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The electric potential energy of the charge will be =3.5*10^8 J
Electric potential energy is the energy that is needed to move a charge against an electric field.
F=Force experienced by the charge = 3.6*10^-4 N
q1 = magnitude of charge producing the electric field
q2 = magnitude of charge experiencing the electric force
r1 = distance between the two charges
Electric force experienced by the charge is given using coulomb's law as
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
[tex]3.6\times 10^{-4}=\dfrac{9\times 10^9 q_1q_2}{(9.8\times 10^{-5})^2}[/tex]
[tex]q_1q_2=3.84\times 10^{-22}[/tex]
Electric potential energy of the charge can be given as
[tex]U=\dfrac{kq_1q_2}{r}[/tex]
[tex]U=\dfrac{(9\times 10^9)q_1q_2}{(9.8\times 10^{-5})^2}[/tex]
[tex]U=3.5\times 10^{-8}\ J[/tex]
Thus Electric potential energy of the charge can be given as [tex]U=3.5\times 10^{-8}\ J[/tex]
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