Equatorial Line or Broadside-on-position or TAN-B position of the electric dipole tends to showing the description, explanation and calculation.

## Equatorial Line or Broadside-on-position or TAN-B

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## Equatorial Line or Broadside-on-position or TAN-B

**Electric Dipole: **The electric dipole consists of a pair of equal and opposite charges at very small distance appart.

**Dipole Moment**: The strength of electric dipole is measured by a vector quantity known as electric dipole moment.

The electric dipole moment is a product between magnitude of charge and distance between them.

Mathematically, it is given by,

\vec{P}=q \times 2\vec{l}## Analytical Treatment for Field intensity Due to Dipole on The Equatorial Line

Let’s consider an electric dipole **AB** of length 2l.

Let OA = OB = l

Let **P** is a observation point at a distance “**r**” from center of dipole **O**, on the equatorial line of the dipole.

So that,

**OP **=** r**

Let E_{1} = electric field intensity at the point P due to charge +q.

So that,

E_{1} = \frac{1}{4\pi\epsilon_0}.\frac{q}{(r^{2}+l^{2})}\:along\:PT --- (1)Let E_{2} = electric field intensity at the point P due to charge -q.

So that,

E_{2} = \frac{1}{4\pi\epsilon_0}.\frac{q}{(r^{2}+l^{2})}\:along\:PS --- (2)Let **E** = electric field intensity at the point P due to electric dipole.

Now according to parallelogram law of vector addition.

\left | \vec{E} \right |= \sqrt{E_{1}^{2}+E_{2}^{2}+2E_{1}E_{2}cos2\Theta} \because \angle OBP = \angle SPK = \Theta (alternate\:angle) and\:\angle OAP = \angle KPT = \Theta (corresponding\:angle) \therefore \angle SPT = \Theta +\Theta = 2\Theta But \left | \vec{E_{1}} \right |=\left | \vec{E_{2}} \right |= \frac{1}{4\pi\epsilon_0}.\frac{q}{(r^{2}+l^{2})}\:---\:(3) \therefore \left | \vec{E} \right |= \sqrt{E_{1}^{2}+E_{2}^{2}+2E_{1}E_{2}cos2\Theta} \left | \vec{E} \right |= \sqrt{E_{1}^{2}+E_{1}^{2}+2E_{1}E_{1}cos2\Theta} \left | \vec{E} \right |= \sqrt{2E_{1}^{2}+2E_{1}^{2}cos2\Theta} \left | \vec{E} \right |= \sqrt{2E_{1}^{2}(1+cos2\Theta)} \left | \vec{E} \right |= \sqrt{2E_{1}^{2}.2cos^{2}\Theta}(As\:1+cos2\Theta)=2cos^2\Theta)

\left | \vec{E} \right |= 2\left | \vec{E_{1}} \right |cos\Theta\:---\:(4)Now in \triangle AOP,

cos\Theta = \frac{OA}{AP}=\frac{l}{\sqrt{r^{2}+l^{2}}}\:---\:(5)By putting the value of \left | \vec{E}_{1} \right | and cos\Theta in equation (4), it turns with

E = 2.\frac{1}{4\pi\epsilon_0}.\frac{q}{(r^{2}+l^{2})}.\frac{l}{\sqrt{r^{2}+l^{2}}} E = 2.\frac{1}{4\pi\epsilon_0}.\frac{q}{(r^{2}+l^{2})}.\frac{l}{\sqrt{r^{2}+l^{2}}} E = \frac{1}{4\pi\epsilon_0}.\frac{2ql}{(r^{2}+l^{2}).{\sqrt{r^{2}+l^{2}}}} E = \frac{1}{4\pi\epsilon_0}.\frac{2ql}{(r^{2}+l^{2})^1.{({r^{2}+l^{2}}})^\frac{1}{2}} E = \frac{1}{4\pi\epsilon_0}.\frac{2ql}{(r^{2}+l^{2})^{1+\frac{1}{2}}}\:along\:PK E = \frac{1}{4\pi\epsilon_0}.\frac{2ql}{(r^{2}+l^{2})^{\frac{3}{2}}} E = \frac{1}{4\pi\epsilon_0}.\frac{P}{(r^{2}+l^{2})^{\frac{3}{2}}}\:along\:PK\:---\:(6)(As\:2ql=P=\:Electric\:dipole\:moment)

If r> > l, then r^2> > l^2

So l^2 can be neglected.

Now equation (6) turns into

\boxed{E = \frac{1}{4\pi\epsilon_0}.\frac{P}{r^3}\:along\:PK}Which is the required expression for the total electric field due to an electric dipole at any point on the equatorial line.

Raaz Sir says

Thanks to all for your valuable quetions. Keep it up.