# Homework Help: Attractive force in an atom near a point charge q

Tags:
1. Aug 5, 2017

### Pushoam

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
Let's consider the primitive model of atom i.e. the nucleus is surrounded by an electron cloud.
Let's say that the nucleus has charge qa and it moves a distance d from the center in the equilibrium position.

$E = \frac {kq} {r^2} \\ = \frac { kq_a d} {r^3} \\ p = αE = q_a d, q_a = α E /d \\F = q_a E = \frac {kq q_a} {r^2} \\= α E^2 /d = \frac α d {\{ \frac {kq} {r^2}}\}^2$
So, I am getting answer in one unknown term i.e. q_a or d.

2. Aug 5, 2017

### Staff: Mentor

This is not the attractive force between the charge and the atom. It is the force you would get between "one charge" in the atom (approximating it as two charges), but there is also the opposite charge of equal magnitude at a different position leading to a force.

3. Aug 7, 2017

### Pushoam

What I calculated was magnitude of force due to q on the nucleus ( taking r+ d ≈ r as d<<r).
Since r << R ( radius of the e-cloud), let's approximate the e - cloud as a point charge at the center of the cloud.
Then under the approximation r+ d ≈ r, net force on the atom i.e. $F_{atom}$ is 0.
So, avoiding this approximation,

$F_ {atom } = \frac {-α\{kq\}^2} d {\{ \frac {1} {r^4} - \left [ \frac {1} { \{ r +d \}^4} = \frac {1 - \frac { 4 d} r} {r^4} \right ] }\} = \frac {-4α \{k q\}^2} {r^5}$

So, because of the polarizability of the atom, the net force on the atom is not zero. Net force falls off as $\frac 1 {r^3}$ i.e. faster than the electric field of the point charge.

I edited the previous post.

#### Attached Files:

File size:
7.3 KB
Views:
69
File size:
8 KB
Views:
69
Last edited: Aug 7, 2017
4. Aug 7, 2017

### Staff: Mentor

That sounds right.

Edit: Why 1/r4?

Last edited: Aug 7, 2017
5. Aug 7, 2017

### Pushoam

Please look at the post #1.
$\\F = q_a E = \frac {kq q_a} {r^2} \\=$ $\alpha E^2 /d$
$= \frac α d \{ { \frac {kq} {r^2}}\}^2$

Please, tell me the mistake in the above latex code as it is not showing in its standard form.

Last edited: Aug 7, 2017
6. Aug 8, 2017

Color tags and $don't mix well. Fixed:$ \\F = q_a E = \frac {kq q_a} {r^2} \\= \alpha E^2 /d = \frac α d \{ { \frac {kq} {r^2}}\}^2 $The first line is again assuming there would just be one charge. 7. Aug 8, 2017 ### Ray Vickson If you have a multi-line formula like yours, you should use  \begin{array}{rcl} F &= & q_a E = \displaystyle \frac{kq q_a}{r^2} \\ &=& \displaystyle \frac{\alpha E^2 }{d} = \frac{\alpha}{d} \left( \frac{kq}{r^2} \right)^2 \end{array}  Just right-click on the formula to see its TeX constructions. The instruction "\array{rcl}" constructs a 3-column array with the first column right-justified ("r"), the second column being centered ("c") and the third column being left-justified ("l"). Columns are separated by "&", so a & b & c means a in column 1, b in column 2 and c in column 3. Note: without the "\displaystyle" instructions we would get  \begin{array}{rcl} F &= & q_a E = \frac{kq q_a}{r^2} \\ &=& \frac{\alpha E^2 }{d} = \frac{\alpha}{d} \left( \frac{kq}{r^2} \right)^2 \end{array}  which does not look as good as the original. 8. Aug 8, 2017 ### TSny I don't understand how you get the second equality above. But, it doesn't appear that you use it. OK. But in the relation$q_a = α E /d$, the left side$q_a$is definitely a constant (independent of the distance$r$between the point charge$q$and the atom). So,$α E /d$in the relation cannot depend on$r$. So, I don't think it's correct to substitute$E = \frac {kq} {r^2}$into$α E /d$and then treat$d$as a constant. But I think this is what you do when you write and then use this as you did in post #3 (where d is treated as a constant). See what you get if you leave the force as$F = \frac {kq q_a} {r^2}$when you do the analysis of post #3. 9. Aug 9, 2017 ### Pushoam Sorry. I am solving it again from the start. Please check this . In the equilibrium, on the nucleus, the electric field due to q is equal and opposite to that due to the negative charge cloud . Now,$ \vec p = \alpha \vec E $The applied external field is different at different points in the sphere. So, I take$\vec E $as electric field at nucleus due to q i.e.$ \vec E_+$.$ p = q_a d = \alpha E_+
\\ q_a = \frac {\alpha} d { E_+} $Now, r is so big compared to the atomic size that I approximate the cloud sphere as a negative point charge at the center.$ E_+ = \frac {kq} { \left ( r + d \right)^2 }$is electric field at nucleus$E_- = \frac {kq} { r ^2 } $is electric field at the center of the negative charge cloud sphere$ F_{atom} = q_a \left (E_+ - E_- \right)  F_{atom} = q_a \left (\frac {kq} { \left ( r + d \right)^2 } - \frac {kq} {r^2} \right) $where q_a is the charge of the nucleus. Taking the approximation,$ \frac 1 { \left (r + d \right )^2 } = \frac {1 - \frac {2d } r} {r^2} $and substituting the expression for$ q_a \begin {align}
F_{atom} & = {\frac α d} { \frac {kq} { \left ( r + d \right)^2 } } \left ( {\frac {kq} { \left ( r + d \right)^2 } } - { \frac {kq} {r^2}} \right)
\\& = \frac {αk^2 q^2} d \{ \frac {-2d} {r^5 } + \frac {4 d^2 } {r^6} \}
\\& = \frac {-2αk^2 q^2} {r^5}
\end {align} $[ignoring the second term as it is very small compared to the first term] Hence,$ \vec F_{atom} = \frac {-2αk^2 q^2} {r^5} ~\hat x $i.e. towards the charge q. #### Attached Files: • ###### upload_2017-8-7_19-18-7-png.png File size: 8 KB Views: 52 Last edited: Aug 9, 2017 10. Aug 9, 2017 ### Pushoam Since α depends on the atomic structure and hence is constant for a given atom. What I conclude from the above arguments is :$E_+/d $must be a constant with appropriate unit. The no. of the eqn. is not showing as 1,2,3. As I edit and save it again, the no. increased every time. How to correct it? Last edited: Aug 9, 2017 11. Aug 9, 2017 ### Pushoam Another way to solve the same problem: Attractive force on atom due to q is equal and opposite to the attractive force on q due to the atom. For r>> atomic size, the atom could be reduced as a dipole. Electric field due to this dipole at q$\vec E_{di} = \frac {2k\vec p}{r^3} \\\vec p = \alpha \vec E_{due~ q} = \alpha \frac {kq} {r^2} \hat x$Attractive force on q :$\vec F_q = \frac{2\alpha k^2 q^2 } {r^5} \hat x##

12. Aug 9, 2017