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 qq at distance rr is E=kqr2E = k\frac{q}{r^2}, where k=14πε0=8.851012Nm2C2k = \frac{1}{4 \pi \varepsilon_0} = 8.85 \cdot 10^{-12} N m^2 C^{-2}.

Hence, for q=e=1.61019Cq = e = 1.6 \cdot 10^{-19}C and r=50cm=0.5mr = 50 cm = 0.5 m, the magnitude of electric field is E=5.71030VmE = 5.7 \cdot 10^{-30} \frac{V}{m}.


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