Charged particles in an electric field

Answer

$F = qE$ — the charge times the field strength. In the direction of the field for a **positive** charge, opposite it for a **negative** charge.

Answer

Two steps: force $F = qE$, then Newton's second law $a = \dfrac{F}{m} = \dfrac{qE}{m}$.

Answer

Because $a = \dfrac{qE}{m}$ and the electron's **mass m is tiny** (9.1 × 10⁻³¹ kg), so even a modest force gives an acceleration of order 10¹⁴ m s⁻².

Answer

A **parabola** — like a projectile. Constant velocity along the plates, constant acceleration across them.

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A **positive** charge accelerates **along** the field; an **electron** (negative) accelerates **opposite** to the field.

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**Constant velocity** — there is no force along the plates, so the horizontal speed never changes.

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$s = \tfrac{1}{2}at^{2}$ (starting from rest sideways) — NOT s = vt, because the sideways motion is accelerated.

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Yes — the booklet gives $E = \dfrac{F}{q}$; rearranged that is F = qE.

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F = qE = (1.6 × 10⁻¹⁹)(2.0 × 10⁴) = 3.2 × 10⁻¹⁵ N.

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a = F ÷ m = (8.0 × 10⁻¹⁶) ÷ (9.1 × 10⁻³¹) ≈ 8.8 × 10¹⁴ m s⁻².

Answer

Because the sideways motion is **accelerated** (constant force qE), not at constant velocity. Use s = ½at² instead.