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# Two electrons lying 10cm apart are released. What will be their speed when they are 20cm apart?

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#### Let their velocities at 20cm a part be v. Then ΔKETotal=2[Me2v2−m2(0)2]=Mev2 Potential energy 10 cm (U Initial ) =−1ke210100=10ke2Potential Energy 20 cm (U final ) =ke220/100=5ke2ΔUTotal=Ufinal−Uinitial=5ke2−10ke2=−5ke2As electrostatics force is conservative, ΔKETotal+ΔPETotal=0⇒mev2−5ke2=0⇒v=√5ke2me=35.58m/s

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