Problem 87 - Master Resource Book in JEE Main Physics

Question

A circular ring of radius \(R\) with uniformly distributed charge \(q\) is placed in their \(y z\)-plane with its centre at the origin. Select the correct statement(s) out of the following. (a) The electric intensity is maximum at \(x=\pm \sqrt{2} R\) (b) The electric intensity is maximum at \(x=\pm \frac{\sqrt{2}}{2} R\) (c) The maximum intensity has a magnitude \(\frac{q}{6 \sqrt{3} \pi \varepsilon_{0} R^{2}}\) (d) The maximum intensity is \(\frac{q}{6 \sqrt{6} \pi \varepsilon_{0} R^{2}}\)

Step-by-Step Solution

Answer

The correct statements are (b) and (c).

Step 1: Understanding the Problem

We have a circular ring of radius \( R \) uniformly charged with total charge \( q \). This ring is placed in the \( yz \)-plane with its center at the origin. We need to find where the electric intensity is maximum along the x-axis, a perpendicular line through the center of the ring, and calculate its value.

Step 2: Electric Field Expression

The electric field due to a charged ring along the axial line at a point on the x-axis a distance \( x \) from the origin is given by the formula:\[E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{qx}{(x^2 + R^2)^{3/2}}\]where \( \varepsilon_0 \) is the permittivity of free space.

Step 3: Maximizing Electric Intensity

To find the maximum electric field \( E \) with respect to \( x \), we take the derivative of \( E \) with respect to \( x \) and set it to zero:\[\frac{dE}{dx} = \frac{1}{4 \pi \varepsilon_0} \cdot \left[ \frac{q(x^2 + R^2)^{3/2} - 3qx^2(x^2 + R^2)^{1/2}}{(x^2 + R^2)^3} \right] = 0\]Simplifying and solving this equation leads to \( x = \pm \frac{R}{\sqrt{2}} \). Thus, the maximum electric intensity occurs at these points.

Step 4: Calculating Maximum Electric Intensity

Substituting \( x = \pm \frac{R}{\sqrt{2}} \) into the electric field expression, we get:\[E_{max} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q \cdot \frac{R}{\sqrt{2}}}{\left(\frac{R^2}{2} + R^2\right)^{3/2}}\]\[= \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q \cdot \frac{R}{\sqrt{2}}}{\left(\frac{3R^2}{2}\right)^{3/2}}\]\[= \frac{q}{6 \sqrt{3} \pi \varepsilon_0 R^2}\]

Step 5: Selecting the Correct Statements

Based on the analysis and calculations, the maximum electric field occurs at \( x = \pm \frac{R}{\sqrt{2}} \), which corresponds to option (b). The value of the maximum intensity matches option (c). Thus, the correct answers are (b) and (c).

Key Concepts

Electrostatics

Electrostatics is the branch of physics that deals with the study of electric charges at rest. It explores how charged objects interact with each other through the electric forces they exert. This field is crucial because it helps us understand the fundamentals of electricity and magnetism, laying the groundwork for various technologies.
Some core concepts within electrostatics include:

• Coulomb's Law: Explains the force between two point charges—proportional to the product of their magnitudes and inversely proportional to the square of their distance.
• Electric Field: Describes how a charged particle would experience force in space, calculated using electric field lines that represent the field's strength and direction.
• Potential Energy: Refers to the work done to bring a charge from infinity to a point in space against the electric force.

In the context of this problem, electrostatics is the foundational principle behind understanding how a charged ring creates an electric field. It helps us derive the value and position of maximum electric field intensity in relation to the ring's structure.

Maximum Electric Field

The concept of a maximum electric field in this scenario involves determining the point along the x-axis where the electric field intensity reaches its highest value. For a charged ring, the electric field intensity varies along this axis due to the symmetry and distribution of the charge.
To find the maximum electric field:

• Use Analytical Methods: Derive the mathematical expression representing the electric field along the axis.
• Differentiate and Solve: Take the derivative of the electric field expression concerning distance x, set it to zero to find critical points.
• Substitute Values: Solve for particular values of x to identify the location of maximum intensity.

Using these steps allows us to ascertain that the electric field is maximized at a distance of \( x = \pm \frac{R}{\sqrt{2}} \) from the ring, demonstrating the application of calculus in physics to find optimal points. This insight is crucial in designing systems like particle accelerators and electron beam devices where precise control over electric fields is necessary.

Permittivity of Free Space

The permittivity of free space, represented by \( \varepsilon_0 \), is a physical constant that describes how electric fields interact in a vacuum. It's a measure of the ability of space to permit electric field lines, affecting the force between charges. This constant appears in almost all formulas involving electromagnetism, including the expression for the electric field of charged objects.
Key points about \( \varepsilon_0 \):

• Its value is approximately \( 8.85 \times 10^{-12} \text{ F/m} \) (farads per meter).
• Essential for calculating the force between charges in Coulomb’s law: \( F = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1q_2}{r^2} \).
• Influences the behavior of electric fields in various mediums; in this problem, it helps us determine the field produced by a charged ring.

Recognizing the role of the permittivity of free space is essential for understanding how electric fields propagate in vacuums or materials, affecting technologies in telecommunications, capacitors, and more complex electric systems.