Electric Field Equation Calculator

A point charge $Q$ sets up an electric field whose magnitude falls off as the square of the distance:

$E = \\frac{kQ}{r^2} \\qquad k = 8.988 \\times 10^{9}\\ \\text{N}\\cdot\\text{m}^2/\\text{C}^2$

Symbols and SI units:

• $E$ — electric field magnitude, newtons per coulomb (N/C), identical to volts per metre (V/m)
• $Q$ — the source charge, coulombs (C)
• $r$ — distance from the source charge to the field point, metres (m)
• $k$ — Coulomb constant, equal to $1/(4\\pi\\varepsilon_0)$

The operating assumption: this is electrostatics in vacuum or air — a stationary point charge, or a uniformly charged sphere seen from outside, with no dielectric or conductor redistributing the charge. Inside a material of relative permittivity $\\varepsilon_r$ the field is smaller by that factor.

The mistake people make: forgetting that $E$ is a vector. With several charges you add the fields component by component, not as bare magnitudes, and the direction points away from a positive source and toward a negative one.

The force between two point charges is the Coulomb force equation:

$F = \\frac{k\\,q_1 q_2}{r^2}$

with $F$ in newtons (N) and both charges in coulombs. Substitute magnitudes to get the size of the force, then decide the direction from the signs: like charges repel, opposite charges attract.

The field and the force are linked by

$F = qE \\qquad \\Longrightarrow \\qquad E = \\frac{F}{q}$

where $q$ is the test charge placed in the field. This is the definition of the field: force per unit charge. It is also why a uniform field between parallel plates can be written $E = V/d$, volts per metre.

The operating assumption: both objects are small compared with $r$, so they behave as points. Two spheres almost touching polarise each other and the inverse-square result no longer holds.

The mistake people make: leaving charges in microcoulombs. A $\\mu$C is $10^{-6}$ C and a nC is $10^{-9}$ C; substituting the raw number inflates the answer by twelve orders of magnitude.

• Not squaring the distance — $r$ appears as $r^2$. Halving the separation quadruples both field and force.
• Leaving the distance in centimetres — $5.0$ cm is $0.050$ m, and $r^2$ turns a factor of $100$ into a factor of $10{,}000$.
• Confusing the source charge with the test charge — $Q$ in $E = kQ/r^2$ creates the field; $q$ in $F = qE$ feels it.
• Adding fields as scalars — superposition is vector addition. Two equal charges either side of a point can give zero field, not double.
• Mixing up $E$ and $V$ — the field is N/C or V/m, the potential is volts. In a uniform field they differ by the plate separation.
• Using $k$ with a permittivity already included — use either $k$ or $1/(4\\pi\\varepsilon_0)$, never both.