Approximating the force on a dipole Taylor series

In summary, the magnitude of the net force exerted on one dipole by the other dipole is given approximately by ##F_{net}≈\frac {6q^2s^2k} {r^4}## for ##r\gg s##, where r is the distance from one dipole to the other dipole and s is the distance across one dipole. This equation was derived by using a taylor series to approximate the net force equation, and getting it into the form of 1±α by suitable division.
  • #1
Zack K
166
6

Homework Statement


Show that the magnitude of the net force exerted on one dipole by the other dipole is given approximately by:$$F_{net}≈\frac {6q^2s^2k} {r^4}$$
for ##r\gg s##, where r is the distance from one dipole to the other dipole, s is the distance across one dipole. (Both dipoles are of equal length and both have charges of magnitude q).

Homework Equations


##F=\frac {kq_1q_2} {r^2}##
##f(x)=\sum_{n=0}^\infty \frac {f^{(n)}(0)} {n!} x^n##

The Attempt at a Solution


I worked out the net force that a dipole would be acting on another as: $$F_{net}=kq^2(\frac {1} {(r-s)^2}+\frac {1} {(r+s)^2}-\frac {2} {r^2})$$ This equation is right because I plugged in values for r and s given r is much greater than s, and got the same value for if I used the approximated equation at the top.

I just lack the knowledge of using a taylor series to approximate my equation into the desired one.
 
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  • #2
Zack K said:
I just lack the knowledge of using a taylor series to approximate my equation into the desired one.
First step is to get everything inside parentheses into the form 1±α, by suitable division, where α<<1.
 
  • #3
haruspex said:
First step is to get everything inside parentheses into the form 1±α, by suitable division, where α<<1.
Ok, so then I get $$F_{net}=kq^2(\frac {1} {(1-α)^2}+\frac {1} {(1+α)^2}-\frac {2} {r^2})$$ Do I then expand both the parentheses into polynomials?
 
Last edited:
  • #4
Zack K said:
Ok, so then I get $$F_{net}=kq^2(\frac {1} {(1-α)^2}+\frac {1} {(1+α)^2}-\frac {2} {r^2})$$ Do I then expand both the parentheses into polynomials?
I wrote "get it into the form, by suitable division", not "arbitrarily change it". It still has to follow from the equation you wrote in post #1.
(Or did you make a mistake in typing out the post?)
 

Related to Approximating the force on a dipole Taylor series

1. What is a dipole?

A dipole is a pair of equal and opposite charges separated by a small distance. It can be thought of as a tiny magnet with a north and south pole.

2. How is the force on a dipole approximated using a Taylor series?

The force on a dipole can be approximated using a Taylor series expansion of the electric potential. This involves expressing the potential as a sum of terms, with each term representing the contribution of a different order of the dipole moment. The first term in the series represents the force due to the dipole's own electric field, while higher order terms account for the influence of external electric fields.

3. Why is approximating the force on a dipole using a Taylor series useful?

Using a Taylor series allows us to account for the effects of higher order terms in the dipole moment, which may be significant in certain situations. It also allows for a more accurate calculation of the force on the dipole compared to simpler approximations.

4. What are the limitations of using a Taylor series to approximate the force on a dipole?

A Taylor series is only accurate for small values of the dipole moment. If the dipole moment becomes too large, higher order terms in the series may become significant and the approximation may no longer be accurate. Additionally, the Taylor series may not be appropriate for highly non-linear systems.

5. Are there other methods for approximating the force on a dipole?

Yes, there are other methods such as using numerical integration or solving the equations of motion. These methods may be more accurate for certain systems, but they may also be more computationally intensive.

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