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Two charges q and 2q are placed along the x -axis in front of a grounded infinite conducting plane as shown in the figure.They are located respectively at a distance of 0.5m and 1.5m from the plane.The force acting on the charge q is: (a) (1)/(4 pi varepsilon_(0))(7q^(2))/(2) (b) (1)/(4 pi varepsilon_(0))q^(2) (c) (1)/(4 pi varepsilon_(0))q^(2) (d) (1)/(4 pi varepsilon_(0))(q^(2))/(2) (a) (1)/(4 pi varepsilon_(0))(q^(2))/(2)?
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Two charges q and 2q are placed along the x -axis in front of a ground...
Problem Statement:
Two charges q and 2q are placed along the x-axis in front of a grounded infinite conducting plane as shown in the figure. They are located respectively at a distance of 0.5m and 1.5m from the plane. Find the force acting on the charge q.
Solution:
We know that the force acting on a point charge due to another point charge is given by Coulomb's law:
F = (1/4πε0)(q1q2/r2)
where q1 and q2 are the two charges, r is the distance between them and ε0 is the permittivity of free space.
The force on charge q due to charge 2q is given by:
F1 = (1/4πε0)(q(2q)/(1.5-0.5)2) = (1/4πε0)(7q2/2)
Similarly, the force on charge q due to the grounded conducting plane is given by:
F2 = -2qE2
where E2 is the electric field due to the grounded conducting plane at the position of charge q. Since the plane is grounded, its potential is zero. Therefore, the potential at the position of charge q due to charge 2q and the grounded plane is:
V = (1/4πε0)((2q)/(1.5) - q/(0.5)) = (1/4πε0)(3q/2)
The electric field at the position of charge q is given by:
E = -dV/dx = -3q/(8πε0)
Therefore, the force on charge q due to the grounded conducting plane is:
F2 = -2qE2 = (3q2/(4πε0))
Thus, the total force acting on charge q is:
F = F1 + F2 = (1/4πε0)(7q2/2) + (3q2/(4πε0)) = (1/4πε0)(5q2/2)
Therefore, the correct option is (d) (1/4πε0)(q
Two charges q and 2q are placed along the x-axis in front of a grounded infinite conducting plane as shown in the figure. They are located respectively at a distance of 0.5m and 1.5m from the plane. Find the force acting on the charge q.
Solution:
We know that the force acting on a point charge due to another point charge is given by Coulomb's law:
F = (1/4πε0)(q1q2/r2)
where q1 and q2 are the two charges, r is the distance between them and ε0 is the permittivity of free space.
The force on charge q due to charge 2q is given by:
F1 = (1/4πε0)(q(2q)/(1.5-0.5)2) = (1/4πε0)(7q2/2)
Similarly, the force on charge q due to the grounded conducting plane is given by:
F2 = -2qE2
where E2 is the electric field due to the grounded conducting plane at the position of charge q. Since the plane is grounded, its potential is zero. Therefore, the potential at the position of charge q due to charge 2q and the grounded plane is:
V = (1/4πε0)((2q)/(1.5) - q/(0.5)) = (1/4πε0)(3q/2)
The electric field at the position of charge q is given by:
E = -dV/dx = -3q/(8πε0)
Therefore, the force on charge q due to the grounded conducting plane is:
F2 = -2qE2 = (3q2/(4πε0))
Thus, the total force acting on charge q is:
F = F1 + F2 = (1/4πε0)(7q2/2) + (3q2/(4πε0)) = (1/4πε0)(5q2/2)
Therefore, the correct option is (d) (1/4πε0)(q
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Two charges q and 2q are placed along the x -axis in front of a grounded infinite conducting plane as shown in the figure.They are located respectively at a distance of 0.5m and 1.5m from the plane.The force acting on the charge q is: (a) (1)/(4 pi varepsilon_(0))(7q^(2))/(2) (b) (1)/(4 pi varepsilon_(0))q^(2) (c) (1)/(4 pi varepsilon_(0))q^(2) (d) (1)/(4 pi varepsilon_(0))(q^(2))/(2) (a) (1)/(4 pi varepsilon_(0))(q^(2))/(2)?
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Two charges q and 2q are placed along the x -axis in front of a grounded infinite conducting plane as shown in the figure.They are located respectively at a distance of 0.5m and 1.5m from the plane.The force acting on the charge q is: (a) (1)/(4 pi varepsilon_(0))(7q^(2))/(2) (b) (1)/(4 pi varepsilon_(0))q^(2) (c) (1)/(4 pi varepsilon_(0))q^(2) (d) (1)/(4 pi varepsilon_(0))(q^(2))/(2) (a) (1)/(4 pi varepsilon_(0))(q^(2))/(2)? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Two charges q and 2q are placed along the x -axis in front of a grounded infinite conducting plane as shown in the figure.They are located respectively at a distance of 0.5m and 1.5m from the plane.The force acting on the charge q is: (a) (1)/(4 pi varepsilon_(0))(7q^(2))/(2) (b) (1)/(4 pi varepsilon_(0))q^(2) (c) (1)/(4 pi varepsilon_(0))q^(2) (d) (1)/(4 pi varepsilon_(0))(q^(2))/(2) (a) (1)/(4 pi varepsilon_(0))(q^(2))/(2)? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two charges q and 2q are placed along the x -axis in front of a grounded infinite conducting plane as shown in the figure.They are located respectively at a distance of 0.5m and 1.5m from the plane.The force acting on the charge q is: (a) (1)/(4 pi varepsilon_(0))(7q^(2))/(2) (b) (1)/(4 pi varepsilon_(0))q^(2) (c) (1)/(4 pi varepsilon_(0))q^(2) (d) (1)/(4 pi varepsilon_(0))(q^(2))/(2) (a) (1)/(4 pi varepsilon_(0))(q^(2))/(2)?.
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