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Two charges q and - 3q are placed fixed on x-axis separated by distance'd'. Where should a third charge 2q be placed such that it will not experience any force ? |
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Answer» Solution :Suppose distance AC = x and AB = d Repulsive force on 2q by q, `E_(q) = (K(q)(2q))/x^(2)`……(1) Attractive force on 2q by `-3q` `E_(-3q)= (k(2q)(3q))/(x+d)^(2)`………..(2) Resultant force on 2q. `F = FQ + F_(-3q)` (`therefore` No force on 2q) `therefore Fq =-F_(-3q)` `k(2q^(2))/x^(2) = +k(6Q^(2))/(x+d)^(2)` `therefore 1/x^(2) =3/(x+d)^(2)` By TAKING square root on both sides, `1/x = sqrt(3)/(x|d) = 1.732/(x+d)` `therefore x+d = 1.732 x` `therefore x +d = 1.732 x` `therefore x = 1.366 d` `=d/2(1+sqrt(3))` |
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