Coloumb's Law and Electric Dipoles question

[John Yi]

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?

 

[gneill]

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!

 

[John Yi]

Ah I am 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|?

 

[gneill]

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?

 

[HalfLostAndSearching]

Your attempt at a solution is correct, but you have not taken into account the electric dipole moment of the HCl molecule in question 2. In addition to the individual partial charges on the H and Cl atoms, the molecule as a whole has a dipole moment due to the separation of the charges. This dipole moment will also contribute to the overall electric field at the position of the sodium ion.

To calculate the electric field at the position of the sodium ion, you will need to use the equation for the electric field due to an electric dipole, which is given by:

E = (1/4πε0) * (p/r^3)

Where ε0 is the permittivity of free space, p is the dipole moment, and r is the distance between the dipole and the point where the electric field is being measured.

In this case, the dipole moment p will be equal to the product of the partial charge and the bond length, which is 0.127 nm. So the equation becomes:

E = (1/4πε0) * (q * d / r^3)

Substituting in the values given in the problem, we get:

E = (9x10^9) * (1.602e-19 * 0.127e-9 / 1.6e-9^3)

Simplifying, we get an electric field of approximately 1.37e-11 N/C, which is the same as your previous answer. This is the correct electric field at the position of the sodium ion due to the electric dipole moment of the HCl molecule.