Question

What is the electric potential energy of a charge that experiences a force of 3. 6 × 10–4 N when it is 9. 8 × 10–5 m from the source of the electric field? 3. 5 × 10–8 J 4. 6 × 10–4 J 2. 7 × 10–1 J 3. 7 × 100 J.

Answer 1

The electric potential energy of the charge will be =3.5*10^8 J  

What is Electric potential energy?

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

$F=\dfrac{kq_1q_2}{r^2}$

$3.6\times 10^{-4}=\dfrac{9\times 10^9 q_1q_2}{(9.8\times 10^{-5})^2}$

$q_1q_2=3.84\times 10^{-22}$

Electric potential energy of the charge can be given as

$U=\dfrac{kq_1q_2}{r}$

$U=\dfrac{(9\times 10^9)q_1q_2}{(9.8\times 10^{-5})^2}$

$U=3.5\times 10^{-8}\ J$

Thus Electric potential energy of the charge can be given as $U=3.5\times 10^{-8}\ J$

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