Relationship between dipole to Eq (3-49) in Cheng's textbook

The problem is as shown in Figure 1.

Figure 1

To obtain the electric field, we can use direct formula from Equation (3-22) given by

$\inline \vec{E}=\frac{1}{4{\pi}{{\epsilon}_0}}\sum_{k=1}^{n}{\frac{q_k{|\vec{R}-\vec{{R_k^'}|}}}{|\vec{R}-\vec{{R_k^'}|^3}}}$

The disadvantage is that this will result in adding vectors of different magnitudes and directions.

$\inline \vec{E}=\frac{q}{4{\pi}{{\epsilon}_0}}{\left(\frac{{\vec{R}-\frac{\vec{d}}{2}}}{|\vec{R}-\frac{\vec{d}}{2}|^3}}-{\frac{{\vec{R}+\frac{\vec{d}}{2}}}{|\vec{R}+\frac{\vec{d}}{2}|^3}}\right)$

Upon further manipulation (with $\inline \vec{p}=q\vec{d}$ , we will have the final expression as given by Eq (3-31)

$\inline \vec{E}=\frac{p}{4{\pi}{{\epsilon}_0}{R}^3}({\hat{a_R}2\cos \theta}+{\hat{a_{\theta}}2\sin \theta})$

The other approach is to use Equation (3-49) is given as follows:

$\inline V=\frac{1}{4{\pi}{{\epsilon}_0}}\sum_{k=1}^{n}{\frac{q_k}{|\vec{R}-\vec{{R_k^'}|}}}$

in which this is a lot easier as the potential difference is in scalar rather than vector. It can be written directly as
$\inline V=\frac{q}{4{\pi}{{\epsilon}_0}}\left(\frac{1}{R_+}-\frac{1}{R_-}\right)$

Upon further simplifications, we have

$\inline V=\frac{qd\cos \theta}{4{\pi}{{\epsilon}_0}{R^3}}$
and

$\inline V=\frac{\vec{p}\vec{a_R}}{4{\pi}{{\epsilon}_0}{R^2}}$

Therefore, taking the minus gradient of the scalar potential yields

$\inline \vec{E}=-{\nabla}V=-\hat{a_R}\frac{\partial V}{\partial R}-\hat{a_{\theta}}\frac{\partial V}{R\partial {\theta}} =\frac{p}{4{\pi}{{\epsilon}_0}{R}^3}({\hat{a_R}2\cos \theta}+{\hat{a_{\theta}}2\sin \theta})$

Relationship between dipole to Eq (3-49) in Cheng's textbook

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