A charged ball B hangs from a silk thread S, which makes an angle theta with a large charged conduct

Q: Why is the electric field $E = \sigma/\varepsilon_0$ and not $\sigma/2\varepsilon_0$?

For a single non-conducting sheet of charge, the field is $\sigma/2\varepsilon_0$. However, for a thick conducting sheet, charges reside on both outer surfaces, effectively creating a field of $\sigma/\varepsilon_0$ outside the conductor. Even if you used $\sigma/2\varepsilon_0$, the proportionality $\tan\theta \propto \sigma$ would still hold true!

Q: Why do we divide the two force equations?

Dividing the equations is a standard mathematical trick to eliminate the unknown tension $T$. It directly gives us a trigonometric ratio ($\tan\theta$) that relates the known horizontal and vertical forces.

Q: Does the length of the silk thread affect the angle $\theta$?

Surprisingly, no! If you look at the final equation $\tan\theta = \frac{q\sigma}{mg\varepsilon_0}$, the length of the thread ($L$) does not appear anywhere. The angle depends only on the mass, charge, and the external electric field.

Q: What if the sheet was negatively charged?

If the sheet was negatively charged, the electric force would be attractive instead of repulsive. The ball would swing towards the sheet, and the angle $\theta$ would simply be on the other side of the vertical axis.

Q: Why is the ball considered to be in equilibrium?

Equilibrium means the net force acting on the object is zero, resulting in zero acceleration. Since the ball has settled at a constant angle and is hanging stationary, it is in static equilibrium.

Question

A charged ball $B$ hangs from a silk thread $S$, which makes an angle $	heta$ with a large charged conducting sheet $P$, as shown in the figure. The surface charge density $\sigma$ of the sheet is proportional to

Options:

(A) $\cos	heta$

(B) $\cot	heta$

(C) $\sin	heta$

(D) $	an	heta$

Sigma Percentile · Accepted Answer

### The Setup: A Dance of Forces

Imagine a delicate dance between a charged ball and a massive conducting sheet. The ball, suspended by a silk thread, is pushed away by the electrostatic repulsion from the sheet. It doesn't fly away forever; instead, it settles gracefully at a specific angle, $	heta$, where all the forces perfectly balance each other out. This state of perfect balance is what physicists call **equilibrium**.

To understand this dance, we must first understand the invisible partner: the electric field.

### Unveiling the Electric Field

A large conducting sheet with a surface charge density $\sigma$ creates a uniform electric field in the space around it. According to Gauss's Law, the magnitude of this electric field $E$ is given by:

$$E = \frac{\sigma}{\varepsilon_0}$$

Notice that the electric field is directly proportional to the surface charge density $\sigma$. The more densely packed the charges are on the sheet, the stronger the outward push.

### The Free Body Diagram: Resolving the Tension

Now, let's zoom in on the ball itself and draw a Free Body Diagram (FBD). There are three main actors pulling and pushing on our ball:

1. **Gravity ($mg$)**: Pulling relentlessly downwards.
2. **Electric Force ($qE$)**: Pushing horizontally away from the sheet.
3. **Tension ($T$)**: The silk thread pulling diagonally upwards and towards the pivot.

Because the tension is acting at an angle, it's a bit tricky to work with directly. We need to break it down into a language that gravity and the electric force understand: vertical and horizontal components.

Using basic trigonometry, the upward vertical component of the tension is $T\cos	heta$, and the horizontal component pulling towards the sheet is $T\sin	heta$.

### The Final Connection

Since the ball is in equilibrium, the opposing forces must perfectly cancel each other out.

Vertically, the upward pull of the thread must balance the downward pull of gravity:

$$T\cos	heta = mg$$

Horizontally, the thread's inward pull must balance the electric field's outward push:

$$T\sin	heta = qE$$

We have a beautiful system of two equations. To find the relationship involving the angle $	heta$, we can eliminate the unknown tension $T$ by dividing the horizontal equation by the vertical equation:

$$\frac{T\sin	heta}{T\cos	heta} = \frac{qE}{mg}$$

This simplifies elegantly to:

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

Finally, we substitute our expression for the electric field $E = \frac{\sigma}{\varepsilon_0}$ into this equation:

$$	an	heta = \frac{q\sigma}{mg\varepsilon_0}$$

Look closely at this final equation. The charge of the ball ($q$), its mass ($m$), the acceleration due to gravity ($g$), and the permittivity of free space ($\varepsilon_0$) are all constants. Therefore, the tangent of the angle is directly proportional to the surface charge density:

**$$	an	heta \propto \sigma$$**

And there we have it! The geometry of the thread's angle is a direct reflection of the charge density on the sheet.

Animated Solution for Physics - Electrostatics: A charged ball $B$ hangs from a silk thread $S$ , which makes an angle $θ$ with a large charged conducting sheet $P$ , as shown in the figure. The surface charge density $σ$ of the sheet is proportional to

Select Answer:

Visualized Solution

Visualizing the Physical Setup

Electric Field of a Conducting Sheet

Identifying the Forces

Resolving the Tension

Equilibrium Equations

Finding the Angle Relationship

Final Proportionality

The Sigma Insight: Electric Field Lines, Flux and Gauss's Law

The Setup

Unveiling the Electric Field

The Free Body Diagram

The Final Connection

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