Equatorial Line or Broadside-on-position or TAN-B Position of The Electric Dipole

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

\therefore AP = BP = \sqrt{r^{2}+l^{2}}

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.