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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 than 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/(4 pi epsilon_0) (q^2)/(a^2)` The direction of resultant is ALONG AC. The force `F_A = 1/(4 pi epsilon_0) (q^2)/((sqrt(2)a)^2) = 2 xx 1/(4 pi epsilon_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/(4 pi epsilon_0) [(q^2)/(2a^2) + sqrt(2) . (q^2)/(2a^2) + (2Qq)/(a^2)] = 0` `q/(a^2) [q/2 + sqrt(2) q + 2Q] = 0 implies q [(1 + 2sqrt(2))/(2)] + 2Q = 0` `2Q = -q ((1 + 2 sqrt(2))/(2)) implies Q= (-q)/(4) (2sqrt(2) + 1)` Negative sign indicates that the force `vec(F_O)` is opposite to the resultant of `vec(F_A), vec(F_B) and vec(F_D)` i.e., along `bar(CO)`. |
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