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 B

icon
Related questions
Question
QE ✰ ✰ ↓
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
Transcribed Image Text:QE ✰ ✰ ↓ 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 ?

steps

Step by step

Solved in 3 steps with 7 images

Blurred answer