Two fixed charges A and B of 5 mu C each are | vedclass

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(A) Let $q = 5 \mu C = 5 	imes 10^{-6} 	ext{ C}$ and $Q = -5 \mu C = -5 	imes 10^{-6} 	ext{ C}$. The distance $AC = CB = 3 	ext{ m}$.By the princ

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$4 	ext{ m}$

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(A) Let $q = 5 \mu C = 5 	imes 10^{-6} 	ext{ C}$ and $Q = -5 \mu C = -5 	imes 10^{-6} 	ext{ C}$. The distance $AC = CB = 3 	ext{ m}$.By the principle of conservation of energy,the loss in kinetic energy equals the gain in potential energy (or loss in potential energy equals gain in kinetic energy).The potential energy at $C$ is $U_C = 2 	imes \frac{k q Q}{r}$,where $r = 3 	ext{ m}$.The potential energy at $D$ is $U_D = 2 	imes \frac{k q Q}{\sqrt{r^2 + x^2}}$,where $x = CD$.Loss in potential energy = $U_C - U_D = 2kq|Q| \left( \frac{1}{r} - \frac{1}{\sqrt{r^2 + x^2}} ight) = K_{initial} = 0.06 	ext{ J}$.Substituting the values: $2 	imes (9 	imes 10^9) 	imes (5 	imes 10^{-6}) 	imes (5 	imes 10^{-6}) 	imes \left( \frac{1}{3} - \frac{1}{\sqrt{3^2 + x^2}} ight) = 0.06$.$0.45 	imes \left( \frac{1}{3} - \frac{1}{\sqrt{9 + x^2}} ight) = 0.06$.$\frac{1}{3} - \frac{1}{\sqrt{9 + x^2}} = \frac{0.06}{0.45} = \frac{6}{45} = \frac{2}{15}$.$\frac{1}{\sqrt{9 + x^2}} = \frac{1}{3} - \frac{2}{15} = \frac{5-2}{15} = \frac{3}{15} = \frac{1}{5}$.$\sqrt{9 + x^2} = 5 \implies 9 + x^2 = 25 \implies x^2 = 16 \implies x = 4 	ext{ m}$.

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