Two charges of magnitude 5, nC and -2, nC are | vedclass

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(A) The electrostatic potential energy $U$ of a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by the formula:$U = \

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$20$

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(A) The electrostatic potential energy $U$ of a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by the formula:$U = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}$Given:$q_1 = 5 	imes 10^{-9} \, C$$q_2 = -2 	imes 10^{-9} \, C$$U = -0.5 	imes 10^{-6} \, J$Distance $r = |x - 2| 	imes 10^{-2} \, m$Substituting the values into the formula:$-0.5 	imes 10^{-6} = (9 	imes 10^9) 	imes \frac{(5 	imes 10^{-9}) 	imes (-2 	imes 10^{-9})}{(x - 2) 	imes 10^{-2}}$$-0.5 	imes 10^{-6} = \frac{-90 	imes 10^{-9}}{(x - 2) 	imes 10^{-2}}$$-0.5 	imes 10^{-6} = \frac{-9 	imes 10^{-7}}{(x - 2) 	imes 10^{-2}}$$(x - 2) 	imes 10^{-2} = \frac{-9 	imes 10^{-7}}{-0.5 	imes 10^{-6}}$$(x - 2) 	imes 10^{-2} = 18 	imes 10^{-1} = 1.8 \, m$$(x - 2) 	imes 10^{-2} = 180 	imes 10^{-2} \, m$$x - 2 = 180$$x = 182 \, cm$ (Note: If the distance is taken as $r = (x-2) 	imes 10^{-2}$,the calculation yields $182$. However,checking the provided options,if $r = 0.18 \, m = 18 \, cm$,then $x-2 = 18 \Rightarrow x = 20$. Let's re-verify: $U = (9 	imes 10^9) 	imes (-10 	imes 10^{-18}) / (0.18) = -90 	imes 10^{-9} / 0.18 = -500 	imes 10^{-9} = -0.5 	imes 10^{-6} \, J$. Thus,$x = 20$ is correct.)

Two charges of magnitude $5\, nC$ and $-2\, nC$ are placed at points $(2\, cm, 0, 0)$ and $(x\, cm, 0, 0)$ respectively in a region of space where there is no other external field. If the electrostatic potential energy of the system is $-0.5\,\mu J$,find the value of $x$ in $cm$.

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