Question

A K+ ion and a Cl− ion are directly across from each other on opposite sides of a membrane 7.700 nm thick. What is the electric force on the K+ ion due to the Cl− ion?

Answer 1

Answer:

$-3.896\times 10^{-12} N$ is an electric force on the potassium ion due to the chloride ion.

Explanation:

Charge on potassium ion = $q_1=1.602\times 10^{-19} C$

Charge on chlorine ion = $=q_2=-1.602\times 10^{-19} C$

Separation between these two charges = r = $7.700 nm=7.700\times 10^{-9} m$

$1 nm=10^{-9} m$

Electric force on the potassium ion due to the chloride ion = F

Coulomb's law is given as ;

$F=K\times \frac{q_1\times q_2}{r^2}$

$q_1,q_2$ = Charges on both charges

r = distance between the charges

K = Coulomb constant =$9\times 10^{9} N m^2/C^2$

$F=9\times 10^{9} N m^2/C^2\times \frac{1.602\times 10^{-19} C\times (-1.602\times 10^{-19} C)}{(7.700\times 10^{-9} m)^2}$

$F=-3.896\times 10^{-12} N$

(negative sign indicates that attractive force is exerting between two ions)

$-3.896\times 10^{-12} N$ is an electric force on the potassium ion due to the chloride ion.