Answer:
[tex]3.5\times10^{-8} J[/tex]
Explanation:
[tex]F[/tex] = Force experienced by the charge = [tex]3.6\times10^{-4} N[/tex]
[tex]q_{1}[/tex] = magnitude of charge producing the electric field
[tex]q_{2}[/tex] = magnitude of charge experiencing the electric force
[tex]r_{1}[/tex] = distance between the two charges
Electric force experienced by the charge is given using coulomb's law as
[tex]F = \frac{k q_{1} q_{2}}{r^{2} } \\3.6\times10^{-4} =\frac{(9\times10^{9}) q_{1} q_{2}}{(9.8\times10^{-5})^{2} } \\q_{1} q_{2} = 3.8416\times10^{-22}[/tex]
Electric potential energy of the charge can be given as
[tex]U = \frac{k q_{1} q_{2}}{r } \\U = \frac{(9\times10^{9}) q_{1} q_{2}}{(9.8\times10^{-5})} \\U = 3.5\times10^{-8} J[/tex]
