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Finding the electric field associated with a dipole anywhere on a plane????


6 years ago

							Finding the electric field associated with a dipol1. An electric dipole is matched, equal but opposite, charges separated by a fixed distance. Find the electric field associated with an electric dipole. Perform this for three cases:a. Find the electric field perpendicular to the axis between dipole charges. Find the far field approximation.b. Find the electric field along the axis of the dipole charges. Compare the far field approximation for this case with that of a) above.c. Find the electric field perpendicular in a plane which includes the dipole charges  Aw: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, theelectric dipole momentis 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 overallpolarity. It is important in understanding practical matters like the construction of a capacitor.Finding the electric field perpendicular to the axis between dipole charges  Figure 1: An electric dipoleTo determine the electric field which result from the dipole charges in Figure 1 that are separated by a distance 2a, we simply sum the fields from each charge individually. Note that the fields are vectors. ${\vec E}={\vec E_1}+{\vec E_2}$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 $\cos\theta = \frac {adjacent} {hypotenuse}$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 yields: $E = 2E_1 \cos\theta\,\! = \frac{2}{4\pi \epsilon_o}\frac{q}{a^2+x^2}\frac{a}{\sqrt{a^2+x^2}} =\frac{1}{4\pi \epsilon_o}\frac{2aq}{(a^2+x^2)^{\frac{3}{2}}}$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 $E \approx \frac{1}{4\pi \epsilon_o}\frac{2qa}{x^3}$An important question for some analyses might be: over what range is the far field approximation valid? We can find out by using a binomial expansion (review of binomial expansion immediately below this section if you need it). $E =\frac{1}{4\pi \epsilon_o}\frac{2qa}{(a^2+x^2)^{\frac{3}{2}}} =\frac{1}{4\pi \epsilon_o}{\frac {2qa}{x^3}}\frac{1}{(1+{(\frac {a}{x})}^2)^{\frac{3}{2}}}$Where we will expand $\frac{1}{(1+{(\frac {a}{x})}^2)^{\frac{3}{2}}}$ $= (1+{(\frac {a}{x})}^2)^{-\frac{3}{2}} = 1 -\frac{3}{2} {(\frac {a}{x})}^2 + ...$If you want to use the approximation only when it introduces an error less than 1%, then - since the first term is unity, you want the second term to be less than 1/100th of unity, or: $\frac {1}{100} > \frac{3}{2} {\left (\frac {a}{x} \right )}^2$ $\frac {2}{300} > { \left(\frac {a}{x} \right)}^2$ $\sqrt {\frac {2}{300}} > {\left (\frac {a}{x} \right )}$ $x > \sqrt {150} \cdot a$So if x is 12.3 times larger than a, this approximation is valid within 1%.Binomial expansion reviewReview of the binomial expansion: $\mathbf{ (1 + a)^n} = 1 + \frac {n}{1!} \cdot a + \frac {n(n-1)}{2!} \cdot a^2 + \frac {n(n-1)(n-2)}{3!} \cdot a^3 + ...$ $\mathbf{ (1 + a)^{- \frac {3}{2} }} = 1 + \frac {1}{1!} \cdot \frac {-3}{2} \cdot a + \frac {1}{2!} \cdot \frac {-3}{2} \cdot \frac {-5}{2} \cdot a^2 + \frac {1}{3!} \cdot \frac {-3}{2} \cdot \frac {-5}{2} \cdot \frac {-7}{2} \cdot a^3 + ...$informal source linkFinding the electric field along the axis formed by dipole charges  Figure 2 - the layout for calculating the electric field along the line through the dipole charges.The net electric field is the sum of the contributions to the electric field from the dipole charges. ${\vec E}={\vec E_1}+{\vec E_2}$The first thing one always should do when solving a problem is to consider what you know by simple inspection. In this case the magnitude and layout of the charges tells use several things. Since they are equal and we know that the electric field from a point charge falls off as 1 over r2then we know that the nearest charge will always dominate. Further we also know that along the axis that we're calculating the electric field there will be no vertical component of the electric field. $E_{1_y}= E_{2_y}=0$So now we will calculate the magnitude of the electric field from each in the x direction: $E_{1_x}$and $E_{2_x}$. $E_1 = \frac{1}{4\pi \epsilon_0} \frac{q}{{(r-a)}^2}\qquad\qquad E_2 = \frac{1}{4\pi \epsilon_0} \frac{-q}{{(r + a)}^2}$Adding the vector magnitudes $E_{1_x}$and $E_{2_x}$ $E=E_{1_x} + E_{1_x}\,\!$And substituting in the equations above yields: $E=\frac{1}{4\pi \epsilon_0} \frac{q}{{(r-a)}^2} + \frac{1}{4\pi \epsilon_0} \frac{-q}{{(r + a)}^2}= \frac{1}{4\pi \epsilon_0}\left [ \frac{q}{{(r-a)}^2}- \frac{q}{{(r + a)}^2}\right ]=\frac{1}{4\pi \epsilon_0}\ 4aq\ \frac{r}{{(r^2 - a^2)}^2}$The quantity 2qa is defined as the dipole moment (p) and so we can substitute: $p=2qa\,\!$With that substitution we now get: $E=\frac{1}{2\pi \epsilon_0} \frac{pr}{{(r^2 - a^2)}^2}$Consider the far field case (i.e., the case where $r\,\!$>> $a\,\!$)Then one can see that $r^2\,\!$>> $a^2\,\!$and that $r^2 - a^2\,\!$approaches $r^2\,\!$, which allows us to simplify the equation in the far field. $E=\frac{1}{2\pi \epsilon_0} \frac{p}{r^3}$Note that one can use the binomial theorem just as in the section above to find out where this far field approximation (x>>a) is valid.Finding the electric field associated with a dipole anywhere on a planeNow that you've solved for two special cases and determined that the dipole field falls off as $\frac {1} {r^3}$at great distances (referred to as the far field), it is instructive to find the field's behavior off the axis.Consider the case of an electric dipole's field at a point $P\,\!$with the coordinates $(x,y)\,\!$as represented in the figure. Note that we can represent the coordinates either as Cartesian, or as cylindrical.  dipole field in a plane. $r_1=\sqrt{{(y+a)}^2+x^2}\quad\cos\alpha=\frac{y+a}{\sqrt{{(y+a)}^2+x^2}}\quad\sin\alpha=\frac{x}{\sqrt{{(y+a)}^2+x^2}}$ $r_2=\sqrt{{(y-a)}^2+x^2}\quad\cos\beta=\frac{y-a}{\sqrt{{(y-a)}^2+x^2}}\quad\sin\beta=\frac{x}{\sqrt{{(y-a)}^2+x^2}}$In Cartesian coordinates the electric field at any point in the plane from each charge can be found as follows: ${E_r}_1= \frac{1}{4\pi\epsilon_0}\frac{q}{{(y+a)}^2+x^2}\quad{E_r}_2= \frac{1}{4\pi\epsilon_0}\frac{-q}{{(y-a)}^2+x^2}$We can apply the principle of superposition and add the fields from each of the charges, remembering that they are vectors so that the vertical components add to vertical and horizontal add to horizontal.First finding the horizontal coordinates labeled as $E_x\,\!$: ${{E_r}_1}_x= \frac{1}{4\pi\epsilon_0}\frac{+q}{{(y+a)}^2+x^2}\cos\alpha=\frac{1}{4\pi\epsilon_0}\frac{+q}{{(y+a)}^2+x^2}\frac{y+a}{\sqrt{{(y+a)}^2+x^2}}=\frac{1}{4\pi\epsilon_0}\frac{+q(y+a)}{{\left [ {(y+a)}^2+x^2\right ]}^{\frac{3}{2}}}$ ${{E_r}_2}_x= \frac{1}{4\pi\epsilon_0}\frac{-q}{{(y-a)}^2+x^2}\cos\beta=\frac{1}{4\pi\epsilon_0}\frac{-q}{{(y-a)}^2+x^2}\frac{y-a}{\sqrt{{y-a)}^2+x^2}}=\frac{1}{4\pi\epsilon_0}\frac{-q(y-a)}{{\left [ {(y-a)}^2+x^2\right ]}^{\frac{3}{2}}}$Next finding the vertical components for each of the two charges labeled as $E_y\,\!$we obtain: ${{E_r}_1}_y= \frac{1}{4\pi\epsilon_0}\frac{q}{{(y+a)}^2+x^2}\sin\alpha=\frac{1}{4\pi\epsilon_0}\frac{q}{{(y+a)}^2+x^2}\frac{x}{\sqrt{{(y+a)}^2+x^2}}=\frac{1}{4\pi\epsilon_0}\frac{qx}{{\left [ {(y+a)}^2+x^2\right ]}^{\frac{3}{2}}}$ ${{E_r}_2}_y= \frac{1}{4\pi\epsilon_0}\frac{-q}{{(y-a)}^2+x^2}\sin\beta=\frac{1}{4\pi\epsilon_0}\frac{-q}{{(y-a)}^2+x^2}\frac{x}{\sqrt{{y-a)}^2+x^2}}=\frac{1}{4\pi\epsilon_0}\frac{-qx}{{\left [ {(y-a)}^2+x^2\right ]}^{\frac{3}{2}}}$Summing the horizontal components yields: $E_x=\frac{1}{4\pi\epsilon_0}q \left \lbrack \frac{y+a} {{\left [{(y+a)}^2+x^2 \right ] }^{\frac{3}{2}}}-\frac{y-a} {{\left [{(y-a)}^2+x^2 \right ] }^{\frac{3}{2}}}\right \rbrack =\frac{2qa}{4\pi\epsilon_0} \left \lbrack \frac{1} {{\left [{(y+a)}^2+x^2 \right ] }^{\frac{3}{2}}}\right \rbrack$And for large r (either large x, large y, or their combination) this becomes: $E_x=\frac{qa}{2\pi\epsilon_0}\frac{1}{r^3}$Summing the vertical components yields: $E_y=\frac{1}{4\pi\epsilon_0}q \left \lbrack \frac{x} {{\left [{(y+a)}^2+x^2 \right ] }^{\frac{3}{2}}}-\frac{x} {{\left [{(y-a)}^2+x^2 \right ] }^{\frac{3}{2}}}\right \rbrack$Thanks & RegardsRinkoo GuptaAskIITians Faculty

6 years ago
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