Derive the expression for electric field intensity at a point on equatorial plane due to electric dipole.

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> [!TLDR]
> On the equatorial plane, $E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{(r^2+a^2)^{3/2}}$, and for $r\gg a$, $E=\dfrac{p}{4\pi\varepsilon_0 r^3}$, opposite to $\vec p$.
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> **Setup.** A dipole consists of charges $+q$ and $-q$ separated by $2a$; dipole moment $p=q(2a)$ points from $-q$ to $+q$. Consider a point $P$ on the equatorial line (perpendicular bisector) at distance $r$ from the centre. The distance from each charge to $P$ is
> $$s = \sqrt{r^2 + a^2}$$
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> **Fields due to each charge.** Magnitudes are equal:
> $$E_+ = E_- = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2+a^2}$$
> Each makes an angle $	heta$ with the equatorial line, where $\cos	heta = a/s$.
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> **Resolving components.** The components perpendicular to the axis ($E_\pm\sin	heta$) are equal and opposite → they **cancel**. The components parallel to the dipole axis ($E_\pm\cos	heta$) point the same way (from $+q$ toward $-q$, i.e. **antiparallel to $\vec p$**) → they **add**:
> $$E_{eq} = 2E_+\cos	heta = 2\cdot\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2+a^2}\cdot\frac{a}{\sqrt{r^2+a^2}}$$ …

Q.Derive the expression for electric field intensity at a point on equatorial plane due to electric dipole.

Concept understanding — Electric Dipole in a Field

Electric Dipole in a Uniform Field — First Look

What happens? Two forces, one twist

The torque formula

What does this mean physically?

Q.Derive the expression for electric field intensity at a point on equatorial plane due to electric dipole.

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