Electrostatics — JEE Mains Physics Formulas and Traps

Coulomb's Law and Equilibrium of Charges

Coulomb's law and adding forces

Coulomb's law in vector form

\vec F_{21} = \frac{kq_1q_2}{|\vec r_2 - \vec r_1|^{3}}\,(\vec r_2 - \vec r_1), \qquad F_{\text{medium}} = \frac{F}{K}

Sharing charge by contact

Sharing and the largest force

q' = \frac{q_1 + q_2}{2}, \qquad F \propto q(Q - q) \ \text{is largest at}\ q = \frac{Q}{2}

Charged bodies at rest

Hanging balls and the liquid condition

\tan\theta = \frac{F_e}{mg}, \qquad K = \frac{\rho}{\rho - \sigma}

Common traps

The vector points from the source to the target

The force on q₂ uses r₂ − r₁. Using r₁ − r₂ gives the force on q₁, which is the same size and points the other way. That reversed vector is usually among the options.

K divides the force; √K scales the distance

A medium cuts the force by K. To get the same force in vacuum, the charges must be √K times farther apart, not K times.

Forces add as vectors

Two forces at right angles of 3 N and 4 N give 5 N, not 7 N. Draw each force on the charge first, then add.

Add the signs before halving

Spheres with +6 and −2 share +4, so each gets +2. Halving 6 + 2 = 8 gives the wrong charge and the wrong direction of force.

Order of contact matters

A sphere carries what it picked up into the next contact. Touching A then B gives a different result from B then A whenever A and B differ.

Equal sharing needs identical spheres

Two spheres of different radii end at a common potential, so the larger one takes more charge, in proportion to its radius.

Angle from the vertical, not between the threads

If the threads make an angle 2θ with each other, each makes θ with the vertical. Use θ in tan θ = F/mg.

A liquid changes two forces

The electric force falls by K and the weight falls by the buoyancy. Changing only one of them gives the wrong K.

The net force on a ball at rest is zero

When a question asks for the force on a ball in equilibrium, it means the electric force. The net force is zero by definition.

Electric Field of Charges, Rods, Rings and Sheets

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Field of point charges and null points

Superposition and the null point of like charges

\vec E = \sum_i \frac{kq_i}{r_i^{2}}\,\hat r_i, \qquad x_{\text{null}} = \frac{r}{1 + \sqrt{q_2/q_1}}

Field of rods, arcs, rings and discs

Arc at its centre and ring on its axis

E_{\text{arc}} = \frac{2k\lambda}{R}\sin\frac{\varphi}{2}, \qquad E_{\text{axis}} = \frac{kQz}{(z^{2} + R^{2})^{3/2}}

Infinite sheets and line charges

Sheet and line

E_{\text{sheet}} = \frac{\sigma}{2\varepsilon_0}, \qquad E_{\text{line}} = \frac{\lambda}{2\pi\varepsilon_0 r}

Common traps

Unlike charges: never between them

Between +q and −q both fields point towards −q, so they add. The null point is outside, beyond the smaller charge. A root between the charges must be rejected.

Add arrows, not numbers

Fields from charges at different places point in different directions. Adding their sizes is right only when they lie along one line and point the same way.

Field lines never cross

At any point the field has one direction, so two lines cannot cross there. Electrostatic field lines also never close on themselves.

The arc formula uses half the angle

For an arc spanning φ, the factor is sin(φ/2). A half ring spans 180°, so the factor is sin 90° = 1, not sin 180° = 0.

λ comes from the arc's own length

A half ring of charge Q has λ = Q/πR, not Q/2πR. Use the length that actually carries the charge.

Let symmetry cancel first

On a ring's axis only the axial parts survive. Adding the full kq/r² of every piece overcounts by the factor z/r that the cancelling removes.

One sheet is σ/2ε₀

σ/ε₀ belongs to the surface of a conductor, or to the region between two opposite sheets. A single sheet gives half of that.

A sheet's field does not fall with distance

Only a line charge (1/r) or a point charge (1/r²) weakens with distance. An infinite sheet gives the same field everywhere on one side.

Conducting plates rearrange their charge

Charge given to one plate does not stay on one face. The outer faces always carry equal charges, each half of the total on both plates.

Electric Flux and Gauss's Law

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Flux through a surface and through a cube

Flux and Gauss's law

\phi = \vec E \cdot \vec A, \qquad \phi_{\text{net}} = \oint \vec E \cdot d\vec A = \frac{q_{\text{enc}}}{\varepsilon_0}

Gauss's law for spheres, shells and cylinders

Fields from Gauss's law

E_{\text{sphere, in}} = \frac{\rho r}{3\varepsilon_0}, \qquad E_{\text{cylinder, in}} = \frac{\rho r}{2\varepsilon_0}, \qquad E_{\text{out}} = \frac{kQ}{r^{2}}

Enclosed charge and sharing flux by symmetry

Charge q placed at	Flux through the whole cube	Flux through single faces	How to see it
Centre of the cube	$q/\varepsilon_0$	$q/6\varepsilon_0$ through each face	Six identical faces share it equally
Centre of one face	$q/2\varepsilon_0$	Zero through the face it sits on	A second cube on the other side takes the other half
Middle of an edge	$q/4\varepsilon_0$	Zero through the two faces meeting at that edge	Four cubes share the edge
A corner	$q/8\varepsilon_0$	$q/24\varepsilon_0$ through each of the three far faces; zero through the three faces at the corner	Eight cubes meet at the corner The far faces get q/24ε₀ each, not q/8ε₀: three faces share the cube's eighth.
On the axis of a square of side a, at distance a/2	$q/\varepsilon_0$ through the imagined cube of side a	$q/6\varepsilon_0$ through the square	The square is one face of a cube centred on q
Outside the cube	Zero	Inward on some faces, outward on others	What enters also leaves

Stack copies of the box round the charge until it sits at a centre of symmetry, then divide.

Common traps

Name the normal, not the plane

A surface 'in the yz-plane' or 'parallel to the yz-plane' has its normal along x. Taking the field's y-part for it gives the wrong flux.

Only faces facing the field count

When E points along x, the four faces parallel to x carry no flux. Do not multiply the field by the total area of the cube.

Inward flux is negative

Net flux is flux out minus flux in. Adding the two sizes gives a charge that is far too large.

A charge on the surface counts in part

A charge on a flat face is half inside the box. Counting all of it doubles the flux.

Outside charges do not change the net flux

They change the field at every point of the surface, so the flux through one face can change. The net flux through the closed surface stays q_enc/ε₀.

The flat face of a hemisphere

With the charge at the centre of the flat face, the field runs along that face and its flux is zero. The q/2ε₀ goes through the curved part.

Inside a solid sphere the field grows

For a uniformly charged insulating sphere, E rises linearly from zero at the centre to its peak at the surface. Only outside does it fall as 1/r².

'On the surface' of a shell

Just outside a shell the field is σ/ε₀; just inside it is zero. Read which side the question means.

Varying density needs an integral

When ρ changes with r, the enclosed charge is ∫ρ 4πr² dr. Multiplying ρ(r) by the volume gives the wrong power of r.

Electric Potential, Conductors and Work

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Potential of charges, rings and shells

Point charges and a shell

V = \sum_i \frac{kq_i}{r_i}, \qquad V_{\text{shell}}(r) = \begin{cases} kQ/R, & r \le R \\ kQ/r, & r \ge R \end{cases}

Drops that merge and spheres joined by a wire

Drops and joined spheres

V_{\text{big}} = n^{2/3}V, \qquad \frac{q_1}{q_2} = \frac{R_1}{R_2}, \qquad \frac{\sigma_1}{\sigma_2} = \frac{E_1}{E_2} = \frac{R_2}{R_1}

Field from potential, work and potential energy

Gradient, work and energy

\vec E = -\nabla V, \qquad W_{\text{ext}} = q(V_B - V_A), \qquad U = \sum_{i<j} \frac{kq_iq_j}{r_{ij}}

Common traps

Potential has no direction

Add the potentials of the charges as signed numbers. Resolving them into components, as for fields, is an error.

Inside a charged shell V is not zero

The field inside is zero, so the potential is constant there, equal to its value at the surface.

Outer shells use their own radius

At a point inside a shell, that shell contributes kq/R, not kq/r. Using r for every shell overcounts the outer ones.

Volume is conserved, not radius

n drops make a drop of radius n^(1/3) r, not n r. Using nr makes the potential stay the same.

Joined spheres share potential, not charge

Charge splits in proportion to radius. Splitting it equally is right only for identical spheres.

Less charge, stronger field

The smaller of two joined spheres holds less charge but has the larger surface density and surface field.

Work by the field or by an agent

The two differ by a sign. The agent's work is q(V_B − V_A); the field's work is the negative of that. Read which one is asked.

Keep the minus sign in E = −∇V

The field points towards falling potential. Dropping the sign reverses every component.

Count each pair once

For three charges there are three pairs, not six. Summing over i and j without care doubles the energy.

Electric Dipoles: Field, Torque and Energy

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Dipole moment, field and potential

Short dipole

E_{\text{axial}} = \frac{2kp}{r^{3}}, \qquad E_{\text{equatorial}} = \frac{kp}{r^{3}}, \qquad V = \frac{kp\cos\theta}{r^{2}}

Dipole in an external field: torque, energy and work

Torque, energy and work

\vec\tau = \vec p \times \vec E, \qquad U = -\vec p \cdot \vec E, \qquad W = pE(\cos\theta_1 - \cos\theta_2)

Common traps

Axial is twice equatorial

At the same distance, the axial field is 2kp/r³ and the equatorial field kp/r³. The equatorial field also points opposite to p.

p points from − to +

The dipole moment runs from the negative charge to the positive one, the reverse of the field lines between them.

Net charge makes p depend on the origin

When the charges do not add to zero, Σqr changes with the origin. Use the origin the question names.

Half a turn costs 2pE

From aligned to reversed, the energy goes from −pE to +pE, so the work is 2pE. Writing pE counts only the quarter turn.

The minimum energy is negative

U = −pE cos θ is −pE when aligned. A dipole set along the field is in stable equilibrium, at its lowest energy, not zero.

Unequal masses turn about the centre of mass

In a uniform field the net force is zero, so the centre of mass stays put. Take the moment of inertia about it, not about the midpoint.

Charges Moving in Electric Fields

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Motion in a uniform field

Deflection between plates

y = \frac{qEL^{2}}{2mv^{2}}, \qquad \tan\theta = \frac{qEL}{mv^{2}}

Holding and swinging charged bodies

Balance and small oscillations

qE = mg, \qquad \omega^{2} = \frac{4kQq_0}{ma^{3}}

Orbits round a line charge

Orbit round a line charge

mv^{2} = 2k\lambda q, \qquad T = 2\pi r\sqrt{\frac{m}{2k\lambda q}}

Common traps

Convert the field's units

A field in V/cm is 100 times larger in V/m, and kV/m is 1000 V/m. Most wrong options here come from a missed power of ten.

Only one velocity part changes

The part along the plates stays the same. Using the full speed for the time between the plates is right only when the charge enters along them.

Electrons go against the field

A negative charge accelerates opposite to E. Check the sign before deciding which plate it bends towards.

Which way must E point?

The force on a positive drop is along E, on a negative drop against it. The force must point up in both cases, so the field's direction depends on the sign.

Densities in SI units

A density in g/cm³ is 1000 times larger in kg/m³. Find the drop's mass in kilograms before comparing qE with mg.

Not every direction is stable

Between two like charges, a charge moved along the line is pushed back, but one moved across the line is pushed further away. Check that the force restores before writing ω².

A line charge's field is 2kλ/r

Using kλ/r², as for a point charge, makes the speed depend on r. The 1/r field is what makes every orbit's speed the same.

Same speed, different period

The speed does not depend on r, but the period does: it grows in proportion to r.

A cylinder acts through its charge per length

Outside a charged cylinder, use λ = ρπR², the charge on one metre of it. Putting ρ in place of λ gives the wrong units.

Capacitance and Capacitor Combinations

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Capacitance of plates and spheres

Plates and spheres

C = \frac{K\varepsilon_0 A}{d}, \qquad C_{\text{sphere}} = 4\pi\varepsilon_0 R, \qquad C_{\text{spherical}} = \frac{4\pi\varepsilon_0 R_1R_2}{R_2 - R_1}

Series, parallel and networks

Series and parallel

\frac{1}{C_{s}} = \sum_i \frac{1}{C_i}, \qquad C_{p} = \sum_i C_i

Common traps

C does not depend on Q or V

Charging a capacitor further raises Q and V together. Only the geometry and the medium change C.

Hollow or solid, the same C

A conductor's charge sits on its surface, so a hollow sphere and a solid one of the same radius have the same capacitance.

Unequal plate charges

Only half the difference of the two charges sits on the inner faces. Using q₁ alone for Q in V = Q/C gives a voltage that is too large.

Decide by the nodes, not by the drawing

Two capacitors drawn one after the other are in parallel if both connect the same two nodes. Label the nodes before calling anything series.

A shorted capacitor stores nothing

If a wire joins a capacitor's two plates, both are on one node, so it holds no charge and drops out of the network.

Steady state means no capacitor current

Once the currents settle, a capacitor branch carries none. Remove the capacitors, find the currents, then read each capacitor's voltage from its two nodes.

Dielectric Slabs and Partly Filled Capacitors

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Slabs and layers across the gap

Slab across the gap

C = \frac{\varepsilon_0 A}{d - t + \dfrac{t}{K}}, \qquad \frac{1}{C} = \sum_i \frac{t_i}{K_i\varepsilon_0 A}

Dielectrics side by side and stair plates

Dielectrics side by side

C = \frac{\varepsilon_0}{d}\sum_i K_iA_i

Common traps

Across the gap means series

A slab covering the whole plate area splits the gap into layers. Add their reciprocals; adding Kε₀A/d terms treats them as side by side.

A metal sheet removes its thickness

There is no field inside the metal, so the gap simply shrinks by t. Using t/K with some large K only approximates this.

Thickness matters, position does not

Moving a slab closer to one plate leaves C unchanged. Only its thickness and K enter the formula.

Side by side means parallel

Dielectrics that each span the whole gap share one voltage. Add their capacitances; adding reciprocals treats them as layers.

Each part uses its own area

A dielectric over half the plates contributes Kε₀(A/2)/d, not Kε₀A/d. Using the full area for every part counts the plates twice.

Read the boundary

If the surface between two dielectrics is parallel to the plates, they are in series; if it is perpendicular, they are in parallel. Look at the figure before choosing.

Energy Stored and Charge Sharing in Capacitors

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Stored energy: battery connected or removed

Stored energy

U = \tfrac{1}{2}CV^{2} = \frac{Q^{2}}{2C}, \qquad V\ \text{fixed: } U \to KU, \qquad Q\ \text{fixed: } U \to \frac{U}{K}

Charge sharing and common potential

Common potential and energy lost

V = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}, \qquad \Delta U = \frac{C_1C_2}{2(C_1 + C_2)}(V_1 - V_2)^{2}

Common traps

Decide what is fixed first

Battery connected: V fixed, use ½CV². Battery removed: Q fixed, use Q²/2C. Using the other form gives the change in the wrong direction.

The battery supplies twice the gain

With V fixed, the battery's work (K − 1)CV² is twice the rise in stored energy. The other half goes into pulling the slab in.

Percentages are squared

Energy goes as Q² or V². A 10% rise in charge is a 21% rise in energy, not 10%.

Unlike plates subtract

Joining positive to negative cancels part of the charge first. Adding C₁V₁ and C₂V₂ here gives a common voltage that is far too high.

Charge is kept, energy is not

The final energy is always less, unless the two voltages were already equal. Setting the energies equal before and after gives a wrong voltage.

Each capacitor's share is CV

After joining, each capacitor holds its own C times the common V. The charge splits equally only when the capacitances are equal.

Electrostatics formulas

Coulomb's Law and Equilibrium of Charges

Coulomb's law and adding forces

Sharing charge by contact

Charged bodies at rest

Common traps

Electric Field of Charges, Rods, Rings and Sheets

Field of point charges and null points

Field of rods, arcs, rings and discs

Infinite sheets and line charges

Common traps

Electric Flux and Gauss's Law

Flux through a surface and through a cube

Gauss's law for spheres, shells and cylinders

Enclosed charge and sharing flux by symmetry

Common traps

Electric Potential, Conductors and Work

Potential of charges, rings and shells

Drops that merge and spheres joined by a wire

Field from potential, work and potential energy

Common traps

Electric Dipoles: Field, Torque and Energy

Dipole moment, field and potential

Dipole in an external field: torque, energy and work

Common traps

Charges Moving in Electric Fields

Motion in a uniform field

Holding and swinging charged bodies

Orbits round a line charge

Common traps

Capacitance and Capacitor Combinations

Capacitance of plates and spheres

Series, parallel and networks

Common traps

Dielectric Slabs and Partly Filled Capacitors

Slabs and layers across the gap

Dielectrics side by side and stair plates

Common traps

Energy Stored and Charge Sharing in Capacitors

Stored energy: battery connected or removed

Charge sharing and common potential

Common traps

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