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Chapter 11: Problem 1535

Two identical charged spheres suspended from a common point by two massless strings of length \(\ell\) are initially a distance d \((d<<\ell)\) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the spheres approach each other with a velocity \(\mathrm{v}\). Then function of distance \(\mathrm{x}\) between them becomes \(\ldots \ldots\) (A) \(v \propto x\) (B) \(\mathrm{v} \propto \mathrm{x}^{(-1 / 2)}\) (C) \(\mathrm{v} \propto \mathrm{x}^{-1}\) (D) \(\mathrm{v} \propto \mathrm{x}^{(1 / 2)}\)

Short Answer

Expert verified
The relationship between the velocity \(v\) and the distance \(x\) between the spheres is given by \(v \propto x^{-1}\). (Option C)

Step by step solution

01

So, the electrostatic force will be \(\text{F} = \frac{k \cdot q^2}{x^2}\). #Step 2: Determine the gravitational force acting on the spheres The gravitational force between two massive spheres is given by Newton's Law of Gravitation, which states that the force \(\text{F}\) = \(\frac{G \cdot m_1 \cdot m_2}{r^2}\), where \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses of the spheres, and \(r\) is the distance between them. In this case, we can represent the masses as \(m\) and the distance as \(x\).

So, the gravitational force will be \(\text{F} = \frac{G \cdot m^2}{x^2}\). #Step 3: Find the effective force on the spheres The effective force on the spheres can be found by subtracting the gravitational force from the electrostatic force.
02

Thus, the effective force on the spheres is \(F_\text{eff} = \frac{k \cdot q^2}{x^2} - \frac{G \cdot m^2}{x^2}\). #Step 4: Determine the acceleration of the spheres using Newton's second law Newton's second law states that Force = mass × acceleration, i.e., \(\text{F} = \text{m} \cdot \text{a}\). In this case, the force acting on the spheres is \(F_\text{eff}\) and the mass is given by \(m\). We will denote the acceleration as "a" and divide the force by the mass to compute it.

So, the acceleration is \(\text{a} = \frac{F_\text{eff}}{m} = \frac{k \cdot q^2}{m \cdot x^2} - \frac{G \cdot m}{x^2}\). #Step 5: Find the relationship between the velocity \(v\) and the distance \(x\) The velocity function can be related to the acceleration using the integral \(\text{v}(\text{x}) = \int \text{a}(x) dx\). We can find the relationship by integrating the acceleration function with respect to \(x\).
03

First, we rewrite the acceleration function as \(\text{a} = A \cdot \frac{1}{x^2} - B \cdot \frac{1}{x^2}\), where \(A = \frac{k \cdot q^2}{m}\) and \(B = G \cdot m\).

Now, we integrate the function \(\int ( A \cdot \frac{1}{x^2} - B \cdot \frac{1}{x^2} ) dx\).
04

The result is \(\int ( A \cdot \frac{1}{x^2} - B \cdot \frac{1}{x^2} ) dx = A \cdot (-\frac{1}{x}) - B \cdot (-\frac{1}{x}) + C\), where \(C\) is the integration constant.

So, the relationship between velocity and distance is \(v(x) \propto \frac{1}{x}\), which corresponds to the option (C) \(v \propto x^{-1}\).

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Most popular questions from this chapter

Iwo points are at distances a and b \((a

Three charges, each of value \(Q\), are placed at the vertex of an equilateral triangle. A fourth charge \(q\) is placed at the centre of the triangle. If the charges remains stationery then, \(q=\ldots \ldots \ldots\) (A) \((\mathrm{Q} / \sqrt{2})\) (B) \(-(\mathrm{Q} / \sqrt{3})\) (C) \(-(Q / \sqrt{2})\) (D) \((\mathrm{Q} / \sqrt{3})\)

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