Question

Show that E (axial) = -2*E(equal)

Answer 1

Given:

A short Electrostatic dipole has been provided.

To show:

E_(axial) = -2 × E_(equatorial)

Proof:

We will assume the Electrostatic dipole to be short such that inter-charge distance will be insignificant as compared to the the distance between the dipole and the point at which the electric field intensity is measured.

At axial position:

$\therefore \: \vec E_{axial} = \dfrac{2kp}{ {r}^{3} } \: \hat{r}$

At equatorial position:

$\therefore \: \vec E_{equatorial} = \dfrac{kp}{ {r}^{3} } \: (- \hat{r})$

$\hat{r}$ represents the direction of the dipole.

Comparing the two Equations:

$\boxed{ \bold{ \vec E_{axial} = - 2 \times ( \vec E_{equatorial})}}$

Answer 2

The ratio between axial and equatorial electric field lines is 1:2, therefore $E_(axial) = 2E_(equatorial)$

Given:

$E_(axial) = 2E_(equatorial)$

To find:

Proving that the ratio between the axial and equatorial line in an electric field is 1:2

Solution:

The electrical dipole is where two opposite charges are separated by a certain distance. Here the charges have equal magnitude. The formulas to find the electric field is different for both axial and equatorial,

For axial,

$E(axial)=(1/4\pi E_o) (2p/r^3)$

For equatorial

$E_(equatorial) = (1/4\pi E_o)(p/r^3)$

In some problems the value of r is different. Since we need to prove the ratio we will take r as the same value and a constant.

Let us find the ratio by dividing $E_(equatorial)$ it by $E_(axial)$

$E_(equatorial)/ E_(axial) = (1/4\pi E_o)(p/r^3) / (1/4\pi E_o)(2p/r^3)$

Cancel out all the common constants and values.

$E_(equatorial)/ E_(axial) = p/2p$

$E_(equatorial)/ E_(axial) = 1/2$

After cross-multiplication

$E_(axial) = 2E_(equatorial)$

We can say that the ratio between $E_(axial)$ and $E_(equatorial)$ is 1:2, therefore $E_(axial) = 2E_(equatorial)$

#SPJ6