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Question

$(a)$ What is the direction and magnitude of an electric field that supports the weight of a free electron near the surface of Earth? $(b)$ Discuss what the small value for this field implies regarding the relative strength of the gravitational and electrostatic forces.

Solution

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Answered 1 year ago

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1 of 8Given and known data: $m_{e} = 9.11 \cdot 10^{-31}\, \mathrm{kg}$ $g = 9.81\, \mathrm{m/s^{2}}$ $e = 1.6 \cdot 10^{-19}\, \mathrm{C}$ $r = 1\, \mathrm{m}$ $E \approx 6 \cdot 10^{-11}\, \mathrm{N/C}$

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