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Two identical charged spheres suspended from a common point by two massless strings of lengths l, are initially at a distance d (d << l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. The v varies as a function of the distance x between the spheres, as
medium
Electric Charges and Fields
2016
physics
v ∝ x − 1 / 2 v \propto x^{-1/2} v ∝ x − 1/2
v ∝ x − 1 v \propto x^{-1} v ∝ x − 1
v ∝ x 1 / 2 v \propto x^{1/2} v ∝ x 1/2
v ∝ x − 1 / 2 v \propto x^{-1/2} v ∝ x − 1/2
Explanation To solve this problem, we need to analyze the forces and motion of the charged spheres. \\
Given: \\
• Two identical charged spheres suspended from a common point by massless strings of length l . l. \\
l . • Initial separation between the spheres is d d d (where d ≪ l ) . d \ll l). \\
d ≪ l ) .
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• Charges leak at a constant rate, causing the spheres to approach each other with velocity
We need to find how
varies as a function of the distance
between the spheres.
Step 1: Analyze the forces acting on the spheres.
The electrostatic force between the spheres is given by Coulomb's law:
F = k q 2 x 2 \\
F = \frac{k q^2}{x^2} \\
F = x 2 k q 2 where
is Coulomb's constant and
is the charge on each sphere.
Step 2: Consider the equilibrium of forces.
At equilibrium, the electrostatic force is balanced by the tension in the strings.
For small angles, the horizontal component of tension
is approximately
T sin θ ≈ T x 2 l . T \sin \theta \approx T \frac{x}{2l}. \\
T sin θ ≈ T 2 l x . Thus,
k q 2 x 2 = T x 2 l . \frac{k q^2}{x^2} = T \frac{x}{2l}. \\
x 2 k q 2 = T 2 l x . Step 3: Relate the rate of change of charge to velocity.
As the charges leak, the force decreases, causing the spheres to move closer.
The rate of change of force is related to the velocity
by:
d d t ( k q 2 x 2 ) = d d t ( T x 2 l ) . \\
\frac{d}{dt} \left( \frac{k q^2}{x^2} \right) = \frac{d}{dt} \left( T \frac{x}{2l} \right). \\
d t d ( x 2 k q 2 ) = d t d ( T 2 l x ) . Step 4: Simplify the expression.
Assuming
decreases at a constant rate,
d q d t = − C \frac{dq}{dt} = -C d t d q = − C (a constant).
Thus,
d d t ( k q 2 x 2 ) = 2 k q x 2 d q d t − 2 k q 2 x 3 d x d t . \frac{d}{dt} \left( \frac{k q^2}{x^2} \right) = \frac{2k q}{x^2} \frac{dq}{dt} - \frac{2k q^2}{x^3} \frac{dx}{dt}. \\
d t d ( x 2 k q 2 ) = x 2 2 k q d t d q − x 3 2 k q 2 d t d x . Substitute
d q d t = − C \frac{dq}{dt} = -C d t d q = − C and
d x d t = v : 2 k q ( − C ) x 2 − 2 k q 2 v x 3 = 0. \frac{dx}{dt} = v: \\
\frac{2k q (-C)}{x^2} - \frac{2k q^2 v}{x^3} = 0. \\
d t d x = v : x 2 2 k q ( − C ) − x 3 2 k q 2 v = 0. Step 5: Solve for
v . 2 k q 2 v x 3 = 2 k q C x 2 v = C x q . v. \\
\frac{2k q^2 v}{x^3} = \frac{2k q C}{x^2} \\
v = \frac{C x}{q}. \\
v . x 3 2 k q 2 v = x 2 2 k qC v = q C x . Since
is decreasing linearly with time,
q ∝ x . q \propto x. \\
q ∝ x . Thus,
v ∝ x − 1 / 2 . v \propto x^{-1/2}. \\
v ∝ x − 1/2 . Therefore, the velocity
varies as
which corresponds to Option 1.