Two identical charged spheres suspended from a common point by two massless strings of length l are

Q: Why do we use $\tan\theta \approx \sin\theta$?

The problem explicitly states that the distance $d$ is much smaller than the length $l$ ($d \ll l$). This geometric constraint ensures that the angle $\theta$ remains very small throughout the motion. For small angles, the Taylor series expansions of $\tan\theta$ and $\sin\theta$ are nearly identical, allowing us to safely approximate $\tan\theta$ with $\sin\theta$ to simplify the geometry.

Q: Why is $\frac{dx}{dt}$ taken as velocity $v$?

Velocity is defined as the rate of change of position. Here, $x$ is the separation distance between the two spheres. Therefore, the rate at which this distance changes over time, $\frac{dx}{dt}$, represents the relative velocity at which the spheres are approaching each other.

Q: What happens to the negative sign in $v = -\frac{dx}{dt}$?

The negative sign simply indicates direction—it tells us that the distance $x$ is decreasing over time as the spheres approach each other. Since the options ask for the proportionality of the magnitude of velocity $v$ with respect to $x$, the negative sign is absorbed into the proportionality constant and doesn't affect the final power of $x$.

Q: Why did we differentiate $q^2$ instead of $q$?

You absolutely can differentiate $q = \text{constant} \cdot x^{3/2}$ directly! It will give you $\frac{dq}{dt} = \text{constant} \cdot \frac{3}{2} x^{1/2} \frac{dx}{dt}$. Substituting $\frac{dx}{dt} = -v$ and knowing $\frac{dq}{dt}$ is constant immediately yields $v \propto x^{-1/2}$. Differentiating $q^2$ is just an alternative algebraic path that avoids dealing with fractional powers early on.

Q: Can we assume the electrostatic force is constant?

No! The electrostatic force $F_e = \frac{kq^2}{x^2}$ depends on both the charge $q$ and the distance $x$. Since the charge is leaking (changing) and the spheres are moving (distance $x$ is changing), the force is highly dynamic. That's exactly why we had to establish a relationship between $q$ and $x$ at any arbitrary instant before differentiating.

Question

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, is

Options:

(A) $v \propto x^{-1}$

(B) $v \propto x^{1/2}$

(C) $v \propto x$

(D) $v \propto x^{-1/2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine two identical charged spheres hanging from a common point via massless strings of length $l$. Because they carry the same charge $q$, they repel each other with an electrostatic force $F_e$. The system settles into a dynamic equilibrium where the distance between them is $x$.

Let's draw the free body diagram for one of the spheres. It experiences three primary forces: a downward gravitational force $mg$, an outward electrostatic repulsion $F_e$, and a tension $T$ directed along the string.

### The Master Equation

Since the sphere is in equilibrium at any given instant, we can balance the forces by resolving the tension $T$ into its vertical and horizontal components. The vertical component balances gravity, while the horizontal component balances the electrostatic repulsion:

$$T \cos	heta = mg$$
$$T \sin	heta = F_e$$

By dividing these two equations, we elegantly eliminate the tension $T$, yielding a direct relationship between the angle and the forces:

$$	an	heta = \frac{F_e}{mg}$$

We know from Coulomb's Law that the electrostatic force is given by $F_e = \frac{kq^2}{x^2}$. Substituting this into our equation gives:

$$	an	heta = \frac{kq^2}{x^2 mg}$$

### The Small Angle Approximation

Here is where the physics intuition kicks in. The problem states that the initial distance $d$ is much smaller than the string length $l$ ($d \ll l$). This implies that the angle $	heta$ is very small. For small angles, we can use the approximation $	an	heta \approx \sin	heta$.

Looking at the geometry of the setup, $\sin	heta$ is simply the opposite side divided by the hypotenuse:

$$\sin	heta = \frac{x/2}{l} = \frac{x}{2l}$$

Substituting this back into our force balance equation, we get:

$$\frac{x}{2l} = \frac{kq^2}{x^2 mg}$$

Rearranging this to isolate $q^2$, we find a beautiful relationship between the charge and the separation distance:

$$q^2 = \frac{mg}{2kl} x^3$$

### The Dynamics of Leakage

The problem introduces a dynamic element: the charge is leaking at a constant rate. This means $\frac{dq}{dt}$ is a constant. To find how the velocity $v$ depends on $x$, we must differentiate our equation with respect to time $t$:

$$2q \frac{dq}{dt} = \frac{3mg}{2kl} x^2 \frac{dx}{dt}$$

We know that the velocity of approach is $v = -\frac{dx}{dt}$ (the negative sign indicates that the distance $x$ is decreasing). We also need to substitute $q$ back into the equation. From our earlier relation, $q = \left( \frac{mg}{2kl} ight)^{1/2} x^{3/2}$.

Plugging these in, we get:

$$2 \left( \frac{mg}{2kl} ight)^{1/2} x^{3/2} \frac{dq}{dt} = \frac{3mg}{2kl} x^2 (-v)$$

### Final Calculation

Since $\frac{dq}{dt}$, $m$, $g$, $k$, and $l$ are all constants, we can group them together. The equation simplifies to a proportionality:

$$x^{3/2} \propto x^2 v$$

Rearranging to solve for $v$, we divide both sides by $x^2$:

$$v \propto \frac{x^{3/2}}{x^2}$$

$$v \propto x^{3/2 - 2}$$

**$$v \propto x^{-1/2}$$**

This elegant result tells us that as the spheres get closer (as $x$ decreases), their velocity of approach actually increases!

Select Answer:

Visualized Solution

Free Body Diagram

Balancing Forces

Eliminating Tension

Small Angle Approximation

Charge-Distance Relation

Differentiating with respect to time

Substituting $q$ and $v$

Velocity-Distance Relation

What if?

The Sigma Insight: Coulomb's Law

Analyzing the Setup

The Master Equation

The Small Angle Approximation

The Dynamics of Leakage

Final Calculation

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