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:
[tex]KE=\frac{kQq}{r}\ \ \ \ \ \ \ ...i[/tex]
[tex]k[/tex] is Coulomb's constant.
#The electric field,[tex]E[/tex] at radius r is expressed as:
[tex]E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii[/tex]
From i and ii, we have:
[tex]KE=Eqr[/tex]
[tex]r=(KE)/Eq[/tex]
#Substitute actual values in our equation:
[tex]r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m[/tex]
Hence, the distance between the charge and the source of the electric field is 2.2 meters
