Question

Four equally charged particles with charge q are placed at the corners of a square with side length L, as shown in the figure below. A fifth charged particle with charge Q is placed at the center of the square so that the entire system of charges is in static equilibrium. What are the magnitude and sign of the charge Q? (Use any variable or symbol stated above as necessary.) magnitude Q =

Answer 1

Force at the center of uniform charge distribution is always zero. As square with equal charges at corners, so the net force at the center is zero. So whatever be the magnitude and direction of Q, it will be at equilibrium.

Let us consider any charge at the corners, let us pick charge 1.

Force between two charges is

Force on charge 1 due to 2 and 4 is same in magnitude as distance between then is same, and it is

as the forces are perpendicular to each other, so net force on 1 due to 2 and 4 is obtained by vector sum of the two forces

Now separation between 1 and 3 is equal to diagonal of square which is square root 2 multiplied by length. So force between 1 and 3 is

Net force on any charge at corner due to other charges at corner is thus,

Force on a corner charge due to charge Q which is at a distance half of diagonal, is

For corner charges to be at equilibrium, the forces must be equal and opposite. So Q has to be negative.