Electric Field Strength (E = F/q)

Coulomb's law describes two charges reaching across a gap to grab each other. Faraday could not accept that, and insisted instead that the first charge changes the space around it, and the second charge responds only to the space it sits in. That change in the space is the electric field, and $E = F/q$ is its definition: put a small test charge somewhere, measure the force on it, divide by the charge, and what is left is a property of the point rather than of the test charge. Move the test charge away and the field is still there.

A worked case. A 2 µC bead in a laboratory field feels 0.05 N. Then $E = 0.05/(2 \times 10^{-6}) = 25{,}000\ \text{V/m}$, or 25 kV/m. Run it the other way and the same field puts $F = qE = 3.2 \times 10^{-4}$ N on a 12.8 nC speck of dust — which is exactly the sum an electrostatic precipitator designer does, and the reason precipitators run at tens of kilovolts across a few centimetres.

The unit deserves a second look, because it appears in two disguises. Force per charge is newtons per coulomb; potential gradient is volts per metre; and $\text{N/C} = (\text{J/m})/\text{C} = (\text{J/C})/\text{m} = \text{V/m}$. They are the same unit, and this site's field type offers both because a physics student and a cable engineer arrive from opposite directions at the identical number.

The classic trap is that a test charge is supposed to be infinitesimal. A real one carries its own field, and if it is big enough to shove the source charges around — polarising a nearby conductor, or dragging a charged sphere on a thread — then the force you measured is not the force the original field would have applied, and the division gives the wrong answer. Second, everything here is a vector. The equation as written is one component along one line; two fields at a point add head to tail, not by arithmetic, and a field of 3 kV/m plus a field of 4 kV/m is 7 only if they point the same way. Third, the sign is information and not an error: divide a force by a negative charge and $E$ comes out with the opposite sign, because a negative charge is pushed against the field. Finally, a field is defined at a point. Quoting one field for a whole region only works where the region is genuinely uniform, which in practice means between large plates and almost nowhere else.