Question

Considering Figure, suppose that q_a=q_d and q_b=q_c . First show that q is in static equilibrium. (You may neglect the gravitational force.) Then discuss whether the equilibrium is stable or unstable, noting that this may depend on the signs of the charges and the direction of displacement of qfrom the center of the square.

Short answer

If all the charges at the edge of the square and the charge q at the center are like charges, the charge q will be in stable equilibrium.

If qhas a charge opposite to any of the pairs of equal charges, the charge will be in unstable equilibrium.

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, Kis the electrostatic force constant, q₁ and q₂ are the charges separated by the distancer

Force at the charge at the center of the square

According to question all charges at the opposie edge of the square are same i.e., $q_a=q_d=Q$and $q_b=q_c=Q^{\text{'}}$. The force on charge q 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,

role="math" localid="1653576615174" $F_b=\frac{Kq{Q}^{\text{'}}}{r^2}$

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

$F_c=\frac{Kq{Q}^{\text{'}}}{r^2}$

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.

Thus, the net force on q will be zero. Hence, the charge qwill be static equilibrium.

When all charges are positive and the displacement is along the diagonal

Consider the charge to be displace along OD. The force on the charge is shown as,

Force on charge q when it is displaced to point P.

The charge q is close to q_d, hence, $F_d>F_a$. The net force is directed along OA. In addition, the charge q is at equal distance from q_b and q_c, which follows that they have equal magnitudes i.e., F_b=F_c and the resultant force due to q_b and q_c acts along the OD.

The resultant of all the forces causes the charge to move towards the midpoint O, where the net force reduces to zero. The charge may continue along OA due to the velocity gain due to acceleration. When it reaches the point along OA which is at the same distance as P from O, it stops and retraces its path. The charge oscillates about the mean position O

When charges at A and D are negative

Consider the charge to be displace along OD. The force on the charge is shown as,

Force on charge q when it is displaced to point P.

The charge q is close to q_d, hence, F_d>F_a. The net force is directed along OD. In addition, the charge q is at equal distance from q_b and q_c, which follows that they have equal magnitudes i.e., F_b = F_c and the resultant force due to q_b and q_c acts along the OD.

The resultant of all the forces causes the charge to move towards D. Thus, the charge will be in unstable equilibrium.