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Q27CQ

Expert-verifiedFound in: Page 663

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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 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.

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, *q _{1}* and

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 *q* will be static equilibrium.

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

The charge ** q** is close to

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

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

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

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

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