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Four identical charges each .q. are placed at four corners of a square of side .a.. Find the charge to be placed at the centre of the square so that the system of charges is in equilibrium |
Answer» Solution : Due to four IDENTICAL charges kept at four corners of a square, null point (E= 0) is formed at CENTRE O. Let .Q. be the charge placed at centre of square 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), F_(D) and F_(O)` be the forces on the charge at .C. due to the charges at A, B, D, and .O. respectively `F_(B) = F_(D) = (1)/(4pi in_(0)) (q^(2))/(a^(2))` Angle between `vec(F)_(B) and vec(F)_(D)" is " 90^(@)` The magnitude of resultant of `vec(F)_(B) and vec(F)_(D) = sqrt2 xx (1)/(4pi in_(0)) (q^(2))/(a^(2))` The direction of resultant is along AC. The force `F_(A) = (1)/(4pi in_(0)) (q^(2))/((sqrt2a)^(2)) = (1)/(4pi in_(0)) (q^(2))/(2a^(2))`. It acts along AC. The force on C due to ..Q.. is `vec(F)_(O) = (1)/(4pi in_(0)) (q^(2))/(((a)/(sqrt2))^(2))= 2 xx (1)/(4pi in_(0))(q^(2))/(a^(2))` The resultant force on charge at C is zero. `vec(F) = vec(F)_(A) + vec(F)_(B) + vec(F)_(D) + vec(F)_(O) = 0` `F= (1)/(4pi in_(0))[(q^(2))/(2a^(2)) + sqrt2. (q^(2))/(a^(2)) + (2Qq)/(a^(2))]=0` `(q)/(a^(2)) [(q)/(2) + sqrt2q + 2Q] = 0 rArr q [(1 + 2 sqrt2)/(2)] + 2Q= 0` `2Q = -q ((1+ 2 sqrt2)/(2)) rArr Q = (-q)/(4) (2 sqrt2 + 1)` Negative sign indicates that the forces `vec(F)_(O)` is OPPOSITE to the resultant of `vec(F)_(A), vec(F)_(B) and vec(F)_(D)` i.e., along `vec(CO)` |
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