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

`-3.896* 10^(-12) N` is an electric force on the potassium ion due to the chloride ion.

Step-by-step explanation:

Charge on potassium ion =
`q_1=1.602* 10^(-19) C`

Charge on chlorine ion =
`=q_2=-1.602* 10^(-19) C`

Separation between these two charges = r =
`7.700 nm=7.700* 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* (q_1* q_2)/(r^2)`

`q_1,q_2` = Charges on both charges

r = distance between the charges

K = Coulomb constant =
`9* 10^(9) N m^2/C^2`

`F=9* 10^(9) N m^2/C^2* (1.602* 10^(-19) C* (-1.602* 10^(-19) C))/((7.700* 10^(-9) m)^2)`

`F=-3.896* 10^(-12) N`

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

`-3.896* 10^(-12) N` is an electric force on the potassium ion due to the chloride ion.