MHT-CET Physics · Electrostatics

Coulomb's Law and Electric Field

Two point charges push or pull along the line joining them with F = kq₁q₂/r², weakened K times in a medium; the electric field is that force per unit positive charge, and fields from several charges add as vectors.

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Why this matters

27 PYQs, eight of them HARD, and every HARD one adds the fields or forces of two or more charges. Four shapes repeat: a force ratio after the charges or the distance change, the point on a line where the field or force vanishes, a symmetric arrangement (square, triangle, hexagon) where most charges cancel, and a charge accelerated or held still by a field.

Concept 1 of 4: Coulomb's Law and Force Ratios

Force grows with each charge and falls with the square of the distance. Almost every question here changes one of those three things and asks for the new force, so the working is a ratio: write the new force over the old one and let k cancel.

Definition

$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2} = \dfrac{k q_1 q_2}{r^2}$ , with $k = 9 \times 10^9\ \text{N m}^2\text{C}^{-2}$ . Like charges repel, unlike attract.
In a medium of dielectric constant $K$ : $F_{\text{medium}} = \dfrac{F_{\text{air}}}{K}$ . The same force at distance $y$ in the medium as at $x$ in air needs $\dfrac{y}{x} = \dfrac{1}{\sqrt{K}}$ .
Moving charge from $+Q$ to $-Q$ shrinks BOTH magnitudes: $+4q, -4q$ with 25% moved become $+3q, -3q$ .
Touching a neutral identical ball to a charged one halves that charge; two identical conductors touched share the total equally.
Square of side $a$ : adjacent corners are $a$ apart, diagonal corners $\sqrt{2}a$ , so $F_{\text{adjacent}} : F_{\text{diagonal}} = 2 : 1$ .

Coulomb's law

F = \frac{1}{4\pi\varepsilon_0 K}\,\frac{q_1 q_2}{r^2}

Worked example

Charges of

2\,\mu

C and

8\,\mu

C are 30 cm apart in air. Find the force, and the force when both are placed in a liquid of

K = 4

at the same separation.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 2 · Q36Moderate

Example 1 · Electrostatics · Coulomb's Law and Electric Field

Two point charges (A and B)

+4q

and

-4q

are placed along a line separated by a distance 'r'. Force acting between them is F. If 25% of charge from point A is transferred to that at point B, the force between the charges now becomes

Adding the moved charge to one side only

When charge moves from the positive body to the negative one, the positive loses it AND the negative's magnitude drops by the same amount.

+4q, -4q

with

q

moved is

+3q, -3q

, so the force is

\frac{9}{16}F

, not

\frac{12}{16}F

Drill 7 more on coulomb's law and force ratios

Concept 2 of 4: Where the Field or Force Is Zero on a Line

Two fields cancel only where they point opposite ways AND are equally strong. For like charges that happens between them, nearer the smaller charge. For unlike charges the fields point the same way everywhere between them, so the cancelling point is outside, beyond the SMALLER charge, where its nearness makes up for its size.

Definition

Like charges $q_1, q_2$ a distance $L$ apart: null point between them at $x = \dfrac{L\sqrt{q_1}}{\sqrt{q_1} + \sqrt{q_2}}$ from $q_1$ .
Unlike charges with $|q_1| > |q_2|$ : null point beyond $q_2$ , at distance $x$ from it where $\dfrac{|q_1|}{(L + x)^2} = \dfrac{|q_2|}{x^2}$ , i.e. $x = \dfrac{L}{\sqrt{|q_1|/|q_2|} - 1}$ .
A third charge placed there feels no force, whatever its sign — the same condition gives the proton-equilibrium questions.
Force on an END charge zero ( $Q$ , $q$ , $Q'$ on a line): set the two forces on it equal and opposite; the middle charge comes out with the opposite sign.
Read the question: 'from the origin' adds $L$ to the distance measured from the second charge.

Unlike charges: distance beyond the smaller one

\frac{|q_1|}{(L+x)^2} = \frac{|q_2|}{x^2} \;\Rightarrow\; x = \frac{L}{\sqrt{|q_1|/|q_2|} - 1}

Worked example

+9q

is at

x = 0

and

-4q

x = L

. Where on the x-axis is the field zero?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q37Moderate

Example 2 · Electrostatics · Coulomb's Law and Electric Field

Two-point charges

+8q

and

-2q

are located at

X=0

(origin) and

X=L

respectively. The net electric field due to these two charges is zero at point P on X-axis. The location of point P from the origin is

Looking between two unlike charges

Between

+8q

and

-2q

both fields point toward

-2q

and can only add. The answer

\frac{L}{4}

sits in the options for students who looked there; the real point is

2L

, beyond the smaller charge.

Drill 6 more on where the field or force is zero on a line

Concept 3 of 4: Superposition: Symmetric Arrangements and Field Lines

At the centre of a regular shape, equal charges on opposite vertices cancel each other's fields. Strip out every cancelling pair and only one or two charges are left to compute. Potential does not cancel this way — it is a plain sum — so a point can have zero field and non-zero potential, or the reverse.

Definition

Opposite vertices of a hexagon (or square) with EQUAL charges cancel at the centre; with opposite charges they ADD.
Equal charges at every vertex of a regular polygon: $E = 0$ at the centre, but $V = \dfrac{nkq}{a}\ne 0$ .
$2q, -q, -q$ on an equilateral triangle: $V = 0$ at the centre (charges sum to zero), $E \ne 0$ .
Uniform semicircular arc of charge density $\lambda$ , radius $r$ : $E = \dfrac{2k\lambda}{r} = \dfrac{\lambda}{2\pi\varepsilon_0 r}$ at the centre, and $V = \dfrac{\lambda}{4\varepsilon_0}$ .
Field lines: start on + and end on − charges; never intersect (one direction per point); crowded where the field is strong; meet a conductor at right angles and do not pass through it.

Field of a point charge

\vec E = \frac{kq}{r^2}\,\hat r, \qquad \vec E_{\text{net}} = \sum_i \vec E_i

Worked example

Equal charges

+q

sit at the four corners of a square of side

a

. Find the field and the potential at the centre.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift II · Q8Moderate

Example 3 · Electrostatics · Coulomb's Law and Electric Field

The point charges

+q, -q, -q, +q, +Q

and

-q

are placed at the vertices of a regular hexagon ABCDEF as shown in figure. The electric field at the centre of hexagon '

O

' due to the five charges at

A,B,C,D

and F is twice the electric field at centre ' O ' due to charge +Q at E alone. The value of Q is

Zero field does not mean zero potential

At the centre of

2q, -q, -q

the potential is zero because the charges sum to zero, but the fields do not cancel. The paper offers all four combinations; decide field and potential separately.

Drill 7 more on superposition: symmetric arrangements and field lines

Concept 4 of 4: A Charge Moving in, or Held by, a Uniform Field

A uniform field puts a constant force

qE

on a charge, so it moves like a body under constant gravity: uniform acceleration

qE/m

. Balance that force against the weight and the charge floats; add it to or subtract it from gravity and a pendulum swings faster or slower.

Definition

$F = qE$ , $a = \dfrac{qE}{m}$ . From rest: $v = \dfrac{qEt}{m}$ , $\text{KE} = \dfrac{q^2E^2t^2}{2m}$ ; after a distance $d$ : $\text{KE} = qEd$ .
Held still between plates: $qE = mg$ with $E = \dfrac{V}{d}$ , so $m = \dfrac{qV}{gd}$ .
Charged pendulum bob in a vertical field: $T = 2\pi\sqrt{\dfrac{L}{g \pm qE/m}}$ — plus when the electric force is downward, minus when upward.

Motion in a uniform field

a = \frac{qE}{m}, \qquad \text{KE} = \frac{q^2E^2t^2}{2m}, \qquad qE = mg \text{ (balance)}

Worked example

An oil drop carrying

3.2 \times 10^{-19}

C floats between horizontal plates 5 mm apart with 400 V across them. Find its mass (

g = 10\ \text{m s}^{-2}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift II · Q34Easy

Example 4 · Electrostatics · Coulomb's Law and Electric Field

A charged particle of mass '

m

' and charge '

q

' is at rest. It is accelerated in a uniform electric field of intensity '

E

' for time '

t

'. The kinetic energy of the particles after time

t

Using the electron's mass for a heavier particle

'Charge

3e

, mass

2m

' gives

a = \frac{3eE}{2m}

. The options include the same expression with the charge and mass swapped; divide the particle's own charge by its own mass.

Drill 3 more on a charge moving in, or held by, a uniform field

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 (4)

Coulomb's Law and Force Ratios
Coulomb's law
$F = \frac{1}{4\pi\varepsilon_0 K}\,\frac{q_1 q_2}{r^2}$
Where the Field or Force Is Zero on a Line
Unlike charges: distance beyond the smaller one
$\frac{|q_1|}{(L+x)^2} = \frac{|q_2|}{x^2} \;\Rightarrow\; x = \frac{L}{\sqrt{|q_1|/|q_2|} - 1}$
Superposition: Symmetric Arrangements and Field Lines
Field of a point charge
$\vec E = \frac{kq}{r^2}\,\hat r, \qquad \vec E_{\text{net}} = \sum_i \vec E_i$
A Charge Moving in, or Held by, a Uniform Field
Motion in a uniform field
$a = \frac{qE}{m}, \qquad \text{KE} = \frac{q^2E^2t^2}{2m}, \qquad qE = mg \text{ (balance)}$

Watch out for (4)

Adding the moved charge to one side only
→ Coulomb's Law and Force Ratios
Looking between two unlike charges
→ Where the Field or Force Is Zero on a Line
Zero field does not mean zero potential
→ Superposition: Symmetric Arrangements and Field Lines
Using the electron's mass for a heavier particle
→ A Charge Moving in, or Held by, a Uniform Field

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