Electrostatics

JEE Main level, single correct, medium

99 of 311

Question 99 of 311

Two particles AA and BB having charges 20μC20\mu C and −5μC-5\mu C respectively are held fixed with a separation of 5cm5cm At what position a third charged particle should be placed so that it does not experience a net electric force ?

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Answer: (A) At 5cm5cm from −5μC-5\mu C on the right side


Charge should be placed near
−5μC-5\mu C at xx distance

K(20)Q(5+x)2=K(5)QX2\dfrac{K(20)Q}{(5+x{)}^{2}}=\dfrac{K(5)Q}{{X}^{2}}
4x2=(5+x)24{x}^{2}=(5+x{)}^{2}
4x2=25+x2+10x4{x}^{2}=25+{x}^{2}+10x
3x2−10x−25=03{x}^{2}-10x-25=0 ⇒x=5\Rightarrow x=5

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