A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the
2 answers:
Answer:
2.2 meters
Explanation:
Potential energy, PE created by a charge, q at a radius r from the charge source, Q, is expressed as:
is Coulomb's constant.
#The electric field, at radius r is expressed as:
From i and ii, we have:
#Substitute actual values in our equation:
Hence, the distance between the charge and the source of the electric field is 2.2 meters
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Answer:
c. 2 MeV.
Explanation:
The computation of the binding energy is shown below
= 2 MeV
As 1 MeV = (1 u) c^2
hence, the binding energy is 2 MeV
Therefore the correct option is c.
We simply applied the above formula so that the correct binding energy could come
And, the same is to be considered