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Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d(d << l) apart because of their mutual repulsion.
The charge begins to leak from both the spheres at a constant rate. As a result charges approach each other with a velocity v. Then as a function of distance x between them,
The charge begins to leak from both the spheres at a constant rate. As a result charges approach each other with a velocity v. Then as a function of distance x between them,
- a)v ∝ x–1
- b)v ∝ x½
- c)v ∝ x
- d)v ∝ x–½
Correct answer is option 'D'. Can you explain this answer?
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Option (d)
When charge begins to leak from both the spheres at a constant rate, then
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However, I will try my best to explain the solution.
Let's consider the situation when the spheres are at rest, hanging from the common point. At this point, the electrostatic force between them is balanced by the tension in the strings.
Let the charge on each sphere be q and the distance between them be d. The electrostatic force between them is given by:
F = (1/4πε) * (q^2 / d^2)
where ε is the permittivity of the medium between the spheres.
Let T be the tension in the strings. Since the strings are massless, the tension is the same throughout their length. Let the angle between the strings be θ.
Resolving the forces along the strings, we get:
2T cos(θ/2) = F
Resolving the forces perpendicular to the strings, we get:
2T sin(θ/2) = mg
where m is the mass of each sphere and g is the acceleration due to gravity.
Dividing these equations, we get:
tan(θ/2) = (F/mg) / 2
Substituting the value of F, we get:
tan(θ/2) = (1/8πε) * (q^2 / md^2g)
Now, let's apply the given perturbation. The spheres are displaced by a small angle α in opposite directions. Let the new angle between the strings be θ'. We need to find the new electrostatic force between the spheres and the new tension in the strings.
The new angle between the strings is given by:
θ' = θ + α
Substituting the value of tan(θ/2), we get:
tan(θ'/2) = tan[(θ/2) + (α/2)]
Using the formula for tan(A+B), we get:
tan(θ'/2) = (tan(θ/2) + tan(α/2)) / (1 - tan(θ/2) * tan(α/2))
Substituting the value of tan(θ/2), we get:
tan(θ'/2) = [(1/8πε) * (q^2 / md^2g) + tan(α/2)] / [1 - (1/8πε) * (q^2 / md^2g) * tan(α/2)]
The new tension in the strings is given by:
2T' sin(θ'/2) = mg
Substituting the value of tan(θ'/2), we get:
2T' = mg / sin(θ'/2) = mg / [sin(θ/2) cos(α/2) + sin(α/2) cos(θ/2)]
Simplifying this expression, we get:
2T' = 2T * [1 + (tan(α/2) / tan(θ/2))]^-1
Substituting the value of tan(θ/2), we get:
2T' = 2T * [1 + (8πε / q^2) * (md^2g / α)]^-1
Therefore, the new electrostatic force between the spheres is:
F' = 2T' cos(θ'/2)
Substituting the value of
Let's consider the situation when the spheres are at rest, hanging from the common point. At this point, the electrostatic force between them is balanced by the tension in the strings.
Let the charge on each sphere be q and the distance between them be d. The electrostatic force between them is given by:
F = (1/4πε) * (q^2 / d^2)
where ε is the permittivity of the medium between the spheres.
Let T be the tension in the strings. Since the strings are massless, the tension is the same throughout their length. Let the angle between the strings be θ.
Resolving the forces along the strings, we get:
2T cos(θ/2) = F
Resolving the forces perpendicular to the strings, we get:
2T sin(θ/2) = mg
where m is the mass of each sphere and g is the acceleration due to gravity.
Dividing these equations, we get:
tan(θ/2) = (F/mg) / 2
Substituting the value of F, we get:
tan(θ/2) = (1/8πε) * (q^2 / md^2g)
Now, let's apply the given perturbation. The spheres are displaced by a small angle α in opposite directions. Let the new angle between the strings be θ'. We need to find the new electrostatic force between the spheres and the new tension in the strings.
The new angle between the strings is given by:
θ' = θ + α
Substituting the value of tan(θ/2), we get:
tan(θ'/2) = tan[(θ/2) + (α/2)]
Using the formula for tan(A+B), we get:
tan(θ'/2) = (tan(θ/2) + tan(α/2)) / (1 - tan(θ/2) * tan(α/2))
Substituting the value of tan(θ/2), we get:
tan(θ'/2) = [(1/8πε) * (q^2 / md^2g) + tan(α/2)] / [1 - (1/8πε) * (q^2 / md^2g) * tan(α/2)]
The new tension in the strings is given by:
2T' sin(θ'/2) = mg
Substituting the value of tan(θ'/2), we get:
2T' = mg / sin(θ'/2) = mg / [sin(θ/2) cos(α/2) + sin(α/2) cos(θ/2)]
Simplifying this expression, we get:
2T' = 2T * [1 + (tan(α/2) / tan(θ/2))]^-1
Substituting the value of tan(θ/2), we get:
2T' = 2T * [1 + (8πε / q^2) * (md^2g / α)]^-1
Therefore, the new electrostatic force between the spheres is:
F' = 2T' cos(θ'/2)
Substituting the value of
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Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d(d << l) apart because of their mutual repulsion.The charge begins to leak from both the spheres at a constant rate. As a result charges approach each other with a velocity v. Then as a function of distance x between them,a)v ∝ x–1b)v ∝ x½c)v ∝ xd)v ∝ x–½Correct answer is option 'D'. Can you explain this answer?
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