Answer:
P = 72 J
Step-by-step explanation:
Given that,
Charges, [tex]q_1=q_2=2\times 10^{-5}\ C[/tex]
These charges are brought from infinity to a separation of 5 cm.
We need to find the increase in the electrical potential energy in the process. The formula for the electric potential energy is given by :
[tex]P=k\dfrac{q_1q_2}{r}\\\\P=9\times 10^9\times \dfrac{(2\times 10^{-5})^2}{0.05}\\\\P=72\ J[/tex]
So, the required electric potential energy is 72 J.
Answer: bye
Step-by-step explanation:
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