Coloumb's Law and Electric Dipoles question

  • Thread starter John Yi
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  • #1
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Homework Statement


1)HCl consists of one H and Cl atom separated by 0.127 nm, the bond length. The Cl atom has a partial charge of -0.177e and the H atom has a partial charge of +0.177e, where e = 1.602x10-19 C is the electron charge.

2)Now the HCl molecule is placed near a sodium ion Na+ with charge +1e. As shown in the figure, the distance between the Cl and Na atoms is d = 1.6 nm.

Homework Equations


I had finished question 1, but am stuck on question 2.

The Attempt at a Solution


I started with k(q1)(q2)/r^2 ----> (9x10^9) x (2.83554e-20) x (1.602e-19) / (1.727e-9)^2

I end up with an answer with 1.37e-11, but it is saying it is wrong. What am i doing incorrectly?
 

Answers and Replies

  • #2
gneill
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What are the questions associated with 1) and 2). What are you supposed to calculate?

The diagram referenced would be helpful: What are the relative positions of the charges? Is the Na ion on the axis of the dipole formed by the H and Cl atoms?

Note that fields and forces are vector quantities: they have both magnitude and direction. They also have units associated with them. Don't forget to include units on all results!
 
  • #3
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Ah im sorry. The question I am supposed to answer is :

What is the magnitude of the total force on the HCl molecule due to the Na+ ion |FHCl|?
 
  • #4
gneill
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You should make a sketch of the arrangement and draw in the force vectors operating on the H and Cl atoms due to the Na atom. Also indicate their X and Y components. Can you tell what the sum of those vectors will be from the sketch?
 

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