Electrostatics

A hollow conducting sphere of radius $R$ carries charge $Q$. At a point halfway to the centre, what are $E$ and $V$?

Everyone gets $E=0$. Then most write $V=0$ as well.

$V=kQ/R$ — the same value as on the surface. Zero field means constant potential, not zero potential.

$E=-\frac{dV}{dr}=0\Rightarrow V=\text{constant, not }V=0$

That single distinction is worth more marks in this chapter than any formula in it. Three facts organise everything that follows:

• Coulomb's law plus superposition is all of electrostatics. Every result below could be got by adding pairwise forces.
• Field is a vector, potential is a scalar. Whenever a route through $V$ exists, take it — scalars add without components.
• Inside a conductor, $E=0$. That one condition generates surface charge, shielding, and perpendicular field lines.

1. Charge and Coulomb's Law

Quantised — every charge is an integer multiple of $e=1.6\times 10^{-19}$ C. Conserved — total charge of an isolated system never changes. Additive — charges add as signed scalars. Invariant — unlike mass, charge does not change with the speed of the frame. This is why a moving atom stays exactly neutral.

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2},k=\frac{1}{4\pi {\epsilon }_0}=9\times 10^9{\text{N m}}^2{\text{C}}^{-2}$

with ${\epsilon }_0=8.85\times 10^{-12}$ C² N⁻¹ m⁻².

Water has $K\approx 80$, so the attraction holding an ionic lattice together is cut eightyfold — which is why salts dissolve in it and not in oil.

Superposition: the force between two charges is unaffected by the presence of others. Compute each pair as if alone, then add as vectors.

Illustration 1

Two identical conducting spheres carry $+7q$ and $-q$ and attract with force $F$. They are touched together and returned to the same separation. Find the new force.

Touching equalises the potential, so identical spheres share the total charge equally:

$q_{each}=\frac{7q+(-q)}{2}=3q$

$\frac{F^{\prime }}{F}=\frac{|3q\cdot 3q|}{|7q\cdot (-q)|}=\frac{9}{7}\Rightarrow F^{\prime }=\frac{9}{7}F,\text{now repulsive}$

Both the magnitude and the sense change. The force grew even though the total charge fell.

2. The Electric Field

$\vec{E}=\frac{\vec{F}}{q_0},E=\frac{kq}{r^2}[{\text{N C}}^{-1}\equiv {\text{V m}}^{-1}]$

The field exists whether or not a test charge is there. That is the whole point of introducing it — the source creates a field, and the field acts on the second charge, so the interaction becomes local.

For a continuous distribution, split it into elements and treat each as a point charge. Symmetry or Gauss's law usually removes the need to integrate.

Field on the axis of a charged ring

The one continuous distribution JEE expects without derivation, because it lacks the symmetry Gauss's law needs.

Each element $dq$ sits at distance $\sqrt{x^2+R^2}$. Perpendicular components cancel in pairs; only the axial component survives, with $\cos \theta =x/\sqrt{x^2+R^2}$:

$E=\int \frac{kdq}{x^2+R^2}\cdot \frac{x}{\sqrt{x^2+R^2}}\Rightarrow E=\frac{kQx}{(x^2+R^2{)}^{3/2}}$

Zero at the centre (every element cancelled by its opposite). Falls as $kQ/x^2$ far away. A quantity vanishing at both ends of a range must peak inside it — and it does, at

$x=\frac{R}{\sqrt{2}}$

Motion of a charged particle in a uniform field

$a=qE/m$, constant. So the constant-acceleration equations apply unchanged, and a charge projected across the field traces a parabola — a projectile with $qE/m$ playing the role of $g$.

Trap. Gravity is almost always negligible here. For an electron in a field of just 1 N C⁻¹, the electric force beats its weight by about $10^{11}$.

Illustration 2

An electron enters midway between two plates of length $L$ and separation $d$, held at potential difference $V$, moving parallel to the plates at speed $u$. Find the vertical deflection at exit.

$E=\frac{V}{d},a=\frac{eV}{md},t=\frac{L}{u}$

$y=\frac{1}{2}a{t}^2\Rightarrow y=\frac{eV{L}^2}{2md{u}^2}$

The electron escapes the plates only if $y<d/2$. This is exactly the cathode-ray-tube deflection calculation.

3. Electric Field Lines

• Start on positive charge, end on negative charge, or run to infinity.
• Never intersect — the field has one direction at each point, and a crossing would give it two.
• Density represents strength; the tangent gives direction.
• Meet a conductor's surface perpendicularly — any parallel component would drive surface charge until it vanished.
• None exist inside a conductor in equilibrium.

Trap. Field lines are not particle trajectories. The field gives the direction of acceleration, not velocity. A particle follows a curved line only if released from rest on a straight one; otherwise inertia carries it off the line at once.

Illustration 3

In a field-line diagram, 12 lines are drawn leaving charge $A$ and 4 lines are drawn entering charge $B$, with no other charges present. Find the ratio $q_A/q_B$, the sign of each, and say where the remaining lines go.

The number of lines drawn from a charge is proportional to its magnitude, so

$\frac{|q_A|}{|q_B|}=\frac{12}{4}=3$

Lines leave positive charge and enter negative charge, so $A$ is positive and $B$ negative, giving $q_A/q_B=-3$.

Four of $A$'s twelve lines terminate on $B$, and the other eight have nowhere to end, so they run to infinity. They must: the pair carries a net positive charge of $2|q_B|$, and a distant observer sees exactly that much charge and exactly that many escaping lines.

4. The Electric Dipole

Two equal and opposite charges separated by $2a$, with moment $\vec{p}=q(2a)$ directed from $-q$ to $+q$.

Axial is exactly twice equatorial at the same distance — the single most-asked fact in the chapter.

Both fall as $1/r^3$, not $1/r^2$: from far away the two charges nearly cancel, and the residual falls off faster than either alone.

In a uniform field the net force is zero — both charges feel equal and opposite forces — but the torque is not:

$\vec{\tau }=\vec{p}\times \vec{E},\tau =pE\sin \theta ,U=-\vec{p}\cdot \vec{E}=-pE\cos \theta$

So the dipole rotates without translating. Stable at $\theta =0$, unstable at $\theta =180°$.

Trap. A net force appears only in a non-uniform field. That is why a charged rod attracts neutral paper: it polarises the paper, then pulls the nearer induced charge harder than it pushes the farther one.

Illustration 4

A dipole sits aligned with a uniform field $E$. How much work is needed to rotate it through $60°$, and through $180°$?

$W=U_f-U_i=pE(\cos {\theta }_i-\cos {\theta }_f)$

$W_{60}=pE(1-\frac{1}{2})=\frac{pE}{2},W_{180}=pE(1-(-1))=2pE$

Three times the angle, but four times the work — because the torque $pE\sin \theta$ is itself growing over most of that sweep.

5. Electric Flux and Gauss's Law

Flux counts field lines through a surface, the dot product picking out only the perpendicular component:

${\Phi }_E=\oint \vec{E}\cdot d\vec{A}=\frac{q_{enc}}{{\epsilon }_0}$

Only enclosed charge contributes to the total flux. External charges do contribute to $\vec{E}$ at every point of the surface — moving one changes the field everywhere on it — and still leave $\Phi$ untouched.

Trap. "Flux is zero" therefore never means "field is zero". They are different statements about different things.

Gauss's law is always true but useful only when symmetry lets you take $E$ outside the integral, which needs constant $|E|$ over the surface with $\vec{E}$ either along or across it everywhere. Exactly three geometries qualify: spherical, cylindrical, planar.

Illustration 5

A point charge $q$ sits at the centre of one face of a cube. Find the flux through the cube.

The charge is not enclosed — it is on the boundary, so $q_{enc}$ is ambiguous. Complete the symmetry: place a second identical cube on the other side of that face. Now the charge is at the centre of the combined block:

${\Phi }_{block}=\frac{q}{{\epsilon }_0}\Rightarrow {\Phi }_{cube}=\frac{q}{2{\epsilon }_0}$

Building the symmetric figure the charge does sit at the centre of is the standard move for every partial-flux question.

6. Standard Results from Gauss's Law

The technique in full, for an infinite line of density $\lambda$. Take a coaxial cylinder of radius $r$, length $L$. The flat ends have $\vec{E}$ running along them, so they carry nothing:

$E(2\pi rL)=\frac{\lambda L}{{\epsilon }_0}\Rightarrow E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$

The shell results match the gravitational shell theorem exactly, and for the same reason — both forces go as $1/r^2$.

The sheet result deserves a pause. Moving away weakens each element's contribution, but brings proportionally more of the sheet into view. For a genuinely infinite sheet the two effects cancel exactly.

Trap. Sheet or plate? A thin non-conducting sheet gives $\sigma /2{\epsilon }_0$. A conducting plate carries charge on both faces and gives $\sigma /{\epsilon }_0$ just outside.

Illustration 6

A solid non-conducting sphere of radius $R$ carries charge $Q$ spread uniformly. At what interior radius does $E$ equal its value at $r=2R$?

$E_{2R}=\frac{kQ}{(2R{)}^2}=\frac{kQ}{4R^2},E_{in}(r)=\frac{kQr}{R^3}$

$\frac{kQr}{R^3}=\frac{kQ}{4R^2}\Rightarrow r=\frac{R}{4}$

Note this fails for a conducting sphere, where the interior field is zero throughout and no such radius exists.

7. Electric Potential

Work done per unit charge to bring a test charge from infinity, moved slowly so no kinetic energy is gained.

$V=\frac{kq}{r},E=-\frac{dV}{dr}$

A scalar — and that is its entire practical advantage. Several charges contribute potentials that add as signed numbers, with no components to resolve.

The field points along the steepest decrease of potential. A positive charge released from rest moves toward lower $V$; a negative charge toward higher.

• Equipotential surfaces are everywhere perpendicular to field lines, and no work is done moving along one. Concentric spheres for a point charge; parallel planes for a uniform field.
• A conductor is one equipotential throughout its volume and surface — any difference would drive current until it vanished.
• A dipole has $V=kp\cos \theta /r^2$, falling one power more slowly than its field. Zero on the equatorial plane, where the field is emphatically not zero.

Illustration 7

Three charges $+q$ sit at the corners of an equilateral triangle of side $a$. Find $V$ and $\vec{E}$ at the centroid.

Each corner is a distance $a/\sqrt{3}$ from the centroid. Potentials are numbers, so they simply add:

$V=3\cdot \frac{kq}{a/\sqrt{3}}=\frac{3\sqrt{3}kq}{a}$

The three field vectors are equal in magnitude and set $120°$ apart, so they cancel:

$\vec{E}=0\text{ with }V\ne 0$

The shell in the opening question is the same lesson from the other side: $E=0$ there with $V\ne 0$. Neither quantity determines the other at a point.

8. Potential Energy of a System of Charges

$U=\frac{k{q}_1{q}_2}{r}\text{per pair},U=qV\text{for a charge at known potential}$

Sum over every distinct pair, not just neighbouring ones: 3 charges give 3 terms, 4 give 6, $n$ give $n(n-1)/2$.

Positive $U$ for like charges, negative for unlike. A negative total means the system is bound — work must be supplied to pull it apart to infinity.

Illustration 8

Four charges, each $+q$, sit at the corners of a square of side $a$. Find the total potential energy.

Six pairs: four sides at separation $a$, two diagonals at $a\sqrt{2}$.

$U=4\cdot \frac{k{q}^2}{a}+2\cdot \frac{k{q}^2}{a\sqrt{2}}\Rightarrow U=\frac{k{q}^2}{a}\left(4+\sqrt{2}\right)$

Counting only the four sides — the commonest error here — loses the diagonals and about a quarter of the answer.

9. Conductors, Dielectrics and Polarisation

Inside a conductor in electrostatic equilibrium, $E=0$ exactly. If it were not, free charges would move and the situation would not be static. Everything else follows:

• All excess charge sits on the surface — interior charge would produce an interior field by Gauss's law.
• Field just outside is $\sigma /{\epsilon }_0$, perpendicular to the surface.
• Charge density is highest where curvature is sharpest — hence points, and hence lightning conductors.
• A cavity is completely shielded from external fields. This is electrostatic shielding: the Faraday cage, the metal instrument enclosure, the car in a thunderstorm.

A dielectric has no free charge, but its molecules polarise — permanent dipoles align, or dipoles are induced. The polarisation sets up an internal field opposing the applied one:

$E_{net}=\frac{E_0}{K}$

A dielectric weakens the field; a conductor cancels it. That is the essential difference.

Illustration 9

A point charge $+q$ sits at the centre of a cavity inside an uncharged conducting shell. Find the charge on the inner and outer surfaces and the field outside at distance $r$. Then move the charge off-centre and see what changes.

Draw a Gaussian surface inside the metal, where $E=0$ at every point, so the charge it encloses must be zero:

$q_{inner}=-q$

The shell carries no net charge, so its outer surface must hold $+q$, and outside the shell

$E=\frac{kq}{r^2}$

Now move the charge off-centre. The inner surface charge redistributes to follow it, bunching up on the near side, but its total stays exactly $-q$. The outer surface stays uniformly $+q$, so the external field does not change at all.

The metal hides where the charge is but cannot hide how much. That asymmetry is the real content of electrostatic shielding: a Faraday cage protects the inside from the outside, and does not protect the outside from the inside.

10. Capacitors

$C=\frac{Q}{V},C=\frac{{\epsilon }_{0}A}{d},C_{diel}=\frac{K{\epsilon }_{0}A}{d}$

Capacitance depends only on geometry and medium — never on the charge stored or the voltage applied.

Trap. These are the reverse of the resistor rules. Sanity check: two parallel capacitors sit side by side and make one bigger plate, so capacitance must rise.

$U=\frac{1}{2}C{V}^2=\frac{Q^2}{2C}=\frac{1}{2}QV,u=\frac{1}{2}{\epsilon }_0{E}^2$

The energy-density form locates the energy in the field itself, not on the plates.

Inserting a dielectric: the two cases

Everything turns on whether the battery is still connected.

Read the last row. With the battery connected it supplies extra charge and the stored energy rises. With it disconnected no charge can enter, the capacitor does work pulling the slab in, and the energy falls.

Write down which quantity is fixed before writing a single equation. Almost every wrong answer in this topic comes from assuming the wrong one.

Sharing charge between two capacitors

$V_{common}=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2},\Delta U=\frac{C_1{C}_2(V_1-V_2{)}^2}{2(C_1+C_2)}$

Charge is conserved; energy is not. The loss appears as heat in the wires and as radiation, and is independent of the wire's resistance — halving $R$ halves the time but doubles the current, dissipating the same total. It vanishes only when the two started at equal potential, which the squared numerator makes immediate.

Illustration 10

A slab of dielectric constant $K$ and thickness $t<d$ is inserted into a parallel plate capacitor. Find $C$, and then the limit $K\to \infty$.

The gap is now two capacitors in series — air of thickness $d-t$, dielectric of thickness $t$:

$\frac{1}{C}=\frac{d-t}{{\epsilon }_{0}A}+\frac{t}{K{\epsilon }_{0}A}\Rightarrow C=\frac{{\epsilon }_{0}A}{(d-t)+t/K}$

As $K\to \infty$ the slab becomes a conductor and $C=\frac{{\epsilon }_{0}A}{d-t}$ — the metal slab acts purely by shortening the gap, and its position does not matter at all.

Illustration 11

Five capacitors form a Wheatstone bridge: $C_1=2$ and $C_2=4$ µF in the left arms, $C_3=3$ and $C_4=6$ µF in the right arms, and $C_5=5$ µF bridging the midpoints. Find $C_{eq}$ across the supply.

Check the balance condition before anything else:

$\frac{C_1}{C_2}=\frac{2}{4}=\frac{1}{2},\frac{C_3}{C_4}=\frac{3}{6}=\frac{1}{2}\Rightarrow \text{balanced}$

The two midpoints sit at the same potential, so no charge collects on $C_5$ and it can simply be removed from the diagram.

Left branch: 2 and 4 in series give $8/6=4/3$ µF. Right branch: 3 and 6 in series give 2 µF. These are in parallel:

$C_{eq}=\frac{4}{3}+2=\frac{10}{3}\mu \text{F}$

The bridge capacitor's value never entered the answer, which is exactly why the balance test comes first.

Illustration 12

A dielectric of constant $K$ fills half of a parallel plate capacitor. Compare filling half the gap with filling half the area.

Half the gap, with the slab parallel to the plates, makes two capacitors in series, each of thickness $d/2$:

$\frac{1}{C}=\frac{d/2}{{\epsilon }_{0}A}+\frac{d/2}{K{\epsilon }_{0}A}\Rightarrow C=\frac{2K}{K+1}\cdot \frac{{\epsilon }_{0}A}{d}$

Half the area, with the slab standing between the plates, makes two capacitors in parallel, each of area $A/2$:

$C=\frac{{\epsilon }_0(A/2)}{d}+\frac{K{\epsilon }_0(A/2)}{d}\Rightarrow C=\frac{K+1}{2}\cdot \frac{{\epsilon }_{0}A}{d}$

For $K=3$ these give $1.5C_0$ and $2C_0$. Same slab, same capacitor, different answers — because series and parallel weight the two halves differently. Read the geometry before choosing the combination rule.

Summary

• Coulomb plus superposition is the whole chapter. Gauss adds no physics, only convenience.
• Charge is quantised, conserved, additive and invariant with speed.
• $F=k{q}_1{q}_2/r^2$, divided by $K$ in a medium.
• Ring on axis: $E=kQx/(x^2+R^2{)}^{3/2}$, peaking at $x=R/\sqrt{2}$.
• A charge in a uniform field is a projectile problem with $a=qE/m$; gravity is negligible.
• Field lines never cross, meet conductors perpendicularly, and are not trajectories.
• Dipole: axial field is twice equatorial, both as $1/r^3$; $\vec{\tau }=\vec{p}\times \vec{E}$, $U=-\vec{p}\cdot \vec{E}$. No net force in a uniform field.
• Flux counts enclosed charge only, though external charges do change $\vec{E}$ on the surface.
• Gauss is useful only for spherical, cylindrical and planar symmetry.
• Sheet $\sigma /2{\epsilon }_0$ (distance-independent); conducting plate $\sigma /{\epsilon }_0$, charged on both faces.
• Shell: $E=0$ inside but $V=kQ/R$ there. Solid sphere: $E=kQr/R^3$ inside.
• $V=kq/r$ is a scalar; $E=-dV/dr$. Zero $V$ never implies zero $E$, and the reverse fails too.
• $U$ sums over every distinct pair — $n(n-1)/2$ of them.
• Inside a conductor $E=0$: charge on the surface, one equipotential, cavity shielded.
• A dielectric divides the field by $K$; a conductor cancels it.
• Capacitors add in parallel, add reciprocally in series — the reverse of resistors.
• Dielectric with the battery connected: $V$ fixed, $U$ rises. Disconnected: $Q$ fixed, $U$ falls.
• Charge sharing always loses energy, independent of wire resistance.