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Question

Physics Question on Electrostatic potential

Potential energy of two charges 10nC10 \, nC each separated by a distance of 0.09 m in air is

A

10μJ10 \mu J

B

1μJ1 \mu J

C

10mJ10 m J

D

10J10 J

Answer

10μJ10 \mu J

Explanation

Solution

Potential energy of two charge system is given by
U=14πe0er×q1q2rU= \frac{1}{4 \pi e_{0} e_{r}} \times\frac{q_{1}q_{2}}{r}
Here, q1=q2=10nC=10×109Cq_{1 } = q_{2} = 10 nC = 10\times 10^{-9}C
14πe0=9×109Nm2C2r=r=0.09m\frac{1}{4 \pi e_{0} } = 9 \times 10^{9} N m^{2} C^{-2} r = r = 0.09 m
U=9×109×10×109×10×1090.09=10μJ\therefore \, U = 9 \times 10^{9} \times \frac{10 \times 10^{-9} \times 10 \times 10^{-9}}{0.09} = 10 \mu J