[/tex]
q = Charge of particle = [tex]1.6\times 10^{-19}\ C[/tex]
Gap = [tex]0.1\times 10^{-3}\ m[/tex]
r = Distance = 5 cm
The charge is given by
[tex]q=ne\\\Rightarrow q=5.6\times 10^{11}\times 1.6\times 10^{-19}\times 0.1\times 10^{-3}\\\Rightarrow q=8.96\times 10^{-12}\ C[/tex]
Charge entering axon is [tex]8.96\times 10^{-12}\ C[/tex]
Electric field is given by
[tex]E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 8.96\times 10^{-12}}{5\times 10^{-2}}\\\Rightarrow E=1.611008\ N/C[/tex]
The electric field is 1.611008 N/C
The charges in the question are all positive so the electric field is directed away from the axon