Two small balls having equal positive charges Q (coulomb) on each are suspended by two insulating st

Q: Why is the angle exactly $180^\circ$?

Because there is no gravity to pull the balls downwards. The electrostatic repulsion pushes them as far apart as possible, which is a straight line ($180^\circ$).

Q: Why is the distance $2L$ and not $L$?

Each string has a length $L$. Since they are stretched out in opposite directions from the common hook, the total distance between the balls is $L + L = 2L$.

Q: What happens to the tension if the charge $Q$ is doubled?

According to Coulomb's law, the force (and thus the tension) is proportional to the product of the charges ($Q^2$). If $Q$ is doubled, the tension becomes four times greater.

Q: Does the mass of the balls affect the answer?

In a state of weightlessness, the mass does not contribute to any weight ($mg = 0$). However, if the setup were accelerating, the mass would matter due to pseudo forces.

Q: Why don't the strings break?

We assume the strings are ideal and strong enough to withstand the tension $T = \frac{Q^2}{16\pi\varepsilon_0 L^2}$. If the charge is too large, a real string might break!

Question

Two small balls having equal positive charges $Q$ (coulomb) on each are suspended by two insulating strings of equal length $L$ (metre) from a hook fixed to a stand. The whole setup is taken in a satellite into space where there is no gravity (state of weightlessness). The angle between the strings is ...... and the tension in each string is ...... newton.

Sigma Percentile · Accepted Answer

### The Weightless Environment

Imagine taking a classic physics experiment and launching it into space. We have two small balls, each carrying a positive charge $Q$, hanging from a common hook by insulating strings of length $L$. Normally, on Earth, gravity pulls these balls downwards, creating a V-shape. But here, we are in a satellite orbiting in space. There is absolutely no gravity!

Without gravity to pull them down, what forces are acting on the balls? The only force at play is the **electrostatic repulsion** between the two positive charges.

### The Battle of Forces

Because like charges repel, the two balls will push each other as far apart as physically possible. The maximum separation occurs when the strings are stretched out in a straight horizontal line. This makes the angle between them exactly **$180^\circ$**.

Now, let's zoom in on one of the charges and analyze the forces acting on it. The string pulls the ball inwards towards the hook with a tension $T$. Simultaneously, the electrostatic force $F_e$ from the other charge pushes it outwards. Since the charges are at rest relative to the hook, these two forces must perfectly balance each other out.

### Calculating the Tension

We can express this equilibrium mathematically as:

$T = F_e$

To find the electrostatic force, we use **Coulomb's Law**. Notice that because the strings are stretched out in a straight line, the total distance between the two charges is $L + L = 2L$.

Substituting this into Coulomb's Law, we get:

$F_e = \frac{1}{4\pi\varepsilon_0} \frac{Q \cdot Q}{(2L)^2}$

### The Final Result

Let's simplify the denominator. Squaring the distance $2L$ gives us $4L^2$. Multiplying this $4$ by the $4\pi\varepsilon_0$ gives us our final, elegant expression for the tension:

$T = \frac{Q^2}{16\pi\varepsilon_0 L^2}$

It is a beautiful result of symmetry and physics! The tension in each string is entirely determined by the electrostatic repulsion, completely independent of the mass of the balls.

Visualized Solution

$I ni t ia l S e t u p$

$A n g l e b e tw ee n s t r in g s$

$F or ceso n e a c h c ha r g e$

$C a l c u l a t in g T e n s i o n$

$F ina l E x p r ess i o n$

$W ha t i f g r a v i t y w a s p r ese n t ?$

The Sigma Insight: Coulomb's Law

The Weightless Environment

The Battle of Forces

Calculating the Tension

The Final Result

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Visualized Solution

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