JEE Main (Engineering) · Physics (JEE & NEET)

Electrostatics

Coulombs law, electric field and potential, Gauss law, capacitors, dielectrics and energy stored in electric fields.

Nine concepts, from a single pair of charges to a capacitor full of dielectric. Electrostatics rewards two habits above all: superpose vectors and scalars differently, and always know which quantity the problem is holding fixed.

JEE Main (Engineering)
Medium level
9 concepts
5 practice questions

1Coulomb's law and superposition

Two point charges push or pull along the line joining them with F = kq₁q₂/r², where k = 1/4πε₀ = 9 × 10⁹ N m² C⁻². The law is an inverse square in the separation and linear in each charge, and it obeys superposition: with several charges present, the force on one is the vector sum of the forces each other charge would exert on its own. Nothing screens it — a third charge sitting in between does not weaken the original pair's attraction.

Figure. The three arrows are drawn to one scale, so their lengths are the forces themselves: the 3 μC charge pushes harder simply because it is bigger, and the leftover 0.4 N points back toward the weaker charge.

How it works

One pair at a timeCompute the Coulomb force between the charge of interest and each other charge separately.
Direction from signLike charges repel, unlike attract — read the direction off the signs, then work with magnitudes.
Add as vectorsResolve into components and sum. Only for collinear charges is it plain arithmetic.

The charge in the middle

Charges q₁ = +2 μC at A and q₂ = +3 μC at B are 30 cm apart. A charge q₃ = +1 μC is placed at the midpoint. Find the net force on it. Use k = 9 × 10⁹.

F₁ = 9×10⁹ × (2×10⁻⁶)(1×10⁻⁶)/(0.15)²0.80 N, toward B
F₂ = 9×10⁹ × (3×10⁻⁶)(1×10⁻⁶)/(0.15)²1.20 N, toward A
F_net = 1.20 − 0.80 (opposite directions)0.40 N, toward A

Pro tip. At equal distances the forces are in the ratio of the charges alone, so the middle charge is always pushed toward the weaker one. It sits at rest only at 13.5 cm from the 2 μC charge — the null point lies nearer the smaller charge, never at the midpoint.

Two charges in vacuum exert a force F on each other. Both charges are doubled and the separation is also doubled. The new force is

F, unchanged
2F
F/4

The product of the charges rises four-fold and r² also rises four-fold, so the two factors cancel exactly. Changing charge and distance together is the standard trap.

2The electric field

The field is what a charge leaves behind in space: E = F/q₀, measured with a test charge small enough not to disturb the source. A point charge gives E = kq/r², directed away from positive charge and toward negative, and the fields of several charges superpose as vectors. Field lines are the curves everywhere tangent to E: they begin on positive charge, end on negative, never cross — the field at a point has exactly one direction — and crowd together where the field is strong.

Figure. This is the field sampled at points, not drawn as continuous lines. The four inner arrows sit at one radius and the four outer ones at twice that radius, and they are drawn to scale — double the distance and the arrow is a quarter as long, which is the inverse square made visible. Continuous field lines carry the same information a different way, by crowding: the closer the lines, the stronger the field.

How it works

The source makes itEvery charge fills space with E = kq/r², whether or not anything is there to feel it.
Superpose the vectorsThe field of a group is the vector sum; contributions cancel only if they are equal and opposite.
Then read the forceDrop a charge q into the field and it feels F = qE — along E if q is positive, against it if negative.

Where does the field vanish?

Charges of +9 μC and +1 μC are 40 cm apart. Find the point on the line joining them where the resultant field is zero.

Equal magnitudes: 9/x² = 1/(0.4 − x)²3(0.4 − x) = x
1.2 = 4xx = 0.30 m from the 9 μC
Check: 9×10⁹ × 9×10⁻⁶/(0.30)² and 9×10⁹ × 1×10⁻⁶/(0.10)²9 × 10⁵ N C⁻¹ each ✓

Pro tip. For two like charges the null point lies between them, nearer the smaller one. For unlike charges it lies outside the pair, on the far side of the smaller charge — because only there can the closer, weaker charge win.

3Potential and its gradient

Potential is energy per unit charge, V = kq/r for a point charge, and it is a scalar — add the contributions with their signs and never with components. Field and potential are two views of one thing: E = −dV/dr, so E points down the steepest slope of V, and equipotential surfaces are everywhere perpendicular to the field. Moving a charge between two points costs W = qΔV, whatever path it takes.

The energy of a group of charges is the work needed to assemble it: one term U = kq₁q₂/r for every pair, so n charges give n(n − 1)/2 terms.

Figure. The potential of a point charge falls as 1/r, and the field is minus the slope of this curve. Near the charge the curve is steep, so E is large; far away it flattens, and E dies faster than V does — as 1/r² against 1/r.

How it works

Add potentials, not vectorsSum kq/r over the charges keeping each sign — far shorter than summing field vectors.
Differentiate for the fieldE = −dV/dr. A flat V means no field; a steep V means a strong one.
Count pairs for energyOne U = kq₁q₂/r per pair. Four charges at the corners of a square make six terms, not four.

Assembling a triangle

Three charges of +1 μC each sit at the corners of an equilateral triangle of side 10 cm. How much work was needed to bring them there from infinity?

Pairs to count: 12, 23, 313
U₁₂ = 9×10⁹ × (10⁻⁶)²/0.109 × 10⁻² J
U = 3 × 0.09 (all pairs identical)0.27 J

Pro tip. Bringing in the first charge is free, the second costs one pair and the third costs two — 0 + 1 + 2 = 3 pairs, which is the same count as n(n − 1)/2. The order of assembly never changes the total.

At a certain point the electric field is exactly zero. The potential there

Must also be zero
May be non-zero
Must be a maximum

E measures the slope of V, not its value. Midway between two equal positive charges the fields cancel while both potentials add, so V is large and positive with E = 0.

4The dipole in a field

A dipole is two equal and opposite charges a short distance apart, described by p = q(2a) pointing from the negative charge to the positive. In a uniform field the two forces are equal and opposite, so there is no net force — but they act along different lines, so there is a torque τ = pE sin θ that turns p toward E. The stored energy is U = −pE cos θ: least at θ = 0, which is stable, and greatest at θ = 180°, which is not.

The dipole's own field falls faster than a single charge's, as 1/r³ — it is 2kp/r³ along the axis and kp/r³ on the equator, pointing the opposite way.

Figure. The two qE arrows are the same length and point opposite ways, which is why the dipole does not drift. They act along different lines, and that offset is the whole torque. The couple turns p until it lines up with E, and then everything is balanced.

How it works

Forces cancelEach charge feels qE, one along the field and one against it, so in a uniform field the net force is zero.
The couple survivesThe two lines of action are offset by 2a sin θ, leaving a couple of magnitude pE sin θ.
Energy from angleTurning from θ₁ to θ₂ costs W = pE(cos θ₁ − cos θ₂), independent of how fast you turn it.

Turning a dipole right round

Two charges of 2 μC and −2 μC are 2 cm apart in a uniform field of 5 × 10⁵ N C⁻¹. Find the dipole moment, the maximum torque and the work needed to rotate the dipole from θ = 0° to θ = 180°.

p = q(2a) = 2×10⁻⁶ × 0.024 × 10⁻⁸ C m
τ_max = pE (at θ = 90°) = 4×10⁻⁸ × 5×10⁵2 × 10⁻² N m
W = pE(cos 0° − cos 180°) = 2pE4 × 10⁻² J

Pro tip. Torque is greatest at 90°, where the energy is zero; energy is greatest at 180°, where the torque is zero. A quarter turn from the stable position costs pE, and the full flip costs exactly twice that.

A small electric dipole is placed in a uniform electric field. It experiences

A torque but no net force
A net force but no torque
A net force pulling it toward the stronger field

The forces on the two charges are equal and opposite, so they cancel. Being offset, they still produce a couple. A net force appears only in a non-uniform field.

5Gauss's law and the right surface

The total flux out of any closed surface is the charge inside it divided by ε₀: Φ = ∮E·dA = q_enc/ε₀. Charges outside contribute nothing to the net flux — their lines enter the surface and leave it again. The law is always true, but it is only useful when symmetry lets you take E outside the integral, which needs a surface on which E has one constant magnitude and is either along dA or flat against it.

Three results come out in a line each: λ/2πε₀r for an infinite line, σ/2ε₀ for an infinite sheet, and σ/ε₀ just outside a conductor.

Figure. Side view of a cylinder wrapped around a long charged line. The field is radial, so it crosses the curved wall squarely and skims along the two flat ends, which therefore carry no flux at all. Only the curved area 2πrL survives, and E = λ/2πε₀r drops straight out.

How it works

Match the symmetryA sphere for a point charge or ball, a cylinder for a line or wire, a flat pillbox for a sheet or a conductor's face.
Split the surfaceSeparate the parts where E lies flat against the surface — those carry zero flux — from the parts it crosses squarely.
Equate and solveE × (the area it actually crosses) = q_enc/ε₀ gives E immediately.

Flux through one face of a cube

A charge of 6 μC sits at the centre of a cube of side 10 cm. Find the flux through one face. What if the charge is moved to a corner of the cube instead?

Φ_total = q/ε₀ = 6×10⁻⁶/8.85×10⁻¹²6.8 × 10⁵ N m² C⁻¹
One of six identical faces1.13 × 10⁵ N m² C⁻¹
At a corner: one-eighth enclosed, shared by the 3 far faces2.8 × 10⁴ N m² C⁻¹

Pro tip. The side length 10 cm never entered the working, because flux depends only on the charge enclosed. At a corner the three faces touching the charge get nothing at all — the field lies flat along them — so the eighth of the flux is split between the other three.

A point charge is placed just outside a closed surface. The net flux through that surface is

Zero, even though E is non-zero at every point of it
q/ε₀, since the field reaches the surface
Half of q/ε₀

Every line that enters the surface leaves it again, so the inward and outward flux cancel exactly. Flux counts enclosed charge; it says nothing about how strong the field on the surface is.

6Conductors in equilibrium

Leave a conductor alone and its free charges rearrange until nothing pushes them any more. So inside the material E = 0, the whole conductor sits at one potential, and every bit of excess charge lives on the surface. Just outside that surface the field is σ/ε₀ and perpendicular to it — twice an isolated sheet's field, because a conductor's charge layer has field on one side only. An empty cavity inside the metal has zero field too: that is electrostatic shielding.

Shielding works one way. Put a charge +q inside the cavity and it induces −q on the cavity wall and +q on the outer surface, so the outside world still sees its field.

Figure. The external field arrows stop dead at the surface and start again on the far side; nothing gets through the metal, and nothing reaches the cavity. At the surface itself the field arrives perpendicular, because any sideways component would slide the surface charge along.

How it works

The field cancels insideAny surviving field would move the free electrons, so equilibrium means it has already been cancelled.
Charge goes to the skinApply Gauss's law to a surface drawn just inside the metal: E = 0 on it, so the charge it encloses is zero.
The surface sets the fieldImmediately outside, E = σ/ε₀ normal to the surface — so a sharp point, where σ piles up, has the strongest field.

Inside and outside a charged sphere

A solid conducting sphere of radius 10 cm carries 2 μC. Find the field just outside it, the potential at its surface, and the field and potential at its centre.

σ = q/4πR² = 2×10⁻⁶/(4π × 0.01)1.59 × 10⁻⁵ C m⁻²
E just outside = σ/ε₀ = 1.59×10⁻⁵/8.85×10⁻¹²1.8 × 10⁶ N C⁻¹
V at the surface = kq/R = 9×10⁹ × 2×10⁻⁶/0.101.8 × 10⁵ V
At the centre: E = 0, and V is unchanged1.8 × 10⁵ V

Pro tip. The same 1.8 × 10⁶ N C⁻¹ comes out of kq/R², because outside the sphere the field is exactly that of a point charge at its centre. Inside, the field dies but the potential does not — it stays flat at its surface value, which is what E = −dV/dr demands.

A charge +q is placed inside the hollow cavity of an uncharged, isolated conductor. The field outside the conductor is

Zero — the conductor shields the outside from +q
The same as that of +q placed at the conductor's centre
Zero inside the metal and inside the cavity alike

The cavity charge induces −q on the cavity wall and +q on the outer surface, which spreads out symmetrically. Shielding protects the cavity from outside fields, not the outside from the cavity.

7Capacitance and stored energy

A capacitor is a pair of conductors holding equal and opposite charge, Q = CV. Its capacitance depends on geometry and the medium only — never on how much charge you have put on it. For parallel plates, E = σ/ε₀ between them and V = Ed, which gives C = ε₀A/d. Charging costs work because every further slice of charge must be pushed against the charge already sitting there, and that is why the energy is ½QV = ½CV² = Q²/2C rather than QV.

The energy is really stored in the field: u = ½ε₀E² per unit volume, and multiplying by the volume Ad between the plates returns ½CV² exactly.

Figure. The arrows between the plates are all the same length because the field there is uniform — each plate contributes σ/2ε₀ and the two add to σ/ε₀ inside while cancelling outside. Widen d with the charge fixed and the field is unchanged — σ has not changed — so it is V = Ed that grows. Leave the battery connected instead and V is what is held: charge drains off the plates and E falls as 1/d.

How it works

Geometry fixes CPlate area, separation and the medium set C = Kε₀A/d, and nothing else touches it.
Pick the energy formUse ½CV² when the voltage is held fixed and Q²/2C when the charge is trapped.
Track what is constantA connected battery fixes V; a disconnected one fixes Q. Settle that before changing anything.

Pulling the plates apart

A 10 μF capacitor is charged to 200 V. Find the charge and the stored energy. The battery is then disconnected and the plate separation is doubled — find the new energy and the work done in separating the plates.

Q = CV = 10×10⁻⁶ × 2002 × 10⁻³ C
U = ½CV² = ½ × 10⁻⁵ × 4×10⁴0.2 J
Battery off, d doubled: C′ = 5 μF, U′ = Q²/2C′0.4 J
Work done by the hand = U′ − U0.2 J

Pro tip. Leave the battery connected and the same experiment gives the opposite answer: V is then fixed, C halves, and U = ½CV² falls from 0.2 J to 0.1 J. Identical plates, identical motion, opposite sign — the question is always which quantity is held fixed.

8Dielectrics: what is held fixed

A dielectric carries no free charge, but its molecules polarise, and the sheet of bound charge that appears on each face opposes the applied field. The field inside falls to E/K and the capacitance rises to KC. What happens to everything else depends entirely on one question: is the battery still connected?

With the battery on, V is pinned, so Q and U both rise K-fold and the battery supplies the difference. With the battery off, Q is trapped, so V, E and U all fall by K. Either way the slab is pulled inward — the force on it never reverses.

Figure. The same capacitor twice, drawn with the field arrows to one scale and K taken as 4. Filling the gap makes the arrows a quarter as long, because the bound charge on the slab's faces — negative next to the positive plate — cancels three quarters of the applied field.

How it works

Ask what is fixedBattery connected pins V; battery removed pins Q. Everything else is then forced.
Scale C firstC always becomes KC. Propagate with Q = CV if V is fixed, or with U = Q²/2C if Q is fixed.

Inserting a slab of constant K
Quantity	Battery connected (V fixed)	Battery removed (Q fixed)
Capacitance C	× K	× K
Charge Q	× K	unchanged
Voltage V	unchanged	÷ K
Field E between plates	unchanged	÷ K
Stored energy U	× K	÷ K

The slab that lowers the energy

A 5 μF capacitor is charged to 100 V and the battery is then removed. A slab of dielectric constant K = 4 is slid in to fill the gap. Find the new voltage and the new stored energy.

Q = CV = 5×10⁻⁶ × 100 (fixed from here on)5 × 10⁻⁴ C
C′ = KC = 4 × 5 μF20 μF
V′ = Q/C′ = 5×10⁻⁴/2×10⁻⁵25 V
U′ = Q²/2C′, against U = ½CV² = 25 mJ before6.25 mJ

Pro tip. The energy fell to a quarter, and that missing 18.75 mJ is the work the field does dragging the slab in. Had the battery stayed connected instead, the same slab would have raised the energy four-fold to 100 mJ, with the battery supplying 150 mJ of which only half ends up stored.

A dielectric slab is inserted into a parallel-plate capacitor while the battery remains connected. The charge on the plates

Stays the same, since the plates are unchanged
Rises K-fold, the battery supplying the extra
Falls K-fold, since the field is weakened

With the battery connected V is fixed, and Q = CV with C multiplied by K. The charge can only stay fixed if the battery has been disconnected first.

9Series and parallel combinations

Capacitors in parallel share one potential difference and their charges add, so C_eq = ΣC. In series every capacitor carries the same charge — the inner plates only redistribute charge they already had — while the voltages add, so 1/C_eq = Σ1/C. A series combination is therefore always smaller than its smallest member, and within it the smallest capacitor takes the largest share of the voltage.

Redraw before you compute. Points joined by plain wire are one node, and a capacitor branch carries no steady current, so its plates simply sit at the potential difference of whatever two points they connect.

Figure. Series: same charge, voltages add → 1/C_eq = 1/2 + 1/3 so C_eq = 1.2 μF. Parallel: same V, charges add → C_eq = 5 μF. The bars carry the arithmetic the missing capacitor glyph cannot.

How it works

Label the nodesMerge every pair of points joined by bare wire, and redraw until the series and parallel groups are obvious.
Reduce inward firstCollapse the innermost group, then work outward until one C_eq remains.
Come back for the splitGet the total charge from Q = C_eq V, then share it: equal Q in series, equal V in parallel.

A mixed network

A 2 μF and a 3 μF capacitor are joined in series, and that pair is connected in parallel with a 6 μF capacitor across a 12 V supply. Find the equivalent capacitance and the voltage across the 2 μF capacitor.

1/C_s = 1/2 + 1/3 = 5/6C_s = 1.2 μF
C_eq = C_s + 6 (parallel)7.2 μF
Series branch: Q_s = C_s V = 1.2 × 1214.4 μC
V across the 2 μF = Q_s/C = 14.4/27.2 V

Pro tip. The 2 μF takes 7.2 V and the 3 μF takes the remaining 4.8 V — in series V is shared in the ratio 1/C, so the smaller capacitor is the one that breaks down first. In parallel the roles invert: the larger capacitor takes the larger charge.

Notes

Coulomb's law and the electric field: Two point charges exert a force F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}. The field of a point charge is E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}, pointing away from positive charge, and \vec{F}=q\vec{E}.
Electric potential and energy: The potential of a point charge is V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}, and \vec{E}=-\nabla V. Work to assemble two charges is U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r}; a dipole in a field has U=-\vec{p}\cdot\vec{E} and torque \tau=pE\sin\theta.
Gauss's law: The flux through any closed surface equals the enclosed charge over \varepsilon_0, \oint\vec{E}\cdot d\vec{A}=\dfrac{q_{enc}}{\varepsilon_0}. It quickly gives E=\dfrac{\lambda}{2\pi\varepsilon_0 r} for a line, \dfrac{\sigma}{2\varepsilon_0} for a sheet, and \dfrac{\sigma}{\varepsilon_0} just outside a conductor.
Capacitance and dielectrics: A capacitor stores charge Q=CV; a parallel-plate capacitor has C=\dfrac{\varepsilon_0 A}{d}, increased K-fold by a dielectric. Energy stored is U=\tfrac12 CV^2=\dfrac{Q^2}{2C}.
Combinations of capacitors: In parallel capacitances add, C_{eq}=\sum C_i; in series reciprocals add, \dfrac{1}{C_{eq}}=\sum\dfrac{1}{C_i}.

Formulas

Coulomb / field: F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2},\quad E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}
Potential and dipole: V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r},\quad U=-pE\cos\theta,\quad \tau=pE\sin\theta
Gauss's law: \oint\vec{E}\cdot d\vec{A}=\dfrac{q_{enc}}{\varepsilon_0}
Field of sheet / line: E_{sheet}=\dfrac{\sigma}{2\varepsilon_0},\quad E_{line}=\dfrac{\lambda}{2\pi\varepsilon_0 r}
Capacitor: C=\dfrac{K\varepsilon_0 A}{d},\quad U=\tfrac12 CV^2=\dfrac{Q^2}{2C}
Combinations: parallel C_{eq}=\sum C_i, series \dfrac{1}{C_{eq}}=\sum\dfrac{1}{C_i}

Exam traps & shortcuts

When a battery stays connected, V is fixed; when it is disconnected, Q is fixed. Decide which is constant first, then track how inserting a dielectric changes the others.
The field of an infinite sheet (\sigma/2\varepsilon_0) is independent of distance, so between two oppositely charged plates the fields add to \sigma/\varepsilon_0 and cancel outside.
For a charged conducting sphere, the potential and field outside are those of a point charge at the centre, while the field inside is zero and the potential is constant.

Reference tables

Every line here should be reconstructible from the concept above it, not merely recalled.

Formula sheet
Quantity	Relation	Watch for
Coulomb force	F = kq₁q₂/r², k = 9 × 10⁹	Convert cm to m before squaring
Field of a point charge	E = kq/r²	A vector — superpose, don't add magnitudes
Potential	V = kq/r	Scalar with sign; V ≠ 0 where E = 0
Field from potential	E = −dV/dr	E points from high V to low
Energy of a pair	U = kq₁q₂/r	One term per pair, n(n − 1)/2 of them
Dipole	τ = pE sin θ, U = −pE cos θ	Net force zero only in a uniform field
Gauss's law	Φ = q_enc/ε₀	Outside charges give zero net flux
Capacitance	C = Kε₀A/d	Geometry and medium only
Stored energy	U = ½CV² = Q²/2C	Choose by what is held fixed
Energy density	u = ½ε₀E²	Times the volume Ad gives ½CV²
Combinations	parallel ΣC, series Σ1/C	Series C is below the smallest member

Each of these is one application of Gauss's law, and each appears in exams as a one-line starting point rather than a derivation.

Standard fields worth knowing cold
Configuration	Field	Note
Point charge	kq/r²	Also the field outside any spherical charge
Infinite line, λ	λ/2πε₀r	Falls as 1/r, not 1/r²
Infinite sheet, σ	σ/2ε₀	Does not depend on distance at all
Between two opposite plates	σ/ε₀	The two sheets add inside, cancel outside
Just outside a conductor	σ/ε₀	Perpendicular to the surface
Conducting sphere, r < R	0	V stays flat at its surface value
Dipole, axial and equatorial	2kp/r³ and kp/r³	The equatorial field points the other way

Recap

Read only this the night before.

Superpose: Forces and fields are vectors added pair by pair; potentials and energies are scalars added with their signs.
Gauss: Only the charge inside counts. Symmetry chooses the surface, and the surface does the rest of the work.
Conductors: E = 0 inside, all charge on the skin, σ/ε₀ perpendicular just outside, one potential throughout.
Standard fields: Point kq/r², line λ/2πε₀r, sheet σ/2ε₀ with no r in it, between plates σ/ε₀.
Dipole: No net force in a uniform field, torque pE sin θ, energy −pE cos θ, its own field falling as 1/r³.
Capacitors: C is geometry alone. Ask what is held fixed — V with the battery on, Q with it off — before inserting anything.

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