Two Charged particles A and B, each having a charge q are placed a dis...
Position of the third particle for maximum force:
The third particle should be placed on the perpendicular bisector of AB. This is because the electric field due to particles A and B will be symmetrically distributed on either side of the bisector, resulting in cancellation of forces if the third particle is not placed on the bisector. By placing the third particle on the bisector, we ensure that it experiences the maximum force.
Determination of the maximum force:
To determine the magnitude of the maximum force, we can use Coulomb's law, which states that the electric force between two charged particles is given by:
F = (k * |q1 * q2|) / r^2
Where F is the force, k is the electrostatic constant, q1 and q2 are the charges of the particles, and r is the distance between the particles.
In this case, the force experienced by the third particle due to particle A will be:
F1 = (k * |q * Q|) / (d/2)^2
Similarly, the force experienced by the third particle due to particle B will be:
F2 = (k * |q * Q|) / (d/2)^2
Since the particles are placed on the perpendicular bisector, the forces F1 and F2 will have equal magnitudes but opposite directions. Therefore, the net force experienced by the third particle will be:
F_net = F1 - F2
= 2 * (k * |q * Q|) / (d/2)^2
= 4 * (k * |q * Q|) / d^2
The magnitude of the maximum force is given by |F_net|, which is:
|F_net| = 4 * (k * |q * Q|) / d^2
Conclusion:
To maximize the force experienced by the third particle, it should be placed on the perpendicular bisector of AB. The magnitude of the maximum force is given by |F_net| = 4 * (k * |q * Q|) / d^2, where k is the electrostatic constant, q is the charge of particles A and B, Q is the charge of the third particle, and d is the distance between particles A and B.
Two Charged particles A and B, each having a charge q are placed a dis...
D/2√2
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