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two equal point charges A and Bare R distance apart. a third point charge placed on the perpendicular bisector at a a distance d from the center will experience maximum electrostatic force when |
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Answer» Answer: A third point CHARGE PLACED on the perpendicular bisector at a DISTANCE 'd' from the centre. Let charge on each particle is q and charge on third particle is Q. but only y = R/2√2 , d²F/dy² < 0 and force will be MAXIMUM. so, at y = R/2√2 electrostatic force will be maximum. |
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