Electric field strength and superposition

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The **force per unit charge** on a small positive test charge: $E = \dfrac{F}{q}$. It is a **vector**. Unit: **N C⁻¹**.

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**N C⁻¹** (newtons per coulomb).

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$E = \dfrac{kQ}{r^{2}}$ — Coulomb constant k × charge Q ÷ distance² (derived from Coulomb's law with E = F ÷ q).

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**Outward** — away from the charge (a positive test charge is pushed away).

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**Inward** — toward the charge (a positive test charge is pulled in).

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E falls to a **quarter** — the field is inverse-square ($E \propto 1/r^{2}$).

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**Superposition** — add the field from each charge **as a vector** (same direction → add sizes; opposite → subtract).

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At the **midpoint** — the two equal fields point in opposite directions and cancel (the null point).

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Rearrange $E = \dfrac{F}{q}$ to $F = qE$ — multiply the charge by the field strength.

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A **vector** — it has size and direction (the direction a +test charge is pushed).

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$F = qE = (2.0\times10^{-9})(5.0\times10^{4}) = 1.0\times10^{-4}$ N, along the field.