Question

Equations 22-8 and 22-9 are approximations of the magnitude of the electric field of an electric dipole, at points along the dipole axis. consider a point p on that axis at distance z = 3.00d from the dipole center(where d is the separation distance between the particles of the dipole). let eappr be the magnitude of the field at point p as approximated by equations 22-8 and 22-9. let eact be the actual magnitude. by how much is the ratio eappr/eact less than 1?

Answer 1

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

$E_{actual} =\frac{q}{2\pi e_{0} Z^{3} } \frac{d}{(1-(\frac{d}{2z})^{2} )^{2} }$

if z = 8d

$E_{actual} =\frac{q}{2\pi e_{0}(3d)^{3} } \frac{d}{(1-(\frac{d}{6d})^{2})^{2} } \\E_{actual} =3.92x10^{-2} \frac{q}{2\pi e_{0}d^{2} }$

the approx magnitude of the dipole at a distance Z from the centre is equal to

$E_{approx} =\frac{1}{2\pi e_{0} } \frac{qd}{(Z^{3} } \\E_{approx} =3.7x10^{-2} \frac{q}{2\pi e_{0}d^{2} }$

The ratio is

$E_{approx}/E_{actual} =\frac{3.7x10^{-2} \frac{q}{2\pi e_{0}d^{2} }}{3.92x10^{-2} \frac{q}{2\pi e_{0}d^{2} }} =0.94N/C$

Answer 2

On axis of the dipole the electric field is given by

$E = \frac{kq}{(r - \frac{d}{2})^2} - \frac{kq}{(r + \frac{d}{2})^2}$

$E = \frac{kq(r + d/2)^2 - (r - d/2)^2}{(r^2 - \frac{d^2}{4})^2}$

$E_{act} = \frac{kq*2rd}{(r^2 - \frac{d^2}{4})^2}$

now if we approximate above equation

$E_{eppr} = \frac{kq*2d}{r^3}$

now we will find the ratio of these two

$\frac{E_{eppr}}{E_{act}} = \frac{(r^2 - \frac{d^2}{4})^2}{r^4}$

put r = 3d

$\frac{E_{eppr}}{E_{act}} = \frac{(9d^2 - \frac{d^2}{4})^2}{81d^4}$

$\frac{E_{eppr}}{E_{act}} = \frac{(9- \frac{1}{4})^2}{81}$

$\frac{E_{eppr}}{E_{act}} = 0.945$

so above is the ratio of two field