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Two charge 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 ? |
Answer» Solution :Here, LET us keep the charge 2q at a distance r from A.  Thus, charge 2q will not EXPERIENCE any force. when, force of repulsion on it DUE to q is balanced by force of attraction on it due to -3 q, at B, where AB=d. thus, force of attraction by -3q=force of repulsion by q `implies(2qxxq)/(4piepsi_(0)x^(2))=(2qxx3q)/(4piepsi_(0)(x+d)^(2))` `implies(x+d)^(2)=3X^(2)` ltBrgt `impliesx^(2)+d^(2)+3xd=3x^(2)` `implies=2x^(2)-d^(2)` ltBrgt `therefore2x^(2)-2dx-d^(2)=0` ltBrgt `x=(d)/(2)+-(sqrt(3)d)/(2)` (Negative sigh be between q and -3q and hence is unadaptable.) `x=-(d)/(2)+(sqrt(3)d)/(2)` `=(d)/(2)(1+sqrt(3))` to the left of q. |
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