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27 Jan 2012 12:44:18 IST

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ELECTROSTATICS, ELECTRIC FIELD AND ELECTRIC POTENTIAL

Physics

The work of electric field done during the displacement of a negatively charged particle towards a fixed positively charged particle is 9J. As a result the distance between the charges has been decreased by half. What work is done by the electric field over the first half of this distance?Ans:3Jhow.

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shivansh shrivastava

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Joined: 19 Jan 2012

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27 Jan 2012 13:13:07 IST

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For initial positn of negativ charg,th potential due to positv charg is x/d (whr d is distanc and x is product of al c0nstants)..at half distanc,potential(V) is 2x/d. .so potential diff.(U) is x/d..it is givn as 9J,so x/d=9. . .at half of ths distanc(1/4 of total distnc). .distnc frm positive charg is 3d/4. .initial V=x/d. .final V=4x/3d . .potential diff,U=x/3d. .also x/d=9. .so U=9/3=3J

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