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Two point charges A and B of charges 3.0 × 10−9C and 6 × 10−9C are kept fixed at a distance 3cm in the horizontal plane. Another particle of charge -3 × 10−9C and mass 6.48 × 10−9kg is released from the rest at the middle of the line joining A and B. a. calculate the potential at the mid point of AB b. What is the speed of the particle P after moving a distance of 1 cm?
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Two point charges A and B of charges 3.0 × 10−9C and 6 × 10−9C are kep...
Given:
Charges A and B have charges 3.0 × 10^(-9) C and 6 × 10^(-9) C respectively.
Distance between charges A and B is 3 cm.
Another particle P has charge -3 × 10^(-9) C and mass 6.48 × 10^(-9) kg.
P is released from rest at the midpoint of the line joining charges A and B.
To Find:
a. Potential at the midpoint of charges A and B.
b. Speed of particle P after moving a distance of 1 cm.
Answer:
1. Calculation of Potential at the Midpoint of AB:
The potential at a point due to a point charge is given by the formula:
V = k * (q / r)
where V is the potential, k is the electrostatic constant (9 × 10^9 Nm^2/C^2), q is the charge, and r is the distance from the point charge.
The potential due to charge A at the midpoint of AB is:
V_A = k * (q_A / r_A)
where q_A = 3.0 × 10^(-9) C (charge of A) and r_A = 1.5 cm (distance from A to the midpoint).
Similarly, the potential due to charge B at the midpoint of AB is:
V_B = k * (q_B / r_B)
where q_B = 6 × 10^(-9) C (charge of B) and r_B = 1.5 cm (distance from B to the midpoint).
The total potential at the midpoint is the sum of the potentials due to charges A and B:
V_midpoint = V_A + V_B
2. Calculation of Speed of Particle P after Moving 1 cm:
The force experienced by particle P due to the electric field created by charges A and B can be calculated using Coulomb's law:
F = k * (q_P * q_A) / r^2 + k * (q_P * q_B) / (d - r)^2
where F is the force, k is the electrostatic constant, q_P is the charge of particle P, q_A and q_B are the charges of A and B respectively, r is the distance between P and A, and d is the distance between charges A and B.
The force experienced by P is equal to the mass of P multiplied by its acceleration:
F = m * a
From the above equations, we can equate the forces and solve for acceleration:
k * (q_P * q_A) / r^2 + k * (q_P * q_B) / (d - r)^2 = m * a
Once we have the acceleration, we can calculate the final speed of P using the equation of motion:
v^2 = u^2 + 2 * a * s
where v is the final speed, u is the initial speed (which is 0 as P is released from rest), a is the acceleration, and s is the distance traveled by P (1 cm in this case).
Conclusion:
a. The potential at the midpoint of charges A and B can be calculated by summing the potentials due to A and B individually.
b. The speed of particle P after
Charges A and B have charges 3.0 × 10^(-9) C and 6 × 10^(-9) C respectively.
Distance between charges A and B is 3 cm.
Another particle P has charge -3 × 10^(-9) C and mass 6.48 × 10^(-9) kg.
P is released from rest at the midpoint of the line joining charges A and B.
To Find:
a. Potential at the midpoint of charges A and B.
b. Speed of particle P after moving a distance of 1 cm.
Answer:
1. Calculation of Potential at the Midpoint of AB:
The potential at a point due to a point charge is given by the formula:
V = k * (q / r)
where V is the potential, k is the electrostatic constant (9 × 10^9 Nm^2/C^2), q is the charge, and r is the distance from the point charge.
The potential due to charge A at the midpoint of AB is:
V_A = k * (q_A / r_A)
where q_A = 3.0 × 10^(-9) C (charge of A) and r_A = 1.5 cm (distance from A to the midpoint).
Similarly, the potential due to charge B at the midpoint of AB is:
V_B = k * (q_B / r_B)
where q_B = 6 × 10^(-9) C (charge of B) and r_B = 1.5 cm (distance from B to the midpoint).
The total potential at the midpoint is the sum of the potentials due to charges A and B:
V_midpoint = V_A + V_B
2. Calculation of Speed of Particle P after Moving 1 cm:
The force experienced by particle P due to the electric field created by charges A and B can be calculated using Coulomb's law:
F = k * (q_P * q_A) / r^2 + k * (q_P * q_B) / (d - r)^2
where F is the force, k is the electrostatic constant, q_P is the charge of particle P, q_A and q_B are the charges of A and B respectively, r is the distance between P and A, and d is the distance between charges A and B.
The force experienced by P is equal to the mass of P multiplied by its acceleration:
F = m * a
From the above equations, we can equate the forces and solve for acceleration:
k * (q_P * q_A) / r^2 + k * (q_P * q_B) / (d - r)^2 = m * a
Once we have the acceleration, we can calculate the final speed of P using the equation of motion:
v^2 = u^2 + 2 * a * s
where v is the final speed, u is the initial speed (which is 0 as P is released from rest), a is the acceleration, and s is the distance traveled by P (1 cm in this case).
Conclusion:
a. The potential at the midpoint of charges A and B can be calculated by summing the potentials due to A and B individually.
b. The speed of particle P after
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Two point charges A and B of charges 3.0 × 10−9C and 6 × 10−9C are kept fixed at a distance 3cm in the horizontal plane. Another particle of charge -3 × 10−9C and mass 6.48 × 10−9kg is released from the rest at the middle of the line joining A and B. a. calculate the potential at the mid point of AB b. What is the speed of the particle P after moving a distance of 1 cm?
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Two point charges A and B of charges 3.0 × 10−9C and 6 × 10−9C are kept fixed at a distance 3cm in the horizontal plane. Another particle of charge -3 × 10−9C and mass 6.48 × 10−9kg is released from the rest at the middle of the line joining A and B. a. calculate the potential at the mid point of AB b. What is the speed of the particle P after moving a distance of 1 cm? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Two point charges A and B of charges 3.0 × 10−9C and 6 × 10−9C are kept fixed at a distance 3cm in the horizontal plane. Another particle of charge -3 × 10−9C and mass 6.48 × 10−9kg is released from the rest at the middle of the line joining A and B. a. calculate the potential at the mid point of AB b. What is the speed of the particle P after moving a distance of 1 cm? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two point charges A and B of charges 3.0 × 10−9C and 6 × 10−9C are kept fixed at a distance 3cm in the horizontal plane. Another particle of charge -3 × 10−9C and mass 6.48 × 10−9kg is released from the rest at the middle of the line joining A and B. a. calculate the potential at the mid point of AB b. What is the speed of the particle P after moving a distance of 1 cm?.
Two point charges A and B of charges 3.0 × 10−9C and 6 × 10−9C are kept fixed at a distance 3cm in the horizontal plane. Another particle of charge -3 × 10−9C and mass 6.48 × 10−9kg is released from the rest at the middle of the line joining A and B. a. calculate the potential at the mid point of AB b. What is the speed of the particle P after moving a distance of 1 cm? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Two point charges A and B of charges 3.0 × 10−9C and 6 × 10−9C are kept fixed at a distance 3cm in the horizontal plane. Another particle of charge -3 × 10−9C and mass 6.48 × 10−9kg is released from the rest at the middle of the line joining A and B. a. calculate the potential at the mid point of AB b. What is the speed of the particle P after moving a distance of 1 cm? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two point charges A and B of charges 3.0 × 10−9C and 6 × 10−9C are kept fixed at a distance 3cm in the horizontal plane. Another particle of charge -3 × 10−9C and mass 6.48 × 10−9kg is released from the rest at the middle of the line joining A and B. a. calculate the potential at the mid point of AB b. What is the speed of the particle P after moving a distance of 1 cm?.
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Two point charges A and B of charges 3.0 × 10−9C and 6 × 10−9C are kept fixed at a distance 3cm in the horizontal plane. Another particle of charge -3 × 10−9C and mass 6.48 × 10−9kg is released from the rest at the middle of the line joining A and B. a. calculate the potential at the mid point of AB b. What is the speed of the particle P after moving a distance of 1 cm?, a detailed solution for Two point charges A and B of charges 3.0 × 10−9C and 6 × 10−9C are kept fixed at a distance 3cm in the horizontal plane. Another particle of charge -3 × 10−9C and mass 6.48 × 10−9kg is released from the rest at the middle of the line joining A and B. a. calculate the potential at the mid point of AB b. What is the speed of the particle P after moving a distance of 1 cm? has been provided alongside types of Two point charges A and B of charges 3.0 × 10−9C and 6 × 10−9C are kept fixed at a distance 3cm in the horizontal plane. Another particle of charge -3 × 10−9C and mass 6.48 × 10−9kg is released from the rest at the middle of the line joining A and B. a. calculate the potential at the mid point of AB b. What is the speed of the particle P after moving a distance of 1 cm? theory, EduRev gives you an
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