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Three charges .q. each are at vertices of equilateral triangle of side .r.. How much charge should be placed at the centroid so that the system remains in equilibrium? |
Answer» Solution : Due to three identical charges kept at three CORNERS of a triangle, null point (E=0) is FORMED at centre O. LET .Q. be the charge placed at centre of triangle then the charge .Q. is in equilibrium. For system to be in equilibrium the force on each charge must be zero. Consider the forces on charge at C. Let `F_(A), F_(B) and F_(0)` be the forces on the charge at .C. due to the charge at A. B and .Q. RESPECTIVELY. `F_(A) = F_(B) = (1)/(4pi in_(0))(q^(2))/(a^(2))` Angle between `VEC(F)_(A) and vec(F)_(B) " is " 60^(@)` The magnitude of resultant of `vec(F)_(A) and vec(F)_(B)= sqrt3 xx (1)/(4pi in_(0))(q^(2))/(a^(2))` The direction of resultant is along OC. The force on C due to ..Q.. is `vec(F)_(O) = (1)/(4pi in_(0))(Qq)/(((a)/(sqrt3))^(2))= 3 xx (1)/(4pi in_(0))(Qq)/(a^(2))` (along CO) The resultant force on charge at C is zero. `vec(F)= vec(F)_(A) + vec(F)_(B) + vec(F)_(O)= 0` `sqrt3 (1)/(4pi in_(0))(q^(2))/(cancel(r)^(2)) + (1)/(4pi in_(0)) (3Q cancel(q))/(cancel(r)^(2))= 0` `rArr Q = - (q)/(sqrt3)` Negative sign indicates that the force `vec(F)_(O)` is opposite to the resultant of `vec(F)_(A) and vec(F)_(B)`. |
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