1C · Potential & Dipole — V(r,θ), E from V + Solved PYQs

This page is about electric potential and the electric dipole. We start with the dipole moment vector, then derive the dipole potential $V(r,\theta)$ and its field. After that we prove that the electrostatic field is conservative.

1. Electric dipole and dipole moment

A dipole is a pair of equal and opposite charges $+q$ and $-q$ separated by a small distance $2l$. Daily example: an HCl molecule. The dipole moment vector is defined as

Fig 1.1: dipole geometry. The arrow $\vec{p} = q\cdot 2\vec{l}$ points from $-q$ toward $+q$. Field point $P(r,\theta)$ uses Eq. (2) below. Verified: the arrow direction matches Eq. (1).

2. Potential and field of a dipole

Potential as a line integral of field

Potential difference is the work per unit charge against the field, so it is a line integral:

This is the integral twin of $\vec{E} = -\nabla V$: one differentiates, the other integrates. For a point charge, $V(r) = -\int_\infty^r (Q/4\pi\varepsilon_0 r'^2)\,dr' = Q/4\pi\varepsilon_0 r$.

Potential of a uniformly charged spherical shell

Shell of radius $R$, total charge $Q$. Outside ($r > R$) the field is the point-charge field, so $V = Q/4\pi\varepsilon_0 r$. Inside ($r < R$) the field is zero, so moving a test charge inside costs no work — the potential is constant, equal to its surface value:

Potential of a uniformly charged solid sphere

Solid sphere of radius $R$, total charge $Q$. Outside it is again $V = Q/4\pi\varepsilon_0 r$. Inside, integrate the linear field $E = Qr/4\pi\varepsilon_0 R^3$ inward from the surface:

Derivation sketch: $V(r) - V(R) = -\int_R^r E\,dr' = -\int_R^r (Qr'/4\pi\varepsilon_0 R^3)\,dr' = (Q/8\pi\varepsilon_0 R^3)(R^2 - r^2)$; add $V(R) = Q/4\pi\varepsilon_0 R$ and simplify.

Electric potential is work per unit positive test charge to bring it from infinity to the point. For a point charge, $V = q/4\pi\varepsilon_0 r$. For a dipole at field point $(r,\theta)$ with $r \gg l$:

2.1 Derivation of $V(r,\theta)$

The point $P$ is far away, so $r \gg l$. The distances from $P$ to $+q$ and $-q$ differ only by the projection of $2l$ on the line to $P$.

1. Approximate distances: $r_1 \approx r - l\cos\theta$ (to $+q$) and $r_2 \approx r + l\cos\theta$ (to $-q$) — reason: projection of half-separation $l$ on $\hat{r}$.
2. Total potential (principle of superposition): $V = \dfrac{q}{4\pi\varepsilon_0}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$.
3. Combine the fractions: $\dfrac{1}{r_1} - \dfrac{1}{r_2} = \dfrac{r_2 - r_1}{r_1 r_2} \approx \dfrac{2l\cos\theta}{r^2}$ — reason: $r_1 r_2 \approx r^2$ when $r \gg l$.
4. Insert $p = q\cdot 2l$: $V = \dfrac{p\cos\theta}{4\pi\varepsilon_0 r^2}$ (Eq. 2). Limit: $V \to 0$ as $r \to \infty$. ✓

2.2 Field from potential

From $\vec{E} = -\nabla V$ in spherical coordinates, the radial and angular components are $E_r = -\partial V/\partial r$ and $E_\theta = -(1/r)\partial V/\partial\theta$.

1. $\partial V/\partial r = -\dfrac{2p\cos\theta}{4\pi\varepsilon_0 r^3}$, so $E_r = +\dfrac{2p\cos\theta}{4\pi\varepsilon_0 r^3}$ (reason: minus times minus).
2. $\partial V/\partial\theta = -\dfrac{p\sin\theta}{4\pi\varepsilon_0 r^2}$, so $E_\theta = +\dfrac{p\sin\theta}{4\pi\varepsilon_0 r^3}$ (reason: same).
3. Special cases — on the axis ($\theta = 0$): $E = 2p/4\pi\varepsilon_0 r^3$; on the equator ($\theta = 90°$): $E = p/4\pi\varepsilon_0 r^3$ (opposite to $\vec{p}$).

Solved PYQs

Reference. Brij Lal & Subrahmanyam, Electricity & Magnetism Ch-1 (dipole, potential, conservative field); S.L. Arora Vol-1 Ch-1.