Question

Equations E = 1 2πε0 qd z3 and E = 1 2πε0 P z3 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 = 4.50d 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 E = 1 2πε0 qd z3 and E = 1 2πε0 P z3 (electric dipole). Let Eact be the actual magnitude. By how much is the ratio Eappr/Eact less than 1?

Answer 1

Answer:

The ratio of $E_{app}$ and $E_{act}$ is 0.9754

Explanation:

Given that,

Distance z = 4.50 d

First equation is

$E_{act}=\dfrac{qd}{2\pi\epsilon_{0}\times z^3}$

$E_{act}=\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}$

Second equation is

$E_{app}=\dfrac{P}{2\pi\epsilon_{0}\times z^3}$

We need to calculate the ratio of $E_{act}$ and $E_{app}$

Using formula

$\dfrac{E_{app}}{E_{act}}=\dfrac{\dfrac{P}{2\pi\epsilon_{0}\times z^3}}{\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}}$

$\dfrac{E_{app}}{E_{act}}=\dfrac{(z^2-\dfrac{d^2}{4})^2}{z^3(z)}$

Put the value into the formula

$\dfrac{E_{app}}{E_{act}}=\dfrac{((4.50d)^2-\dfrac{d^2}{4})^2}{(4.50d)^3\times4.50d}$

$\dfrac{E_{app}}{E_{act}}=0.9754$

Hence, The ratio of $E_{app}$ and $E_{act}$ is 0.9754