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


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
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:

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

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)