Answer:
electric potential is 3.31 × V
potential energy is 152 MeV
Explanation:
given data
fragment charge Q = 46 protons = 46 × 1.6 × C
to find out
electric potential and potential energy
solution
we know here distance from fragment d = 2 × m
and constant for electric force k that is 9 × N-m²/C²
so that we can find electric potential = kQ/d
electric potential = 9 × / ( 2 ×
)
electric potential = 3.31 × V
and
we know relation between electric potential and potential
that is V = U/q
so U will be = qV
now put all value
we get potential energy U
potential energy = 46 × 3.31 ×
potential energy = 1.52 × eV
so potential energy = 152 MeV