Uniform fields, parallel plates and potential difference

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A field with the **same strength and direction everywhere** — drawn as **evenly-spaced, parallel** lines. You get one in the gap between two parallel charged plates.

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**Evenly-spaced parallel lines** running from the **+ plate** to the **− plate** (the direction a positive charge is pushed).

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$E = \dfrac{V}{d}$ — voltage across the plates ÷ the gap between them. Given in the data booklet. Unit: V m⁻¹.

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**Volts per metre (V m⁻¹)**, which is the same as **N C⁻¹** (newtons per coulomb).

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E **doubles** — field strength is inversely proportional to the separation d (E = V ÷ d).

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$F = qE$ (rearranged from the data-booklet definition $E = \dfrac{F}{q}$). Bigger charge or stronger field → bigger force.

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$W = qV$ (in joules). This is the energy the charge gains — and for a charge from rest, its kinetic energy. Not in the booklet — memorise it.

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The energy a charge of **e** (1.6 × 10⁻¹⁹ C) gains moving through **1 V**: 1 eV = 1.6 × 10⁻¹⁹ J. A charge e through V volts gains V eV.

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Multiply by 1.6 × 10⁻¹⁹: 250 × 1.6 × 10⁻¹⁹ = 4.0 × 10⁻¹⁷ J.

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E = V ÷ d = 600 ÷ 0.020 = 3.0 × 10⁴ V m⁻¹.

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From the **+ plate to the − plate** — the direction a **positive** charge would be pushed.