Respuesta :
On axis of the dipole the electric field is given by
[tex]E = \frac{kq}{(r - \frac{d}{2})^2} - \frac{kq}{(r + \frac{d}{2})^2}[/tex]
[tex]E = \frac{kq(r + d/2)^2 - (r - d/2)^2}{(r^2 - \frac{d^2}{4})^2}[/tex]
[tex]E_{act} = \frac{kq*2rd}{(r^2 - \frac{d^2}{4})^2}[/tex]
now if we approximate above equation
[tex]E_{eppr} = \frac{kq*2d}{r^3}[/tex]
now we will find the ratio of these two
[tex]\frac{E_{eppr}}{E_{act}} = \frac{(r^2 - \frac{d^2}{4})^2}{r^4}[/tex]
put r = 3d
[tex]\frac{E_{eppr}}{E_{act}} = \frac{(9d^2 - \frac{d^2}{4})^2}{81d^4}[/tex]
[tex]\frac{E_{eppr}}{E_{act}} = \frac{(9- \frac{1}{4})^2}{81}[/tex]
[tex]\frac{E_{eppr}}{E_{act}} = 0.945[/tex]
so above is the ratio of two field
Answer:
The answer is 0.94 N/C
Explanation:
The actual magnitude of the dipole at a distance Z from the centre is equal to
[tex]E_{actual} =\frac{q}{2\pi e_{0} Z^{3} } \frac{d}{(1-(\frac{d}{2z})^{2} )^{2} }[/tex]
if z = 8d
[tex]E_{actual} =\frac{q}{2\pi e_{0}(3d)^{3} } \frac{d}{(1-(\frac{d}{6d})^{2})^{2} } \\E_{actual} =3.92x10^{-2} \frac{q}{2\pi e_{0}d^{2} }[/tex]
the approx magnitude of the dipole at a distance Z from the centre is equal to
[tex]E_{approx} =\frac{1}{2\pi e_{0} } \frac{qd}{(Z^{3} } \\E_{approx} =3.7x10^{-2} \frac{q}{2\pi e_{0}d^{2} }[/tex]
The ratio is
[tex]E_{approx}/E_{actual} =\frac{3.7x10^{-2} \frac{q}{2\pi e_{0}d^{2} }}{3.92x10^{-2} \frac{q}{2\pi e_{0}d^{2} }} =0.94N/C[/tex]
