Two charges 5 text{ nC} and -2 text{ nC} are | vedclass

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(B) Given:$q_1 = 5 	ext{ nC} = 5 	imes 10^{-9} 	ext{ C}$$q_2 = -2 	ext{ nC} = -2 	imes 10^{-9} 	ext{ C}$The distance between the charges is $r =

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$-5 	imes 10^{-7} 	ext{ J}$

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(B) Given:$q_1 = 5 	ext{ nC} = 5 	imes 10^{-9} 	ext{ C}$$q_2 = -2 	ext{ nC} = -2 	imes 10^{-9} 	ext{ C}$The distance between the charges is $r = (23 - 5) 	ext{ cm} = 18 	ext{ cm} = 18 	imes 10^{-2} 	ext{ m}$.The electrostatic potential energy $U$ of a system of two point charges is given by:$U = \frac{k q_1 q_2}{r}$Substituting the values:$U = \frac{(9 	imes 10^9 	ext{ N m}^2/	ext{C}^2) 	imes (5 	imes 10^{-9} 	ext{ C}) 	imes (-2 	imes 10^{-9} 	ext{ C})}{18 	imes 10^{-2} 	ext{ m}}$$U = \frac{9 	imes 5 	imes (-2) 	imes 10^{9-9-9}}{18 	imes 10^{-2}} 	ext{ J}$$U = \frac{-90 	imes 10^{-9}}{18 	imes 10^{-2}} 	ext{ J}$$U = -5 	imes 10^{-7} 	ext{ J}$

Two charges $5 \text{ nC}$ and $-2 \text{ nC}$ are placed at points $(5 \text{ cm}, 0, 0)$ and $(23 \text{ cm}, 0, 0)$ in a region of space where there is no other external field. The electrostatic potential energy of this charge system is

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