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At what separation should two equal charges, 10 C each,he placed so that the force between them equals theweight of a 50 kg person?

Answer»

Given:Magnitude of charges, q1 = q2 = 1 CElectrostatic force between them, F = Weight of a 50 kg person = mg = 50 × 9.8 = 490 NLet the required distance be r.By Coulomb's Law, electrostatic force, F=14πε0q1q2r2⇒490=9×109×1×1r2⇒r2=9×109490

⇒r=√949×108=37×104 m = 4.3×103 m



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