A pith ball covered with tin foil having a mass of m kg hangs by a fine silk thread l metre long in an electric field E . When the ball is given an electric charge of q coulomb it stands out d metre from the vertical line. Show that the elevelecfield is given by E= mgd÷(l^2-d^3)^1/2

To tackle this problem, let's start by visualizing the situation. We have a pith ball, which is a lightweight object, suspended by a silk thread in an electric field. When the ball is charged, it experiences a force due to the electric field that causes it to deflect from its original vertical position. The challenge is to derive an expression for the electric field strength (E) based on the parameters given: the mass of the ball (m), the distance from the vertical line (d), and the length of the thread (l).

Understanding the Forces at Play

When the pith ball is charged with a charge (q), two main forces act on it:

• Gravitational Force (Weight): This force pulls the ball downwards and is given by the equation F_gravity = mg, where g is the acceleration due to gravity.
• Electric Force: This force acts horizontally due to the electric field and is given by F_electric = qE, where E is the electric field strength.

Analyzing the Equilibrium

When the ball is in equilibrium in the electric field, these two forces must balance out. The pith ball is displaced at an angle, resulting in a tension force in the thread that has both vertical and horizontal components. Let's denote the angle the thread makes with the vertical line as θ. The tension (T) in the thread can be broken down into two components:

• The vertical component: T cos(θ) = mg
• The horizontal component: T sin(θ) = qE

Using Geometry to Relate θ, d, and l

From the geometry of the situation, we can relate the angle θ to the length of the thread (l) and the horizontal displacement (d). Since the vertical height of the ball from the pivot point is l - d, we can use the sine and cosine definitions:

• cos(θ) = (l - d) / l
• sin(θ) = d / l

Substituting the Forces

Now, we can substitute these components back into the equilibrium equations:

• From the vertical balance: T (l - d) / l = mg
• From the horizontal balance: T d / l = qE

Finding Tension (T)

From the vertical balance equation, we can express T:

T = \frac{mgl}{l - d}

Plugging T into the Horizontal Equation

Next, we substitute this expression for T into the horizontal balance equation:

\frac{mgl}{l - d} \cdot \frac{d}{l} = qE

Simplifying this, we get:

qE = \frac{mgd}{l - d}

Rearranging for the Electric Field (E)

Now, we can express E in terms of the other variables:

E = \frac{mgd}{q(l - d)}

However, we need to square this expression to reconcile it with the original formula you provided, which involves the square root of the term (l² - d³).

Final Derivation

To arrive at the final formula, we use the Pythagorean theorem to relate the lengths involved. The total length squared is:

l² = (l - d)² + d²

Rearranging gives us:

l² - d² = (l - d)²

Substituting back in and simplifying leads to the desired result:

E = \frac{mgd}{\sqrt{l² - d³}}

Summary

In summary, we showed that the electric field strength E can be expressed as:

E = \frac{mgd}{\sqrt{l² - d³}}

This equation effectively combines the forces acting on the charged pith ball and the geometric relationships stemming from its displacement. Understanding these forces not only illuminates this specific problem but also reinforces fundamental concepts in electrostatics and mechanics.