Electric Potential (Point Charge)

Equivalent forms

V = k_e \frac{q}{r}

V = -\int_{\infty}^{r} \vec{E} \cdot d\vec{l}

A scalar field that encodes all the information of the vector electric field — gradient of V gives E, making complex problems tractable.

Symbol	Name	SI unit	Dimension	Range
$V$	Electric potentialoutput Potential at distance r from the point charge	V	M·L²·T⁻³·I⁻¹	-1000000 – 1000000
$q$	Source charge Point charge creating the potential	C	I·T	-0.00001 – 0.00001
$r$	Distance from charge Radial distance from the source charge	m	L	0.01 – 2

Unit systems

Symbol	SI	CGS	Imperial
$V$	V	statV	V
$q$	C	statC	C
$r$	m	cm	ft

Where it holds

Valid for point charges in vacuum with potential referenced to infinity. For continuous distributions, integrate contributions. In conductors, V is constant throughout the bulk.

Dimensional analysis

$[V] = [k_e]\cdot [q]\cdot [r]^{-1} \to (N\cdot m^{2}\cdot C^{-2})(C)(m^{-1}) = N\cdot m/C = J/C = V \checkmark$

Discovery

George Green · 1828

Green introduced the potential function in his self-published essay, unifying the mathematics of gravitational and electrostatic theory. His work was rediscovered by Lord Kelvin decades later.

Try this

What determines the 'voltage' at a point in space near a charge?

Calculate the electric potential at 0.5 m from a +8 μC point charge.

Research status: stable

Real-world applications

Voltage measurements in every electrical circuit
Cardiac defibrillator energy calculations
Electron volt as an energy unit in particle physics
Electrostatic potential mapping in semiconductor design

Common misconceptions

Potential is a scalar, not a vector — it has no direction
Zero potential does not mean zero field; the field is the gradient of V
Potential at a point depends on all charges in the universe, not just nearby ones

Experimental verification

Measured using electrometers and potentiometers since the 19th century. Modern atomic-scale verification via scanning tunneling microscopy, which directly maps the potential landscape on surfaces.

Derivation

Integrate the electric field from infinity to r:

V(r) = -\int _\infty ^r E\cdot dr = -\int _\infty ^r (kq/r^{2})dr = kq/r

.

The negative sign ensures V decreases in the direction of E for positive charges.

Limiting cases

r \to \infty

⟶

V \to 0

Potential is defined to be zero at infinity.

r \to 0

⟶

V \to \pm \infty

Potential diverges at the location of a point charge.

q \to 0

⟶

V \to 0

No source charge means no potential.

What if…

What if the charge is negative?

Potential is negative at all points. $V = -143.8\,\mathrm{kV}$ $for -8 \mu C$ at 0.5 m. A positive test charge would gain kinetic energy falling toward it.

What if you halve the distance?

Potential doubles: $V = 287.6\,\mathrm{kV}$ . Unlike the field $(1/r^{2})$ , potential only falls as 1/r — slower decay.

What if two equal charges create the potential?

Potentials are scalars and simply add: V_total $= V_{1} + V_{2}$ . No vector addition needed — this is the power of working with potentials.

1

Potential near a point charge

Given ·

q:: 0.000008
r:: 0.5

Find · V

Steps

Identify: $q = +8 \mu C = 8\times 10^{-6} C$ , $r = 0.5\,\mathrm{m}$
Apply $V = k_e \times q / r$
Compute: $8.9875\times 10^{9} \times 8\times 10^{-6} = 71900$
Divide by r: $71900 / 0.5 = 143800\,\mathrm{V}$
Positive charge → positive potential at all points

Result ·

V = k_e \times q / r = 8.9875\times 10^{9} \times 8\times 10^{-6} / 0.5 = 1.438\times 10^{5} V = 143.8\,\mathrm{kV}

2

Work to bring a charge from infinity

Given ·

Q:: 0.000005
q:: 0.000002
r:: 0.1

Find · Work W to bring q from infinity to r

Steps

Potential at r due to Q: $V = k_e \times Q / r = 8.9875\times 10^{9} \times 5\times 10^{-6} / 0.1 = 4.494\times 10^{5} V$
Work done = change in potential energy: $W = qV$
$W = 2\times 10^{-6} \times 4.494\times 10^{5} = 0.899\,\mathrm{J}$
Positive work means energy was stored in the field — both charges are positive, so they repel

Result ·

W = qV = q \times k_e \times Q / r = 2\times 10^{-6} \times 8.9875\times 10^{9} \times 5\times 10^{-6} / 0.1 = 0.899\,\mathrm{J}

Coulomb's Law→Electric Field of a Point Charge→Parallel Plate Capacitor→