Q.(a) Calculate the electric field due to a dipole at a point on the axial line. OR

(a) The axial field of a dipole, found by vector-adding the individual charges' fields, reduces to $E\approx2kp/r^3$ for points far from the dipole; (b) a nuclear reactor's moderator slows neutrons, its control rods regulate the chain reaction, and its cooling system extracts the generated heat. Both alternatives answered below.

(a) Electric field due to a dipole on the axial line

1. Setup. An electric dipole consists of charges $-q$ at A and $+q$ at B, separated by distance $2a$, with dipole moment $p=q(2a)$ directed from $-q$ to $+q$. Consider a point P on the axial line (the line through both charges), at distance $r$ from the centre O of the dipole.

2. Field due to each charge. Distance from $+q$ to P is $(r-a)$; distance from $-q$ to P is $(r+a)$. Both fields are along the axis:

$E_{+q} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(r-a)^2} \quad \text{(away from +q, i.e. towards P)}$

$E_{-q} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(r+a)^2} \quad \text{(towards $-q$, i.e. opposite to P)}$

3. Resultant. Since $E_{+q} > E_{-q}$ (P is closer to $+q$), the net field points from $-q$ to $+q$ (same direction as $\vec p$):

$E = E_{+q} - E_{-q} = \dfrac{q}{4\pi\varepsilon_0}\left[\dfrac{1}{(r-a)^2}-\dfrac{1}{(r+a)^2}\right]$

Simplifying the bracket over a common denominator:

$= \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{(r+a)^2-(r-a)^2}{(r^2-a^2)^2} = \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{4ar}{(r^2-a^2)^2}$

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

4. Far-field (short dipole) approximation. For a point far from the dipole, $r\gg a$, so $a^2$ can be neglected compared to $r^2$:

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

directed along $\vec p$ (from $-q$ to $+q$).

(b) Nuclear reactor components

A nuclear reactor is a controlled setup that sustains a self-propagating nuclear fission chain reaction to release energy at a steady, controllable rate (unlike an uncontrolled reaction, as in a bomb).

Moderator. Fission of $^{235}U$ is most efficiently triggered by slow (thermal) neutrons, but the neutrons released by fission are fast. The moderator (commonly heavy water $D_2O$, ordinary water, or graphite) surrounds the fuel rods and slows down these fast neutrons via repeated elastic collisions, without absorbing them, so they can efficiently trigger further fissions.

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