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Expert-verified Found in: Page 663 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # Considering Figure, suppose that qa=qd and qb=qc . 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 q from the center of the square.

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

See the step by step solution

## Step 1: 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, K is the electrostatic force constant, q1 and q2 are the charges separated by the distance r

## Step 2: 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 qa is directed along OD is given as,

${F}_{a}=\frac{KqQ}{{r}^{2}}$

The force of q due to qd 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 Fa and Fd cancel each other.

The force of q due to qb 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 qc 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 Fb and Fc cancel each other.

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

## Step 3: 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 qd, hence, ${F}_{d}>{F}_{a}$. The net force is directed along OA. In addition, the charge q is at equal distance from qb and qc, which follows that they have equal magnitudes i.e., Fb=Fc and the resultant force due to qb and qc 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

## Step 4: 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 qd, hence, Fd>Fa. The net force is directed along OD. In addition, the charge q is at equal distance from qb and qc , which follows that they have equal magnitudes i.e., Fb = Fc and the resultant force due to qb and qc 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. ### Want to see more solutions like these? 