Q.Derive formula for the electric field due to electric dipole at any point on the equatorial plane. Draw necessary diagram. OR Obtain an expression for the electric field at any point due to a uniformly charged infinite plane sheet with the help of Gauss's law. Draw necessary diagram.

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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