As shown in the figure, an electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0×10⁴ N C⁻¹. Find the acceleration of the electron due to the electric field. [e = 1.6×10⁻¹⁹ C, mₑ = 9.1×10⁻³¹ kg]

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(3) 3.52×10¹⁵ m s⁻². Given: E = 2.0×10⁴ N C⁻¹ (upward in the figure), e = 1.6×10⁻¹⁹ C, mₑ = 9.1×10⁻³¹ kg. Idea: the force on the electron is eE, opposite to ⃗E (downward here), and a = F/(mₑ). F = eE = (1.6×10⁻¹⁹)(2.0×10⁴) = 3.2×10⁻¹⁵ N a = F/(mₑ) = (3.2×10⁻¹⁵)/(9.1×10⁻³¹) = 3.52×10¹⁵ m s⁻² This is more than 10¹⁴ times g, so gravity can be ignored; the distance 1.5 cm is needed only for the time of fall, not for a. So the acceleration is 3.52×10¹⁵ m s⁻².

As shown in the figure, an electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0×10⁴ N C⁻¹. Find the acceleration of the electron due to the electric field. [e = 1.6×10⁻¹⁹ C, mₑ = 9.1×10⁻³¹ kg]

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