Unit 01: Electrostatics • Chapter 01
Electric Field, Field Lines & Electric Dipole
Field mapping, axial vs equatorial derivations, torque in uniform field, and subtle exam traps.
Socho ek heavy bowling ball trampoline par rakhi hai—woh aas-paas ke space ko distort karti hai taaki koi dusri marble paas aate hi roll hone lage. Isi tarah ek charge vacuum ko modify karke apna Electric Field ($\vec{E}$) banata hai. Aur jab do equal opposite charges ek strict separation par lock ho jaate hain (jaise ek water molecule $H_2O$), toh woh ek Electric Dipole banate hain: pure system ka net charge zero hota hai, fir bhi boundary ke bahar iska electric field zero nahi hota kyunki dono charges ki locations spatially alag hoti hain!
Electric Field & Dipole Moment
Electric field definition and dipole moment vector directed from $-q$ to $+q$:
SI unit of dipole moment: $\text{C}\cdot\text{m}$. Vector points from negative to positive charge.
Dipole Fields (Short Dipole: $r \gg a$)
Axial (end-on) and Equatorial (broadside-on) field comparison:
Notice: $|\vec{E}_{\text{axial}}| = 2 |\vec{E}_{\text{eq}}|$ and field drops as $1/r^3$, faster than a point charge's $1/r^2$.
Derivation A: Electric Field on the Axial Line
1. Consider a dipole with charges $-q$ and $+q$ separated by $2a$. Point $P$ lies on the axis at distance $r$ from midpoint $O$.
2. Field due to $+q$: $E_1 = \frac{1}{4\pi\varepsilon_0}\frac{q}{(r-a)^2}$ (directed away from $+q$).
3. Field due to $-q$: $E_2 = \frac{1}{4\pi\varepsilon_0}\frac{q}{(r+a)^2}$ (directed towards $-q$).
4. Net axial field ($E_{\text{axial}} = E_1 - E_2$):
5. Substitute $p = q(2a)$. For a short dipole where $r \gg a$ ($r^2 - a^2 \approx r^2$):
Derivation B: Electric Field on the Equatorial Line (Broadside-on)
1. Let point $P$ be on the perpendicular bisector at distance $r$ from center $O$. Distance to both charges is $AP = BP = \sqrt{r^2 + a^2}$.
2. Individual field magnitudes are identical: $E_+ = E_- = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2 + a^2}$.
3. Normal components ($E\sin\theta$) cancel out symmetrically, while horizontal components ($E\cos\theta$) reinforce each other antiparallel to $\vec{p}$:
4. Setting $p = q(2a)$ and applying the short-dipole limit ($r \gg a$):
Derivation C: Torque & Potential Energy of Dipole in Uniform $\vec{E}$
1. Forces on charges: $\vec{F}_+ = +q\vec{E}$ and $\vec{F}_- = -q\vec{E}$. Total translational force $\vec{F}_{\text{net}} = \vec{F}_+ + \vec{F}_- = \vec{0}$.
2. Perpendicular distance between lines of force $= 2a\sin\theta$.
3. Work done in rotating dipole from $\theta_1$ to $\theta_2$: $W = \int_{\theta_1}^{\theta_2}\tau\,d\theta = -pE(\cos\theta_2 - \cos\theta_1)$.
4. Taking zero energy reference at $\theta = 90^\circ$: $U = -\vec{p}\cdot\vec{E} = -pE\cos\theta$.
Trap 1: The Direction of $\vec{E}_{\text{eq}}$ vs $\vec{p}$
Students consistently miss the negative vector sign in equatorial field! On the axial line, $\vec{E}_{\text{axial}}$ points in the same direction as $\vec{p}$. But on the equatorial line, $\vec{E}_{\text{eq}}$ points antiparallel (180° opposite) to $\vec{p}$.
Trap 2: Uniform vs Non-Uniform Field Trap
In a uniform field: $F_{\text{net}} = 0$ and $\tau \ne 0$ (only rotational motion). But in a non-uniform field: $F_{\text{net}} \ne 0$ and $\tau \ne 0$ (both translation and rotation occur simultaneously).
Trap 3: Stable vs Unstable Equilibrium
At $\theta = 0^\circ$: $\tau = 0$ and $U = -pE$ (Minimum PE $\implies$ Stable Equilibrium). At $\theta = 180^\circ$: $\tau = 0$ but $U = +pE$ (Maximum PE $\implies$ Unstable Equilibrium).
Trap 4: Electric Field Lines Never Form Closed Loops
Electrostatic fields are conservative ($\oint \vec{E}\cdot d\vec{l} = 0$). Hence, field lines originate on positive charges and terminate on negative charges; they never form continuous closed paths (unlike magnetic field lines).
Problem: An electric dipole of length $4\text{ cm}$ when placed with its axis at $60^\circ$ to a uniform electric field experiences a torque of $4\sqrt{3}\text{ N}\cdot\text{m}$. If the charge on dipole is $\pm 8\text{ nC}$, calculate the magnitude of the electric field and work done to rotate it to unstable equilibrium.
Solution:
1. Dipole moment: $p = q(2a) = (8 \times 10^{-9}\text{ C}) \times (0.04\text{ m}) = 3.2 \times 10^{-10}\text{ C}\cdot\text{m}$.
2. Torque formula: $\tau = pE\sin\theta \implies 4\sqrt{3} = (3.2 \times 10^{-10}) \cdot E \cdot \sin 60^\circ$.
Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$:
$$4\sqrt{3} = (3.2 \times 10^{-10}) \cdot E \cdot \frac{\sqrt{3}}{2} \implies E = \frac{8}{3.2 \times 10^{-10}} = 2.5 \times 10^{10}\text{ N/C}$$
3. Work to rotate from $\theta_1 = 60^\circ$ to unstable equilibrium ($\theta_2 = 180^\circ$):
$$W = -pE(\cos 180^\circ - \cos 60^\circ) = -pE\left(-1 - \frac{1}{2}\right) = \frac{3}{2}pE$$
$$W = 1.5 \times (3.2 \times 10^{-10}) \times (2.5 \times 10^{10}) = 12\text{ Joules}$$
Problem: Two short electric dipoles of moments $p_1$ and $p_2$ are placed along the same line at a distance $r$ ($r \gg \text{length of dipoles}$). What is the mutual interaction force between them?
Solution:
1. Axial field created by dipole 1 at the location of dipole 2: $E_1 = \frac{1}{4\pi\varepsilon_0}\frac{2p_1}{r^3}$.
2. Potential energy of dipole 2 in field $E_1$: $U = -p_2 E_1 = -\frac{1}{4\pi\varepsilon_0}\frac{2p_1 p_2}{r^3}$.
3. Interaction force is the spatial gradient of potential energy ($F = -\frac{dU}{dr}$):
$$F = -\frac{d}{dr}\left(-\frac{1}{4\pi\varepsilon_0}\frac{2p_1 p_2}{r^3}\right) = -\frac{1}{4\pi\varepsilon_0}\frac{6p_1 p_2}{r^4}$$
Exam Trap Note: Force between two point charges scales as $1/r^2$. Force between a charge and a dipole scales as $1/r^3$. Force between two dipoles scales as $1/r^4$!
Axial Field
$E = \frac{2kp}{r^3}$
Equatorial Field
$E = \frac{kp}{r^3} \quad (\parallel -\vec{p})$
Torque in $\vec{E}$
$\vec{\tau} = \vec{p} \times \vec{E}$
Potential Energy
$U = -\vec{p} \cdot \vec{E}$