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

The ratio of
`E_(app)` and
`E_(act)` is 0.9754

Step-by-step explanation:

Given that,

Distance z = 4.50 d

First equation is

`E_(act)=(qd)/(2\pi\epsilon_(0)* z^3)`

`E_(act)=(Pz)/(2\pi\epsilon_(0)* (z^2-(d^2)/(4))^2)`

Second equation is

`E_(app)=(P)/(2\pi\epsilon_(0)* z^3)`

We need to calculate the ratio of
`E_(act)` and
`E_(app)`

Using formula

`(E_(app))/(E_(act))=((P)/(2\pi\epsilon_(0)* z^3))/((Pz)/(2\pi\epsilon_(0)* (z^2-(d^2)/(4))^2))`

`(E_(app))/(E_(act))=((z^2-(d^2)/(4))^2)/(z^3(z))`

Put the value into the formula

`(E_(app))/(E_(act))=(((4.50d)^2-(d^2)/(4))^2)/((4.50d)^3*4.50d)`

`(E_(app))/(E_(act))=0.9754`

Hence, The ratio of
`E_(app)` and
`E_(act)` is 0.9754