Question 20.3: (The Dipole Field Along the Dipole Axis) A proton and an ele......

In this problem we have a = 10^{−10}  m,  q  =  e  =  1.6  ×  10^{−19}  C,  x  =  20  ×  10^{−10}  m,  ke  =  (9  ×  10^{9} N.m^{2}/C^{2}) (1.6  ×  10^{−19}  C)  =  1.44  ×  10^{−9}  N.m^{2}/C,  x  −  a  =  19  ×  10^{−10}  m,  and  x  +  a  =  21  ×  10^{−10} m. Using the exact formula given by Eq. 20.11 in the case of x > +a, we have:

=(1.44\times10^{-9}\,\mathrm{N.m^{2}/C})\left[\frac{1}{(19\times10^{-10}\,{\mathrm{m}})^{2}}-\frac{1}{(21\times10^{-10}\,{\mathrm{m}})^{2}}\right]

= (1.44 × 10^{−9}  N.m^{2}/C)[2.770 × 10^{17}  m^{−2}  −  2.268  ×  10^{17}m^{−2}]
= 7.236 × 10^{7} N/C

On the other hand, we have x >> a and we can use the approximate formula given by Eq. 20.13 as follows:

\vec{E}=\left\{\begin{matrix}2 \ k\frac{\vec{P}}{x^{3}}& x>>a\\2 \ k\frac{\vec{P}}{|x|^{3}}& x<<-a \end{matrix} \right. \quad (\vec{p} = 2  a  q  \vec{i}) \qquad    (20.13)

=(1.44\times10^{-9}\,\mathrm{N.m^{2}/C})\frac{(4\times10^{-10}\,\mathrm{m})}{(20\times10^{-10}\,\mathrm{m})^{3}}

=7.200\times10^{7}{\mathrm{~N/C}}

Clearly this calculation is a good approximation when x/a = 20.