Two positive point charges of 12 and 5 microcoulombs, are placed 10 cm apart in air. The work needed to bring them 4 cm closer isA. 2.4 JB. 3.6 JC. 1.6 JD. 6.0 J

Two positive point charges of 12 and 5 microcoulombs, are placed 10 cm

1.	Two positive point charges of 12 and 5 microcoulombs, are placed 10 cm apart in air. The work needed to bring them 4 cm closer isA. 2.4 JB. 3.6 JC. 1.6 JD. 6.0 J
Answer» Correct Answer - B Potential energy of charges `q_(1) and q_(2)`, r distance aprat `U=(1)/(4pi epsilon_(0))(q_(1)q_(2))/(r)` For r = 0.1 m `U_(1)=(1)/(4pi epsilon_(0))(12xx10^(-6)xx5xx10^(-6))/(0.1)` `=(9xx10^(9)xx60xx10^(-12))/(0.1)=5.4J` For r = 0.06 m, `U_(2)=(9xx10^(9)xx60xx10^(-12))/(0.06)=9J` `therefore" Work done"=(9-5.4)J=3.6J`

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Two positive point charges of 12 and 5 microcoulombs, are placed 10 cm apart in air. The work needed to bring them 4 cm closer isA. 2.4 JB. 3.6 JC. 1.6 JD. 6.0 J

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