Three particles, each of mass 1 g and carrying a charge q, are suspended from a common point by insu

Q: Why is the net electrostatic force $\sqrt{3} F_e$?

Each charge experiences two repulsive forces $F_e$ from the other two charges, separated by an angle of $60^\circ$. The resultant of two equal vectors at $60^\circ$ is $2 F_e \cos(60^\circ/2) = \sqrt{3} F_e$.

Q: How do we find the distance $r$ from the centroid to the corner?

In an equilateral triangle of side $a$, the distance from the centroid to a vertex is $a / \sqrt{3}$. You can derive this using basic trigonometry: $r \cos 30^\circ = a/2$.

Q: Why do we use $\cot \theta$ in the force balance equation?

From the equilibrium conditions, $T \sin \theta = mg$ (vertical) and $T \cos \theta = F$ (horizontal). Dividing the horizontal equation by the vertical one gives $F / mg = \cos \theta / \sin \theta = \cot \theta$.

Q: Can we assume $\sin \theta \approx 1$?

Yes! Since $\cos \theta = \sqrt{3}/100 \approx 0.01732$, the angle $\theta$ is very close to $90^\circ$ (specifically $89^\circ$). For such angles, $\sin \theta$ is extremely close to $1$, making $\cot \theta \approx \cos \theta$.

Q: What if the strings were not massless?

If the strings had mass, they would bow outwards due to their own weight and the distributed tension, forming a catenary-like curve. The simple concurrent force model at the particle would no longer be sufficient!

Question

Three particles, each of mass $1 	ext{ g}$ and carrying a charge $q$, are suspended from a common point by insulated massless strings, each $100 	ext{ cm}$ long. If the particles are in equilibrium and are located at the corners of an equilateral triangle of side length $3 	ext{ cm}$, calculate the charge $q$ on each particle. (Take $g = 10 	ext{ m/s}^2$).

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are standing in a room, and from the ceiling, three identical pendulums are hanging. But these aren't just any pendulums—they are charged particles! Because they all carry the same charge $q$, they repel each other. They push each other away until they reach a perfect, symmetrical equilibrium, forming a horizontal equilateral triangle.

This is a classic problem that beautifully marries 3D geometry with electrostatics and Newtonian mechanics. Let's break it down!

### The Master Equation: Electrostatic Force
Focus your attention on just one of the particles. It is being pushed away by the other two particles. 
According to Coulomb's Law, the force from one particle is:
$$F_e = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{a^2}$$

Since the particles form an equilateral triangle, the angle between the two repulsive forces acting on our chosen particle is $60^\circ$. 
Using vector addition, the net horizontal electrostatic force $F$ is:
$$F = 2 F_e \cos(30^\circ) = \sqrt{3} F_e$$

Substituting the known values ($a = 0.03 	ext{ m}$ and $\frac{1}{4\pi\varepsilon_0} = 9 	imes 10^9$):
$$F = \sqrt{3} 	imes \frac{9 	imes 10^9 	imes q^2}{(0.03)^2} = \sqrt{3} 	imes 10^{13} q^2$$

### The Geometry of the Strings
Now, let's look at the suspension strings. The strings are anchored at a common point and extend down to the corners of the triangle. 
Let $r$ be the distance from the centroid of the triangle to any of its corners. From the geometry of an equilateral triangle:
$$r = \frac{a}{\sqrt{3}}$$

If the string makes an angle $	heta$ with the horizontal plane, we can use simple trigonometry in the right-angled triangle formed by the string $l$, the radius $r$, and the vertical height:
$$\cos	heta = \frac{r}{l} = \frac{a}{\sqrt{3}l}$$

Plugging in $a = 3 	ext{ cm}$ and $l = 100 	ext{ cm}$:
$$\cos	heta = \frac{3}{100\sqrt{3}} = \frac{\sqrt{3}}{100} \approx 0.01732$$
This tells us that $	heta \approx 89^\circ$. The strings are almost perfectly vertical!

### Equilibrium: Balancing the Forces
Our particle is at rest, which means the net force acting on it is zero. Let's resolve the forces into vertical and horizontal components.
The tension $T$ in the string has two jobs:
1. Its vertical component balances gravity:
$$T \sin	heta = mg$$
2. Its horizontal component balances the electrostatic repulsion:
$$T \cos	heta = F$$

Dividing the horizontal equation by the vertical equation, the tension $T$ beautifully cancels out:
$$\frac{T \cos	heta}{T \sin	heta} = \frac{F}{mg} \implies F = mg \cot	heta$$

### Final Calculation
We are now ready for the grand finale. Let's substitute everything into our equilibrium equation:
$$\sqrt{3} 	imes 10^{13} q^2 = mg \cot 89^\circ$$

Using $m = 10^{-3} 	ext{ kg}$, $g = 10 	ext{ m/s}^2$, and $\cot 89^\circ \approx 0.01745$:
$$\sqrt{3} 	imes 10^{13} q^2 = (10^{-2}) 	imes 0.01745$$
$$q^2 \approx 1.007 	imes 10^{-17}$$
$$q \approx 3.17 	imes 10^{-9} 	ext{ C}$$

And there we have it! The charge on each particle is $3.17 	imes 10^{-9} 	ext{ C}$. A perfect harmony of forces!

Visualized Solution

\text{System Geometry and Forces}

\text{Net Electrostatic Force } F

\text{Geometric Analysis of the String}

\text{Equilibrium Conditions}

\text{Calculation of Charge } q

The Sigma Insight: Coulomb's Law

Analyzing the Setup Imagine you are standing in a room, and from the ceiling, three identical pendulums are hanging

The Master Equation

The Geometry of the Strings Now, let's look at the suspension strings

Equilibrium

Final Calculation We are now ready for the grand finale

Similar Questions

A charge $q$ is placed at the centre of the line joining two equal charges $Q$ . The system of the three charges will be in equilibrium if $q$ is equal to

Two identical tennis balls each having mass $m$ and charge $q$ are suspended from a fixed point by threads of length $l$ . What is the equilibrium separation when each thread makes a small angle $θ$ with the vertical ?

Three charges $+ Q, q, + Q$ are placed respectively at distance $0, d /2$ and $d$ from the origin on the X-axis. If the net force experienced by $+ Q$ placed at $x = 0$ is zero, then value of $q$ is

If a charge $q$ is placed at the centre of the line joining two equal charges $Q$ such that the system is in equilibrium, then the value of $q$ is

Four charges equal to $- Q$ are placed at the four corners of a square and a charge $q$ is at its centre. If the system is in equilibrium, the value of $q$ is

A charge $Q$ is placed at each of the opposite corners of a square. A charge $q$ is placed at each of the other two corners. If the net electrical force on $Q$ is zero, then $Q / q$ equals

Three charges $- q_{1}, + q_{2}$ and $- q_{3}$ are placed as shown in the figure. The x-component of the force on $- q_{1}$ is proportional to

Five point charges, each of value $+ q$ coulomb, are placed on five vertices of a regular hexagon of side $L$ metre. The magnitude of the force on the point charge of value $- q$ coulomb placed at the centre of the hexagon is ………newton.