Formula of Electric Field: E = F/q and E = kq/r²

The formula of electric field is $\vec{E} = \frac{\vec{F}}{q}$ : place a small positive test charge q at a point, measure the electric force F on it, and divide. The result, called the electric field intensity, is the force per coulomb at that point and does not depend on which test charge you used.

Around a single point charge Q the magnitude of the field is $E = \frac{kQ}{r^2}$ . Here k is Coulomb's constant, roughly nine billion (8.99 × 10⁹) newton square metres per coulomb squared. Both forms come up in Class 12 electrostatics, AP Physics 2 and JEE, and they connect straight to Coulomb's law.

Quick facts

Definition
$E = \frac{F}{q}$
Point charge
$E = \frac{kQ}{r^2}$
Between parallel plates
$E = \frac{V}{d}$
SI unit
N/C, which equals V/m
Dimensional formula
[M¹ L¹ T⁻³ I⁻¹]
Type
Vector: away from positive charges, toward negative ones

Symbols in the formula of electric field and its magnitude

What each letter stands for
Symbol	Meaning	SI unit
$E$	Electric field intensity	N/C
$F$	Electric force on the test charge	N
$q$	Test charge, small and positive	C
$Q$	Source charge creating the field	C
$r$	Distance from the source charge	m

The test charge must be small so that it does not push the source charges around. For a point source, combine $F = \frac{kQq}{r^2}$ with the definition and q cancels:

$E = \frac{F}{q} = \frac{kQ}{r^2} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$

This is the magnitude. The direction is radially outward if Q is positive and inward if it is negative.

Electric field intensity in common setups

Field formulas for standard charge shapes
Source	Magnitude of E
Point charge Q	$\frac{kQ}{r^2}$
Long line, charge per length λ	$\frac{\lambda}{2\pi\varepsilon_0 r}$
Large flat sheet, charge per area σ	$\frac{\sigma}{2\varepsilon_0}$
Just outside a charged conductor	$\frac{\sigma}{\varepsilon_0}$
Between parallel plates with voltage V, gap d	$\frac{V}{d}$
Axis of a dipole, far away	$\frac{2kp}{r^3}$

The line and sheet results use the linear charge density and surface charge density. All of them follow from Gauss's law, which Class 12 students derive.

Worked examples

Example 1: from force and charge

A 4.8 μC charge experiences 1.2 × 10⁻² N at a point. Find the field there.

Convert the charge: 4.8 μC is 4.8 × 10⁻⁶ C.

$E = \frac{1.2\times10^{-2}}{4.8\times10^{-6}} = 2.5\times10^{3}\ \text{N/C}$

Example 2: a point charge in cm and nC

Find the field 25 cm from a 5.0 nC charge.

Here Q = 5.0 × 10⁻⁹ C and the distance is 0.25 m, so r² = 0.0625 m².

$E = \frac{(8.99\times10^{9})(5.0\times10^{-9})}{0.0625} \approx 7.2\times10^{2}\ \text{N/C}$

It points away from the charge, since Q is positive.

Example 3: parallel plates

Two plates 4.0 mm apart are held at a potential difference of 120 V. Find the uniform field between them.

$E = \frac{V}{d} = \frac{120}{4.0\times10^{-3}} = 3.0\times10^{4}\ \text{V/m}$

Units: N/C and V/m are the same

From the definition, field is newtons per coulomb. From the plate formula it is volts per metre. They agree because 1 V = 1 J/C = 1 N m/C, so 1 V/m = 1 N/C. Breaking the newton into base units gives kg m s⁻² and the coulomb gives A s, so

$[E] = \frac{[M L T^{-2}]}{[I T]} = [M L T^{-3} I^{-1}]$

For a step by step version with a check on $E = V/d$ , open the dimensional formula of electric field.

Adding fields from several charges

Fields obey superposition: find the field from each charge separately, as a vector, then add. On the line joining two equal positive charges the fields cancel at the midpoint. For opposite charges they add there. Draw arrows first, then add magnitudes with signs along one axis, or use components when the charges are not in a line.

Field lines start on positive charges and end on negative ones. They never cross, and closer lines mean a stronger field.

Mistakes that cost marks

Using r instead of r² in the point charge formula.
Forgetting to convert cm to m and nC or μC to C before substituting.
Adding fields as plain numbers when they point in different directions.
Thinking the field depends on the test charge. Doubling q doubles F, so E stays put.

The electricity formula hub has the full chapter list, and drift velocity shows how the field drives current in a wire.

Frequently asked questions

What is the formula of electric field intensity?

E = F / q, the force on a small positive test charge divided by that charge. For a point charge Q at distance r it becomes E = kQ / r².

In which units is electric field expressed?

Newton per coulomb (N/C). An equal unit is volt per metre (V/m).

Is electric field a vector?

Yes. It has a magnitude and points in the direction a positive charge would be pushed.

What is the magnitude of an electric field formula for a point charge, and how does it change with distance?

The magnitude is E = kQ / r², an inverse square law. Doubling the distance cuts the field to one quarter.

What is the difference between electric field and electric force?

Force acts on a particular charge and depends on its size. Field is the force per unit charge, a property of the point in space set up by the source charges.