Two particles A and B, each having a charge Q are placed a distance d apart. Where should a particle of charge q be placed on the perpendicular bisector of AB so that it experiences maximum force ? What is the magnitude of the maximum force ?Correct answer is '. 3. a = l(1 + ), the equilibrium will be stable'. Can you explain this answer?

Question

Two particles A and B, each having a charge Q are placed a distance d apart. Where should a particle of charge q be placed on the perpendicular bisector of AB so that it experiences maximum force ? What is the magnitude of the maximum force ?Correct answer is '. 3. a = l(1 + ), the equilibrium will be stable'. Can you explain this answer?

Answer

Solution:

To find the position of a particle of charge q on the perpendicular bisector of AB, we need to find the net force on the particle due to particles A and B.

Let the position of the particle q be at a distance x from A and at a distance (d-x) from B. Let F1 and F2 be the forces on q due to A and B, respectively.

The force F1 on q due to A is given by Coulomb's law as:

F1 = (1/4πε) * (Qq/x^2)

where ε is the electric constant.

Similarly, the force F2 on q due to B is given by:

F2 = (1/4πε) * (Qq/(d-x)^2)

The net force F on q is given by the vector sum of F1 and F2:

F = F1 + F2

= (1/4πε) * Qq * [(1/x^2) + (1/(d-x)^2)]

To find the position of q where F is maximum, we differentiate F with respect to x and equate it to zero:

dF/dx = (1/4πε) * Qq * [-2/x^3 + 2/(d-x)^3] = 0

Simplifying this expression, we get:

x = d/2

This means that the particle q should be placed at the midpoint of AB, i.e., at a distance of d/2 from both A and B, on the perpendicular bisector of AB.

Magnitude of Maximum Force:

To find the magnitude of the maximum force, we substitute x = d/2 in the expression for F:

F = (1/4πε) * Qq * [(1/(d/2)^2) + (1/(d/2)^2)]

= (1/4πε) * Qq * (8/d^2)

= (2Qq/πεd^2)

Therefore, the magnitude of the maximum force is (2Qq/πεd^2).

Stability of Equilibrium:

The equilibrium at x = d/2 is stable because if the particle q is displaced slightly from this position, it experiences a restoring force that tends to bring it back to the equilibrium position.

To see why this is the case, consider the expression for F as a function of x. Near x = d/2, F can be approximated as a quadratic function of x:

F = (1/4πε) * Qq * [(1/(d/2)^2) + (1/(d/2)^2)] + (1/4πε) * Qq * [(2/d^3) - (2/d^3)] * (x - d/2)

= (2Qq/πεd^2) - (4Qq/πεd^3) * (x - d/2)

The coefficient of (x - d/2) is negative, which means that if q is displaced slightly to the left or right of x = d/2, the force on it will be directed back towards x = d/2. This is why the equilibrium at x = d/2 is stable.

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