Q.Two equal and opposite charges placed in air as shown in figure:

The point P is closer to the $+q$ charge (distance $r-a$) than to $-q$ (distance $r+a$), so the field due to $+q$ at P follows the inverse-square law with $(r-a)$ in the denominator, pointing away from $+q$ along the dipole axis.

Figure recap: The dipole has $+q$ at one end and $-q$ at the other, separated by $2a$, with centre $O$. Point $P$ lies on the axial line at distance $r$ from $O$, on the side of the $+q$ charge, so $P$ is only $(r-a)$ away from $+q$ but $(r+a)$ away from $-q$.

Field due to $+q$ at P: Using Coulomb's law for a point charge,

$E(+q) = \frac{q}{4\pi\varepsilon_0 (r-a)^2}$

directed AWAY from $+q$, i.e. along $\hat p$ (the unit vector along the dipole axis, pointing from $-q$ towards $+q$ and beyond, towards P).

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At the same distance r, the axial field is twice the equatorial field. Rotating the dipole 90° about its perpendicular axis makes the observation point equatorial, so the field drops to E/2. Option (c).

Step 1 — recall the two standard dipole fields at distance r (r ≫ dipole size), from NCERT/CBSE Class 12 Physics:

• Axial (end-on): E_axial = (1/4πε₀)(2p/r³)
• Equatorial (broadside): E_equatorial = (1/4πε₀)(p/r³) …

The axial field of a dipole falls off much faster than that of a single point charge — as the inverse cube, not the inverse square, of distance.

For a short electric dipole of moment $p = q(2a)$, the electric field at a point on the axial line at distance $r$ from the centre (for $r \gg a$) is:

$E_{axial} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}$

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The force on a point charge due to a short electric dipole on its axis follows an inverse-cube law. Doubling the distance reduces the force by a factor of 8, so the new force is $F/8$.

The electric field on the axial line of a short dipole falls off as 1/r^3 and is twice the equatorial field at the same distance.

For an electric dipole of dipole moment $p = q(2a)$, the exact field at an axial point at distance $r$ from the centre is

$E_{axial} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2pr}{(r^2-a^2)^2}$

For a short dipole, i.e. when $r \gg a$ (the usual approximation used), this simplifies to …

For a dipole, the potential falls one power of r faster than a single charge (1/r² instead of 1/r), and the field falls one power faster than the potential (1/r³).

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For a short dipole, $E_{axial} = 2E_{equatorial}$ at the same distance, so the ratio is 2:1.

For an electric dipole of moment $p$, at a distance $r$ from the centre (with $r$ much greater than the dipole separation):

Field at an axial (end-on) point:

$E_{axial} = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}$

Field at an equatorial (broadside-on) point:

$E_{equatorial} = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}$

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The axial field of a short electric dipole is E = (1/4πε₀)·2pr/(r²−l²)², matching the Column-B formula involving p (dipole moment) and r (distance).

For an electric dipole of dipole moment $p$ (charges $\pm q$ separated by $2l$), the electric field intensity at a point on its axial line, at distance $r$ from the centre, is:

$E_{axial} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{2pr}{(r^2-l^2)^2}$ …

The axial-line field of a short dipole varies as $1/r^3$, so doubling the distance cuts the force to $1/8$ of its original value.

Step 1 — Field of a short dipole on its axial line

$E_{axial} = \dfrac{2kp}{r^3}$

where $p$ is the dipole moment and $r$ is the distance from the centre of the dipole (valid for $r \gg$ dipole length).

Step 2 — Force on the charge

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