Coulomb's Law, Superposition & Electric Field

PHYCoulomb's Law, Field & Dipole

Coulomb's law gives the electrostatic force between two stationary point charges, and is the foundation of the whole chapter. For more than two charges the net force is found by vector superposition, treating each pair independently.

The electric field $\vec{E}$ is the force per unit positive test charge and lets us describe the influence of a charge distribution at any point in space.

Coulomb's law (magnitude)

F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}

q_1,q_2

are the charges (C),

r

the separation (m),

\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\text{N m}^2\,\text{C}^{-2}

in vacuum

Force in a medium

F_{med} = \dfrac{1}{4\pi\varepsilon_0 K}\dfrac{q_1 q_2}{r^2} = \dfrac{F_{vac}}{K}

K

(or

\varepsilon_r

) is the dielectric constant of the medium; force is reduced by factor

K

Derivation

A material medium has permittivity $\varepsilon = K\varepsilon_0$ , where $K$ is its dielectric constant. Coulomb's law is written with this full permittivity:
$F_{med} = \dfrac{1}{4\pi\varepsilon}\dfrac{q_1 q_2}{r^2} = \dfrac{1}{4\pi K\varepsilon_0}\dfrac{q_1 q_2}{r^2}$
Vacuum is the case $K=1$ . Dividing the two forces, everything cancels except the factor $K$ :
$\dfrac{F_{med}}{F_{vac}} = \dfrac{1}{K} \implies F_{med} = \dfrac{F_{vac}}{K}$

Superposition of forces

\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \cdots + \vec{F}_n

vector sum of forces on a charge due to every other charge taken one at a time

Electric field of a point charge

E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}

E

\text{N C}^{-1}

, directed radially outward for

+q

, inward for

-q

Derivation

The field is the force per unit positive test charge $q_0$ placed at the point:
$E = \dfrac{F}{q_0}$
By Coulomb's law the force on $q_0$ from the source charge $q$ is:
$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q\,q_0}{r^2}$
Dividing by $q_0$ removes the test charge, leaving the field of $q$ alone:
$E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$

Force from a field

\vec{F} = q\vec{E}

q

is the charge placed in field

\vec{E}

; force on a negative charge is opposite to

\vec{E}

Derivation

The electric field is defined as the force experienced per unit charge:
$\vec{E} = \dfrac{\vec{F}}{q}$
Rearranging gives the force on any charge $q$ placed in that field:
$\vec{F} = q\vec{E}$

Coulomb's law is an inverse-square law: $F\propto 1/r^2$ , so halving $r$ multiplies $F$ by $4$ and doubling $r$ divides it by $4$ .
Like charges repel and unlike charges attract; always assign directions, not just signs, when adding force vectors.
Superposition means the force between any two charges is unaffected by the presence of others — compute each pair separately, then add as vectors.
For charges on a straight line the forces are collinear, so add or subtract magnitudes according to direction; for non-collinear charges resolve into components.
$\vec{E}$ points away from positive charge and toward negative charge; field lines never cross and start on $+$ , end on $-$ .
A field can exist where there is no charge to feel it; $\vec{E}=\vec{F}/q_0$ is defined using a vanishingly small positive test charge $q_0\to0$ .
Always convert: $1\,\mu\text{C}=10^{-6}\,\text{C}$ , $1\,\text{nC}=10^{-9}\,\text{C}$ , and distances to metres before substituting.

Where the marks go

Forgetting to convert $\mu\text{C}$ to C and cm to m — a 30 cm gap is $0.30\,\text{m}$ , not $30\,\text{m}$ , giving errors of many powers of ten.
Treating the superposition force as a scalar sum: on the middle charge the two neighbours may push in opposite directions, so signs/directions matter.
Squaring only the distance and not also handling the constant — remember $F=9\times10^{9}\,q_1 q_2/r^2$ , with $r^2$ in the denominator.
Confusing electric field $E$ (units $\text{N C}^{-1}$ ) with force $F$ (units N); $F=qE$ , they are not the same quantity.

How the board asks it

Numericalcoulomb's law magnitude and unit conversion
Two point charges of $+3\,\mu\text{C}$ and $-5\,\mu\text{C}$ are placed $20\,\text{cm}$ apart in vacuum. Calculate the magnitude of the electrostatic force between them and state its nature.
Numericalsuperposition of forces on triangular / collinear charges
Three charges $q_1=+2\,\mu\text{C}$ , $q_2=+2\,\mu\text{C}$ and $q_3=-3\,\mu\text{C}$ are fixed at the corners of an equilateral triangle of side $10\,\text{cm}$ . Find the magnitude and direction of the net force on $q_3$ .
Numericalfield of a point charge and force from a field
Calculate the electric field intensity at a point $30\,\text{cm}$ from a point charge of $+4\,\mu\text{C}$ , and hence find the force experienced by a charge of $-2\,\text{nC}$ placed at that point.
Define / statedefinition of electric field and the test-charge limit
Define electric field intensity at a point. Why is the test charge $q_0$ taken to be vanishingly small in the relation $\vec{E}=\lim_{q_0\to 0}\vec{F}/q_0$ ?
Give reasonsinverse-square dependence and the dielectric medium factor
Account for the following: when the separation between two fixed point charges is halved, the force between them becomes four times as large. How is this force changed when the charges are immersed in a medium of relative permittivity (dielectric constant) $\varepsilon_r$ ?
Applicationfield-line rules and null-point location
Two equal positive point charges are separated by a distance $d$ . Sketch the electric field pattern and find the position on the line joining them at which the resultant electric field is zero.

Practise this topic

NextElectric Dipole

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.