An action force is 50 N to the left. The reaction force must be:
2 answers:
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In this problem we have the electric field intensity E:
E = 6.5 × newtons/coulomb
We have the magnitude of the load:
q = 6.4 × coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 × meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 × )(6.5 ×
)(1.2 ×
)
PE = 5.0 x joules
None of the options shown is correct.