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An atomic nucleus has a net charge of +42e. What is the magnitude of
the electric field (in nN/C) at a distance of 5.14 m from the center of
the nucleus?
nN/C

Expert Solution

Step 1: To know the given

Charge $q equals 42 e equals 42 cross times 1.6 cross times 10 to the power of negative 19 end exponent space C$

where e is the fundamental charge.

Distance $r equals 5.14 space m$

Electric field $E equals ?$

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