Electric potential and work (HL)

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The **potential energy per unit positive charge** at a point: $V = kQ/r$. Unit: **volt** (J C⁻¹). It is a **scalar**.

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It falls off as **1/r** — double r and V halves. (The field E falls off as 1/r².)

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**Yes** — near a negative charge V is negative. Unlike gravity (always negative), charge has two signs.

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At **infinity**. V is the work per coulomb to bring a small +charge from infinity to the point.

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$E_p = kQq/r$ (a scalar for the **pair**, in joules, with the sign of Qq).

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**Like** charges (Qq > 0) ⇒ **positive** E_p; **unlike** charges (Qq < 0) ⇒ **negative** E_p (bound).

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$W = q\,\Delta V$, where $\Delta V = V_{final} - V_{initial}$.

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Because **potential depends only on position**, so W = qΔV uses only the two endpoints.

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$V = kQ/r = (8.99\times10^9)(2.0\times10^{-9})/0.30 = 60$ V.

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**Potential** V is a **scalar** (just add them); **field** E is a **vector** (add by components).

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The **field** did the work — energy is **released**. A positive W means **you** supplied energy.

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Compute each $V = kQ/r$ **with its sign**, then **add** them as numbers (scalars).