Two identical charges +q are fixed at points A Nd B at a distance 2a a...
Problem Statement:
Two charges +q are fixed at points A and B at a distance 2a apart. A third particle with charge -q and mass M is thrown with a speed u at point C. From the midpoint C, we need to find the maximum distance this particle will travel before coming to rest.
Solution:
- Initial Analysis: Initially, the particle at C will experience a force from both charges at A and B, leading to an acceleration. As it moves towards A or B, the force from the opposite charge will decrease until it eventually comes to rest.
- Calculating Force: The force on the particle at C due to the charge at A is k*q*q/(2a)^2 and in the same way due to the charge at B is k*q*q/(2a)^2. The net force can be calculated by summing these two forces.
- Calculating Acceleration: Using Newton's second law, F = m*a, we can find the acceleration of the particle. The acceleration will be maximum at the midpoint C.
- Calculating Distance: To find the maximum distance traveled by the particle, we can use the equation of motion, v^2 = u^2 + 2*a*s, where v is the final velocity (0 m/s), u is the initial velocity, a is acceleration, and s is the distance traveled.
- Substitute Values: Substitute the known values into the equation and solve for s to find the maximum distance the particle will travel before coming to rest.
- Final Answer: The maximum distance the particle will travel can be calculated using the above steps.
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