JEE Mains Physics · Electrostatics

Charges Moving in Electric Fields

A charge in a field feels qE, so its acceleration is qE/m; in a uniform field it moves like a projectile, and where the field balances gravity or pulls towards a centre, the usual mechanics finishes the job.

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Why this matters

Twenty-five PYQs, twenty of them multiple choice, and four from 2026. Eleven send a charge through a uniform field: stopping distances, deflection between plates and the work the field does. Ten hold or swing a charged body against gravity: drops held still, pendulums in a field, and small oscillations between fixed charges. Four put a charge in orbit round a line charge or a charged cylinder.

Concept 1 of 3: Motion in a uniform field

A uniform field gives a constant force qE, so a charge moves exactly as a ball does under gravity, with qE/m in place of g. A charge entering between plates along them keeps its speed along the plates and gains speed across them: its path is a parabola. A positive charge speeds up along the field, a negative one against it.

Definition

$a = \dfrac{qE}{m}$ , constant. Use the equations of uniform acceleration.
Fired against the field, a charge stops after $s = \dfrac{mu^{2}}{2qE}$ .
Between plates of length L, entering along them at speed v: time $t = L/v$ , sideways shift $y = \dfrac{qEL^{2}}{2mv^{2}}$ , exit angle $\tan\theta = \dfrac{qEL}{mv^{2}}$ .
The velocity part perpendicular to the field never changes.
Work done by the field: $qE \times$ (displacement along E). Between plates a distance d apart: $V = Ed$ .
A magnetic force is always perpendicular to the velocity, so it does no work: any change of speed comes from the electric field.

Deflection between plates

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

Worked example

Particles with

q/m = 5 \times 10^{7}

C/kg enter at

2 \times 10^{5}

m/s along plates 4 cm long. The field between the plates is 1000 V/m. Find the sideways shift and the exit angle.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 24 Jan 2023 · Q24Moderate

Example 1 · Electrostatics · Charges Moving in Electric Fields

A stream of a positively charged particles having

\frac{q}{\text{ }m}= 2 \times10^{11}\frac{C}{kg}

and velocity

{\overrightarrow{v}}_{0}= 3 \times\frac{10^{7}\widehat{im}}{s}

is deflected by an electric field

\frac{1.8\widehat{\jmath}kV}{m}

. The electric field exists in a region of

10\text{ }cm

along

x

direction. Due to the electric field, the deflection of the charge particles in the

y

direction___ is

mm

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.

Drill 10 more on motion in a uniform field

Concept 2 of 3: Holding and swinging charged bodies

A drop hangs still when the upward electric force equals its weight. A charged bob in a sideways field settles at an angle, as if gravity were tilted and made stronger. A charge between two fixed equal charges is pushed back towards the middle when moved along the line, so it oscillates, and the restoring force sets the period just as a spring constant would.

Definition

Drop held still: $qE = mg$ , so the number of extra electrons is $n = \dfrac{mg}{eE}$ . A drop's mass is $\tfrac{4}{3}\pi r^{3}\rho$ .
A positive drop needs E pointing up: the lower plate at the higher potential.
Bob in a horizontal field: $\tan\theta = \dfrac{qE}{mg}$ , $T = \sqrt{(mg)^{2} + (qE)^{2}}$ . A pendulum then swings with $g_{\text{eff}} = \sqrt{g^{2} + (qE/m)^{2}}$ .
Energy: the work by the field plus the work by gravity equals the gain in kinetic energy.
Charge $q_0$ between two fixed equal charges Q a distance 2a apart, moved along the line: $\omega^{2} = \dfrac{4kQq_0}{ma^{3}}$ . Moved across the line, when the fixed charges attract it: $\omega^{2} = \dfrac{2kQq_0}{ma^{3}}$ .

Balance and small oscillations

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

Worked example

An oil drop of mass

3.2 \times 10^{-15}

kg is held still between horizontal plates 1 cm apart, with 500 V across them. How many extra electrons does it carry, and which plate is positive? (g = 10 m/s², e = 1.6 × 10⁻¹⁹ C)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 27 July 2022 · Q10Moderate

Example 2 · Electrostatics · Charges Moving in Electric Fields

Two identical positive charges

Q

each are fixed at a distance of '

2a

' apart from each other. Another point charge

q_{0}

with mass '

m

' is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge

q_{0}

executes SHM. The time period of oscillation of charge

q_{0}

will be:

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.

Drill 9 more on holding and swinging charged bodies

Concept 3 of 3: Orbits round a line charge

A long line charge pulls an opposite charge towards it with a force that falls as 1/r. For a circular orbit, that force supplies mv²/r, and the r cancels: every orbit has the same speed. A wider orbit is longer to go round at that speed, so its period grows in proportion to the radius.

Definition

Charge −q of mass m circling a line charge +λ at radius r: $q\dfrac{2k\lambda}{r} = \dfrac{mv^{2}}{r}$ , so $mv^{2} = 2k\lambda q$ .
The speed and the kinetic energy (kλq) do not depend on r.
Period $T = \dfrac{2\pi r}{v} = 2\pi r\sqrt{\dfrac{m}{2k\lambda q}}$ , so $T \propto r$ .
Outside a long charged cylinder of radius R and density ρ, the field is that of a line charge $\lambda = \rho\pi R^{2}$ .
Compare a point charge Q: $v^{2} = \dfrac{kQq}{mr}$ and $T \propto r^{3/2}$ .

Orbit round a line charge

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

Worked example

A particle of mass

3.6 \times 10^{-6}

kg and charge

-1\ \mu\text{C}

circles a long line with

\lambda = +2\ \mu\text{C/m}

at radius 0.5 m. Find its speed and period.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 30 Jan 2024 · Q103Moderate

Example 3 · Electrostatics · Charges Moving in Electric Fields

A particle of charge '

-q

' and mass '

m

' moves in a circle of radius '

r

' around an infinitely long line charge of linear density '

+\lambda

'. Then time period will be given as: (Consider k as Coulomb's constant)

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.

Drill 3 more on orbits round a line charge

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

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

Watch out for (9)

Convert the field's units
→ Motion in a uniform field
Only one velocity part changes
→ Motion in a uniform field
Electrons go against the field
→ Motion in a uniform field
Which way must E point?
→ Holding and swinging charged bodies
Densities in SI units
→ Holding and swinging charged bodies
Not every direction is stable
→ Holding and swinging charged bodies
A line charge's field is 2kλ/r
→ Orbits round a line charge
Same speed, different period
→ Orbits round a line charge
A cylinder acts through its charge per length
→ Orbits round a line charge

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