Answer to Question #98400 in Field Theory for Marvin

Question #98400
find the magnitude of an electron's electric field if the field itself is 50.0 cm away from it
1
Expert's answer
2019-11-11T15:56:29-0500

The magnitude of the electric field, created by a point charge "q" at distance "r" is "E = k\\frac{q}{r^2}", where "k = \\frac{1}{4 \\pi \\varepsilon_0} = 8.85 \\cdot 10^{-12} N m^2 C^{-2}".

Hence, for "q = e = 1.6 \\cdot 10^{-19}C" and "r = 50 cm = 0.5 m", the magnitude of electric field is "E = 5.7 \\cdot 10^{-30} \\frac{V}{m}".


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