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A negative point charge 2q and a positive charge q are fixed at a distance l apart. Where should a positive test charge Q be placed on the line connecting the charge for it to be in equilibrium? What is the nature of the equilibrium with respect to longitudinal motions?
Correct answer is 'a = l(1 + ), the equilibrium will be stable'. Can you explain this answer?
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A negative point charge 2q and a positive charge q are fixed at a dist...
Solution:
To find the position of the test charge Q, let x be the distance of Q from the negative charge 2q.
Let the force on the test charge Q due to the positive charge q be F1 and the force due to the negative charge 2q be F2.
The force F1 is repulsive and the force F2 is attractive.
Hence, the net force on the test charge Q is given by F = F1 - F2.
For equilibrium, the net force must be zero. Therefore, F1 = F2.
Let k be the Coulomb constant and let Q be the magnitude of the test charge.
Then, we have
F1 = k(Qq)/(l-x)^2 (repulsive force)
F2 = k(Q2q)/x^2 (attractive force)
Equating F1 and F2, we get
(Qq)/(l-x)^2 = 2(Q2q)/x^2
Simplifying, we get
x/l = (1 ± sqrt(2))/2
Since x must be less than l, we take the negative sign and get
x/l = (1 - sqrt(2))/2
Therefore, the position of the test charge Q is given by
x = l(1 - sqrt(2))/2
Hence, the test charge Q must be placed at a distance of l(1 - sqrt(2))/2 from the negative charge 2q on the line connecting the charges for equilibrium.
Nature of Equilibrium:
To determine the nature of equilibrium with respect to longitudinal motions, we consider the second derivative of the potential energy U of the test charge Q at the equilibrium position.
If the second derivative is positive, the equilibrium is stable. If the second derivative is negative, the equilibrium is unstable. If the second derivative is zero, further analysis is required.
The potential energy U of the test charge Q is given by
U = k(Qq)/(l-x) - k(Q2q)/x
Substituting x = l(1 - sqrt(2))/2, we get
U = kQqsqrt(2)/(2 - sqrt(2)) - kQ2q(2 - sqrt(2))/2sqrt(2)
Simplifying, we get
U = kQq(2 + sqrt(2))/2 - kQ2q(2 - sqrt(2))/2sqrt(2)
The second derivative of U with respect to x is given by
d2U/dx2 = 2k(Qq)/(l-x)^3 + 2k(Q2q)/x^3
Substituting x = l(1 - sqrt(2))/2, we get
d2U/dx2 = 6kQq/(l(3 - 2sqrt(2))^3) + 6kQ2q/(l(2 - sqrt(2))^3)
Simplifying, we get
d2U/dx2 = (18sqrt(2) - 26)/(4 - 3sqrt(2))^3 * k(Qq+Q2q)/l^3
Since k, Q, q, and l are all positive constants, the second derivative is negative.
Therefore, the equilibrium is unstable with respect to longitudinal motions.
To find the position of the test charge Q, let x be the distance of Q from the negative charge 2q.
Let the force on the test charge Q due to the positive charge q be F1 and the force due to the negative charge 2q be F2.
The force F1 is repulsive and the force F2 is attractive.
Hence, the net force on the test charge Q is given by F = F1 - F2.
For equilibrium, the net force must be zero. Therefore, F1 = F2.
Let k be the Coulomb constant and let Q be the magnitude of the test charge.
Then, we have
F1 = k(Qq)/(l-x)^2 (repulsive force)
F2 = k(Q2q)/x^2 (attractive force)
Equating F1 and F2, we get
(Qq)/(l-x)^2 = 2(Q2q)/x^2
Simplifying, we get
x/l = (1 ± sqrt(2))/2
Since x must be less than l, we take the negative sign and get
x/l = (1 - sqrt(2))/2
Therefore, the position of the test charge Q is given by
x = l(1 - sqrt(2))/2
Hence, the test charge Q must be placed at a distance of l(1 - sqrt(2))/2 from the negative charge 2q on the line connecting the charges for equilibrium.
Nature of Equilibrium:
To determine the nature of equilibrium with respect to longitudinal motions, we consider the second derivative of the potential energy U of the test charge Q at the equilibrium position.
If the second derivative is positive, the equilibrium is stable. If the second derivative is negative, the equilibrium is unstable. If the second derivative is zero, further analysis is required.
The potential energy U of the test charge Q is given by
U = k(Qq)/(l-x) - k(Q2q)/x
Substituting x = l(1 - sqrt(2))/2, we get
U = kQqsqrt(2)/(2 - sqrt(2)) - kQ2q(2 - sqrt(2))/2sqrt(2)
Simplifying, we get
U = kQq(2 + sqrt(2))/2 - kQ2q(2 - sqrt(2))/2sqrt(2)
The second derivative of U with respect to x is given by
d2U/dx2 = 2k(Qq)/(l-x)^3 + 2k(Q2q)/x^3
Substituting x = l(1 - sqrt(2))/2, we get
d2U/dx2 = 6kQq/(l(3 - 2sqrt(2))^3) + 6kQ2q/(l(2 - sqrt(2))^3)
Simplifying, we get
d2U/dx2 = (18sqrt(2) - 26)/(4 - 3sqrt(2))^3 * k(Qq+Q2q)/l^3
Since k, Q, q, and l are all positive constants, the second derivative is negative.
Therefore, the equilibrium is unstable with respect to longitudinal motions.
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A negative point charge 2q and a positive charge q are fixed at a distancelapart. Where should a positive test charge Q be placed on the line connecting the charge for it to be in equilibrium? What is the nature of the equilibrium with respect to longitudinal motions?Correct answer is 'a = l(1 + ), the equilibrium will be stable'. Can you explain this answer?
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