Four charges equal to + Q are placed at the four corners of a square and a charge (−q) is at its centre. If the system is in equilibrium, then the value of −q is
Question
Four charges equal to + Q are placed at the four corners of a square and a charge (−q) is at its centre. If the system is in equilibrium, then the value of −q is
Solution
The system is in equilibrium, which means the net force on each charge is zero. Let's consider the charge at the center (-q).
The force between two charges is given by Coulomb's law: F = k * |q1*q2| / r^2, where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them.
In this case, the charge at the center (-q) experiences a force from each of the four corner charges (+Q). Because the square is symmetrical, the forces from the opposite corner charges will cancel each other out. Therefore, we only need to consider the force from one pair of adjacent corner charges.
The distance between the center charge and each corner charge is the half-diagonal of the square. If a is the side length of the square, then the half-diagonal is r = a*sqrt(2)/2.
The force from one corner charge on the center charge is F1 = k * |Q*(-q)| / r^2 = k * Qq / (a^2/2).
Because there are two such forces (from a pair of adjacent corner charges), the total force on the center charge is F = 2*F1 = 2k * Qq / (a^2/2) = 4k * Qq / a^2.
For the system to be in equilibrium, this force must be zero. Therefore, we have 4k * Qq / a^2 = 0.
Solving for -q, we get -q = 0. This means the charge at the center must be zero for the system to be in equilibrium.
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