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Chapter 17: Problem 2

The maximum field intensity on the axis of a uniformly charged ring of charge \(q\) and radius \(R\) will be (a) \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{3 \sqrt{3} R^{2}}\) (b) \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 q}{3 R^{2}}\) (c) \(\frac{1}{4 \pi E_{0}} \cdot \frac{2 q}{3 \sqrt{3} R^{2}}\) (d) \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3 q}{2 \sqrt{3} R^{2}}\)

Short Answer

Expert verified
Option (c) is correct: \(\frac{1}{4 \pi E_{0}} \cdot \frac{2 q}{3 \sqrt{3} R^{2}}\).

Step by step solution

01

Identifying the Formula for Field Intensity

The electric field intensity on the axis of a charged ring can be calculated using the formula: \[ E = \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{qz}{(R^2 + z^2)^{3/2}} \]where \(q\) is the total charge of the ring, \(R\) is the radius of the ring, and \(z\) is the distance from the center of the ring along its axis.
02

Condition for Maximum Field Intensity

To find the maximum field intensity, we need to determine the value of \(z\) that maximizes \(E(z)\). We do this by taking the derivative of \(E\) with respect to \(z\) and setting it to zero: \[ \frac{dE}{dz} = 0 \]
03

Calculating the Derivative

Compute the derivative of the electric field \(E(z)\): \[ \frac{d}{dz} \left( \frac{qz}{(R^2 + z^2)^{3/2}} \right) \]Applying the quotient rule and simplifying leads to: \[ E'(z) = \frac{q((R^2 + z^2)^{3/2} - 3z^2(R^2 + z^2)^{1/2})}{(R^2 + z^2)^{5/2}} \] Setting \(E'(z) = 0\), we solve for \(z\).
04

Solving for z

Equating the numerator to zero gives:\[ (R^2 + z^2)^{3/2} = 3z^2(R^2 + z^2)^{1/2} \] Dividing both sides by \((R^2 + z^2)^{1/2}\) and simplifying yields:\[ R^2 + z^2 = 3z^2 \] Thus, \( R^2 = 2z^2 \), so \( z = \frac{R}{\sqrt{2}} \).
05

Substituting z to Find Maximum Intensity

Substitute \( z = \frac{R}{\sqrt{2}} \) back into the formula for \(E\):\[ E = \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{q \cdot \frac{R}{\sqrt{2}}}{(R^2 + \left(\frac{R}{\sqrt{2}}\right)^2)^{3/2}} \]Simplifying gives:\[ E = \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{q \cdot \frac{R}{\sqrt{2}}}{\left(\frac{3R^2}{2}\right)^{3/2}} \]On further simplification, this becomes:\[ E = \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{2q}{3\sqrt{3}R^2} \]
06

Selecting the Correct Option

The result \( \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{2q}{3\sqrt{3}R^2} \) matches the option (c): \( \frac{1}{4 \pi E_{0}} \cdot \frac{2 q}{3 \sqrt{3} R^{2}} \), which is the correct answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Charged Ring
A charged ring is a fascinating setup in electromagnetism where charge is distributed evenly along a circular path. Imagine a ring or a loop that has been uniformly charged with a total charge denoted by \(q\). The radius of this ring is represented by \(R\). Now, if you consider a point lying along the axis that passes perpendicularly through the center of this ring, you would find that the electric field has a certain intensity there. This is known as electric field intensity, and the charged ring generates this due to its charge distribution.
  • The electric field created by a charged ring is not uniform but rather varies depending on how far you are from the ring along its axis.
  • Its intensity can be calculated using specific formulas derived from fundamental electrostatic principles.
  • The axial point where we calculate the field is key because it helps in understanding how charges influence regions of space around them.
Calculating field intensity allows us to comprehend the extent of the electric influence the ring has over a certain area, which is crucial for various practical applications in physics and engineering.
Maximum Field Intensity
The point of maximum field intensity on the axis of a charged ring is particularly significant in physics. To determine this point, you first need to derive a formula that gives the field intensity at any given point on the axis of the ring. This involves some calculus, specifically differentiating the field intensity with respect to the distance along the axis, denoted by \(z\).
  • Field intensity depends on the proximity to the charge distribution, with a notable peak point.
  • This peak is where the ring's influence in terms of the electric field is most pronounced.
  • Finding this maximum helps in optimizing designs and applications that rely on electric fields, like sensors and theoretical physics models.
By calculating correctly, you can find that this maximum intensity doesn't exactly occur at the center of the ring (at \(z=0\)), but rather at some distance away, where the combination of all charged elements along the ring most effectively contributes to the field at that point.
Derivative in Calculus
The derivative is a fundamental concept in calculus, often used to find maxima and minima of functions. In this context, the derivative allows us to determine the rate at which the electric field changes as we move along the axis of the charged ring. To find the maximum electric field intensity, we first take the derivative of the electric field function with respect to \(z\), which represents distance from the center of the ring and along its axis.
  • Finding the derivative involves applying rules such as the product rule, chain rule, and quotient rule depending on the complexity of the function.
  • Setting the derivative to zero enables us to find critical points, which might correspond to a maximum, minimum, or saddle point.
  • In this problem, solving the derivative equation gives us the specific \(z\)-value where the electric field is maximized.
By equating the derivative to zero and solving for \(z\), we reveal the point on the axis where the electric field reaches its peak. This is an excellent example of how calculus can be used to solve real-world physics problems and provides critical insights into electromagnetic phenomena.

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Most popular questions from this chapter

The magnitude of electric field \(\mathbf{E}\) in the annual region of a charged cylindrical capacitor (a) is same throughout (b) is higher near the outer cylinder than near the inner cylinder (c) varies as \(-\), where \(r\) is the distance from the axis (d) varies as \(\frac{1}{2^{2}}\), where \(r\) is the distance from the axis

Two charges \(5 \times 10^{-8} \mathrm{C}\) and \(-3 \times 10^{-8} \mathrm{C}\) are located \(16 \mathrm{~cm}\) apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. [NCERT] (a) \(6 \mathrm{~cm}\) from the charge \(-3 \times 10^{-8} \mathrm{C}\) (b) \(6 \mathrm{~cm}\) from the charge \(5 \times 10^{-8} \mathrm{C}\) (c) \(9 \mathrm{~cm}\) from the charge \(-3 \times 10^{-8} \mathrm{C}\) (d) \(9 \mathrm{~cm}\) from the charge \(5 \times 10^{-8} \mathrm{C}\)

A given charge situated at a certain distance from an electric dipole in the end on position, experiences a force \(F\). If the distance of charge is doubled, the force acting on the charge will be (a) \(2 \mathrm{~F}\) (b) \(F / 2\) (c) \(F / 4\) (d) \(F / 8\)

A positive point charge \(q\) is carried from a point \(B\) to a point \(A\) in the electric field of a point charge \(+Q\) at \(O\). If the permittivity of free space is \(\varepsilon_{0}\), the work done in the process is given by (where, \(a=O A\) and \(b=O R\) ) (a) \(\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{1}{a}+\frac{1}{b}\right)\) (b) \(\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{1}{a}-\frac{1}{b}\right)\) (c) \(\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{1}{a^{2}}-\frac{1}{b^{2}}\right)\) (d) \(\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)\)

The electric potential \(V\) at any point \(x, y, z\) (all in metre) in space is given by \(V=4 x^{2}\) volt. The electric field at the point \((1 \mathrm{~m}, 0,2 \mathrm{~m})\) in \(\mathrm{Vm}^{-1}\) is (a) \(-8 \hat{\mathbf{i}}\) (b) \(+8 \hat{i}\) (c) \(-16 \hat{i}\) (d) \(16 \hat{\mathrm{k}}\)
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