Electric Force Calculator

Two point charges attract or repel along the line joining them with a force of magnitude

$F = k\frac{|q_1 q_2|}{r^2}, \qquad k = \frac{1}{4\pi\varepsilon_0} = 8.988 \times 10^{9}\ \text{N·m}^2\text{/C}^2$

• $F$ — force, newtons (N)
• $q_1, q_2$ — charges, coulombs (C); note 1 μC = $10^{-6}$ C and 1 nC = $10^{-9}$ C
• $r$ — separation between the charges, metres (m)
• $\varepsilon_0 = 8.854 \times 10^{-12}$ C²/(N·m²), the permittivity of free space

The sign convention that actually works: take the magnitude from the absolute values, then decide the direction physically — like charges repel, unlike charges attract. Charge is quantised in units of $e = 1.602 \times 10^{-19}$ C, so a net charge is always an integer multiple of $e$.

The assumption people forget: this is an inverse-square law for point charges or uniform spheres in vacuum, and it gives the force from one other charge. With three or more charges, add the individual forces as vectors.

The same $1/r^2$ structure reappears in every electrostatic quantity, so it is worth keeping the four straight:

$E = \frac{F}{q_0} = k\frac{|Q|}{r^2}, \qquad V = k\frac{Q}{r}, \qquad U = k\frac{q_1 q_2}{r}$

• $E$ — electric field, N/C (equivalently V/m); a vector
• $V$ — electric potential, volts (V = J/C); a scalar, and it keeps the sign of $Q$
• $U$ — potential energy of the pair, joules (J)

Notice the pattern: force and field go as $1/r^2$, energy and potential go as $1/r$. The links between them are $F = q_0 E$ and $U = q_0 V$, and the work done moving a charge through a potential difference is

$W = q\,\Delta V$

in joules.

When it applies: the $k Q/r$ forms assume the zero of potential is at infinity, which is the standard choice for isolated point charges but not for parallel plates or circuits.

• Leaving charges in microcoulombs — convert to coulombs first. Forgetting the $10^{-6}$ on both charges is a factor of $10^{12}$ error.
• Using centimetres for $r$ — the separation must be in metres before it is squared.
• Squaring $r$ in the potential — $V$ and $U$ go as $1/r$, only $F$ and $E$ go as $1/r^2$.
• Adding fields as scalars — $E$ is a vector. Two fields of 100 N/C at right angles give 141 N/C, not 200 N/C.
• Dropping the sign in $U$ — an attracting pair has negative potential energy, and that minus sign is what makes the pair bound.
• Assuming zero field means zero potential — midway between two equal positive charges the field cancels but the potential is at a maximum.