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Suppose we have Cl-, Na+, and Ca2+ ions in an aqueous solution (with dielectric constant κ = 80.4—we will discuss this later). Consider the situation where a sodium ion is between a chlorine ion and a calcium ion as shown:
(Cl-) (Na+) (Ca2+)
If the sodium ion is 1.50 nm from the chlorine ion and 3.00 nm from the calcium ion, find the electric force on the Na+ ion.

Respuesta :

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Answer:

a) 1.19 x 10^7 N/C

b) 2 x 10^-12 N

Explanation:

field due to Cl on (9.0xE¡)(1.6x10-19 C) (1.5Å—10-9m) -6.4x10 N/C field due to Ca+3 ion, 2(90x10 (1.6x10-19c) ((4.5-1.5)Å—10-9 m)' magnitude of net field at given point without dielectric E E+E 6.4x108 +3.2x108-9.6x10 N/C magnitude of net field at given point with di electric K 80.4 the force on sodium ion at this point, F-Edq= (1.19å—107 N/C)(1.6å—10-19C)= 2.0Å—10-12 N

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