Question

Using the symmetry of the arrangement, show that the net Coulomb force on the charge q at the center of the square below (Figure 18.46) is zero if the charges on the four corners are exactly equal.

Figure 18.46 Four point chargesq_a, q_b, q_c, and q_d lie on the corners of a square and q is located at its center.

Short answer

The net force on the charge qlocated at the center of the square is zero.

Step-by-step solution

Coulomb law

Coulomb stated that when two-point charges are separated by some distance in space, they attract or repel each other by a force known as electrostatic force.

The expression for the electrostatic force is,

$F=\frac{K{q}_1{q}_2}{r^2}$

Here, role="math" localid="1653573956708" $K$ is the electrostatic force constant, q₁ and q₂are the charges separated by the distance r.

Force at the charge at the center of the square

According to question all charges at the edge of the square are same i.e., . The force on charge located at the center of the square is represented as,

Force acting on the charge q located at the center of the square.

The force of q due to q_a is directed along OD is given as,

$F_a=\frac{KqQ}{r^2}$

The force of q due to q_d is directed along OA is given as,

$F_d=\frac{KqQ}{r^2}$

These two forces are equal in magnitude but opposite and their lines of action meet. Therefore, the forces F_a and F_d cancel each other.

The force of q due to q_b is directed along OC is given as,

$F_b=\frac{KqQ}{r^2}$

The force of q due to q_c is directed along OB is given as,

$F_c=\frac{KqQ}{r^2}$

These two forces are equal in magnitude but opposite and their lines of action meet. Therefore, the forces F_b and F_c cancel each other.

Hence, the net force on qwill be zero.