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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?

User Teen
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1 Answer

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Answer:


-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.

User Joseph Quinsey
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9.1k points
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