User talk:UmptanumMath/sandbox1

Binomial Expansion

$\mathbf {(1+a)^{n}} =1+n\cdot a+n(n-1)\cdot a^{2}+n(n-1)(n-2)\cdot a^{3}+...$

$\mathbf {(1+a)^{\frac {-3}{2}}} =1+{\frac {-3}{2}}\cdot a+{\frac {-3}{2}}\cdot {\frac {-5}{2}}\cdot a^{2}+{\frac {-3}{2}}\cdot {\frac {-5}{2}}\cdot {\frac {-7}{2}}\cdot a^{3}+...$

Dipole Moment

A water molecule. A molecule of water has an electric moment because of the unequal sharing of its electrons in a "bent" structure. A separation of charge is present with negative charge in the middle (red shade), and positive charge at the ends (blue shade).

In physics, the electric dipole moment is a measure of the separation of positive and negative electrical charges in a system of charges, that is, a measure of the charge system's overall polarity. It is important in understanding practical matters like the construction of a capacitor.

Finding the electric field going out perpendicular to the axis between the dipole charges

Figure 1: An electric dipole

To determine the electric field (the dipole moment) which result from the charges in Figure 1 that are separated by a distance 2a, we simply sum the fields from each charge individually.

${\vec {E}}={{\vec {E}}_{1}}+{{\vec {E}}_{2}}$

The distance of each charge (r) from the point where the electric field is measured is:

$r={\sqrt {a^{2}+x^{2}}}$

The magnitude of the electric field (E₁) due to the first charge (q₁) and the magnitude of the electric field (E₂) due to the second charge (q₂) are the same although the directions are different.

$E_{1}=E_{2}={\frac {1}{4\pi \epsilon _{0}}}{\frac {q}{r^{2}}}={\frac {1}{4\pi \epsilon _{0}}}{\frac {q}{a^{2}+x^{2}}}$

The components of the electric field along the x axis $E_{1_{x}}$ and $E_{2_{x}}$ have the same magnitude but opposite directions and cancel each other out.

$E_{1_{x}}+E_{2_{x}}=0$

As a result of this the horizontal components of the electric field vectors cancelling is that summing ${{\vec {E}}_{1}}$ and ${{\vec {E}}_{2}}$ reduces to summing $E_{1_{y}}$ and $E_{2_{y}}$. Further, the magnitude of the vertical components are identical, $E_{1_{y}}=E_{2_{y}}$. Since $\cos \theta ={\frac {opposite}{hypotenuse}}$ finding the vertical component reduces to:

$E_{1_{y}}=E_{2_{y}}=E_{1}\cos \theta \,\!$ or :$E_{y}=2E_{1}\cos \theta \,\!$

But we also know that:

$\cos \theta ={\frac {a}{\sqrt {a^{2}+x^{2}}}}$

So we can substitute this into the equation for $E\,\!$ which yeilds:

$E=2E_{1}\cos \theta \,\!={\frac {2}{4\pi \epsilon }}{\frac {q}{a^{2}+x^{2}}}{\frac {a}{\sqrt {a^{2}+x^{2}}}}={\frac {1}{4\pi \epsilon _{o}}}{\frac {2aq}{(a^{2}+x^{2})^{\frac {3}{2}}}}$

In the region distant from the dipole x is much greater than a (x>>a). This is commonly referred to as the "far field" region. We can find how the field behaves in the far field region by factoring out x.

$E={\frac {1}{4\pi \epsilon _{o}}}{\frac {2qa}{(a^{2}+x^{2})^{\frac {3}{2}}}}={\frac {1}{4\pi \epsilon _{o}}}{\frac {2qa}{x^{3}}}{\frac {1}{(1+{({\frac {a}{x}})}^{2})^{\frac {3}{2}}}}$

By inspection we can conclude that when x>>a then this approaches

$E\approx {\frac {1}{4\pi \epsilon _{o}}}{\frac {2qa}{x^{3}}}$

El producto $2aq\,\!$ se denomina momento $p\,\!$ del dipolo eléctrico. Entonces, se puede volver a escribir la ecuación de $E\,\!$ como:

$E={\frac {1}{4\pi \epsilon }}{\frac {p}{(a^{2}+r^{2})^{\frac {3}{2}}}}$

Y si r>>a, es decir, para puntos distantes a lo largo de la bisectriz del eje del dipolo como:

$E\approx {\frac {1}{4\pi \epsilon _{o}}}{\frac {1}{x^{3}}}$

An important question for some analyses might be: over what range it the far field approximation valid? —Preceding unsigned comment added by 71.94.160.115 (talk) 06:00, 22 September 2010 (UTC)

Quadrupole Moment

one form of electric quadrupole