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Consider the charges q, q and -q placed atvertices of an equilateral triangle as shown it figure. What is the force on each charge ? |
Answer» Solution : Situation given in the statement is depicted in the following figure. Here `|q_(1)| = |q_(2)| = |q_(3)|=Q`  Here since AABC is an equilateral TRIANGLE, we have AB = BC = CA = I Magnitudes of mutual Coulombian forces are equal to F. `F_(12) = F_(21) = F_(23) = F_(13) = F_(31) = F = (kq^(2))/l^(2)`....(1) Resultant Coulombian force on `q_(1)` is: `vecF_(1) = vecF_(12) + vecF_(13)` `therefore F_(1) = sqrt(F^(2) + F^(2) + 2F F cos (120^(@))) (therefore F_(12) = F_(13) =F)` `=sqrt(2F^(2) + 2F^(2) (-1/2))` (Here, `theta = 120^(@)`) `=sqrt(F^(2))` `therefore F_(1) = F` (in the direction parallel to `bar(BC)`).....(2) Resultant Coulombian force on `q_(3)`is, `vecF_(3) = vecF_(31) + vecF_(32)` `therefore F_(3) = sqrt(F^(2) + F^(2) + 2F F cos 60^(@)) (therefore F_(31) = F_(32) = F)` `therefore F_(3) = sqrt(2F^(2) + 2F^(2)(1/2))` (Here `theta = 60^(@)`) `=sqrt(3F^(2))` `therefore F_(3) = sqrt(3F)` (in the direction of `vec(CD)`) where D is the midpoint of `bar(AB)` ) (or along the direction bisecting `angle(ACB)`) `therefore vecF_(B) = {(F cos 60^(@))(-hati) + (F sin 60^(@)hatj)}+ Fcos 60^(@) hati + F sin 60^(@) hatj + sqrt(3)F(-hatj)` `=-F/2hati + sqrt(3)/2 Fhatj + F/2hati + sqrt(3)/2 Fhatj - sqrt(3)Fhatj` `=sqrt(3)Fhatj - sqrt(3)Fhatj` `therefore vecF_(B) = vec0` `therefore vecF_(B) =0` Here, if given point charges have equal mass then resultant gravitational force on this isolated system will also be zero. Another alternative Method (Aliter) : Here, resultant Coulombian force acting on whole system is, `vecF_(B) = vecF_(1) + vecF_(2) + vecF_(3)` `therefore =(vecF_(12) + vecF_(13) + vecF_(21) + vecF_(23) + vecF_(31) + vecF_(32))` `=(vecF_(12) + vecF_(21)) + (vecF_(13) + vecF_(31)) + (vecF_(23) + vecF_(32))` `=vec0 + vec0 + vec0` `=vec0` (`therefore` According to Newton.s third law) `vecF_(12) =-vecF_(21).vecF_(13) =-vecF_(31)` and `vecF_(23) =-vecF_(32)` |
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