Electrostatics

JEE Main level, single correct, medium

30 of 311

Question 30 of 311

Two identical charged particles each having a mass 10g10g and charge 2.0×10−7C2.0\times {10}^{-7}C are placed on a horizontal table with a separation of LL between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.250.25, find the value of LL. [Use g=10ms−2g=10{ms}^{-2} ]

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Answer: (A) 12cm12cm

Normal reaction applied by table on object will be, N=mgN=mg.
Both charges are same, hence a repulsive force will act between them.
Value of force acting will be, kQ2L2\dfrac{k{Q}^{2}}{{L}^{2}}.
Now the maximum value of friction force which table can provide is,
=μN=μmg=\mu N=\mu mg
For equilibrium, kQ2L2=μmg\dfrac{k{Q}^{2}}{{L}^{2}}=\mu mg.
L=kQ2μmg=k0.25mgQ=2kmgQL=\sqrt{\dfrac{k{Q}^{2}}{\mu mg}}=\sqrt{\dfrac{k}{0.25mg}}Q=2\sqrt{\dfrac{k}{mg}}Q
=29×10910×10−3×10(2×10−7)=0.12m=12cm=2\sqrt{\dfrac{9\times {10}^{9}}{10\times {10}^{-3}\times 10}}\left(2\times {10}^{-7}\right)=0.12m=12cm

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