In free space,a particle A of charge 1,mu C i | vedclass

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(D) Given: $q_A = 1\,\mu C = 10^{-6}\,C$,$q_B = 1\,\mu C = 10^{-6}\,C$,$m_B = 4\,\mu g = 4 	imes 10^{-9}\,kg$,$r_1 = 1\,mm = 10^{-3}\,m$,$r_2 = 9\,mm

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None of these

Vedclass · Answer

(D) Given: $q_A = 1\,\mu C = 10^{-6}\,C$,$q_B = 1\,\mu C = 10^{-6}\,C$,$m_B = 4\,\mu g = 4 	imes 10^{-9}\,kg$,$r_1 = 1\,mm = 10^{-3}\,m$,$r_2 = 9\,mm = 9 	imes 10^{-3}\,m$.By the law of conservation of energy,the loss in electrostatic potential energy equals the gain in kinetic energy:$\Delta U = \Delta K$$k q_A q_B \left( \frac{1}{r_1} - \frac{1}{r_2} ight) = \frac{1}{2} m_B v^2$Substituting the values:$9 	imes 10^9 	imes (10^{-6}) 	imes (10^{-6}) \left( \frac{1}{10^{-3}} - \frac{1}{9 	imes 10^{-3}} ight) = \frac{1}{2} 	imes 4 	imes 10^{-9} 	imes v^2$$9 	imes 10^{-3} \left( 1000 - \frac{1000}{9} ight) = 2 	imes 10^{-9} 	imes v^2$$9 	imes 10^{-3} 	imes 1000 \left( 1 - \frac{1}{9} ight) = 2 	imes 10^{-9} 	imes v^2$$9 	imes \frac{8}{9} = 2 	imes 10^{-9} 	imes v^2$$8 = 2 	imes 10^{-9} 	imes v^2$$v^2 = 4 	imes 10^9$$v = \sqrt{40 	imes 10^8} = 2 	imes 10^4 	imes \sqrt{10} \approx 6.32 	imes 10^4\,m/s$.Since this value is not among the options,the correct choice is $D$.

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