Problem 22 - Calculus with Concepts in Calculus

Question

A ring of radius \(a\) carries a uniform electric charge \(Q\). The electric field intensity at any point \(x\) along the axis of the ring is given by $$ E(x)=\frac{Q x}{\left(x^{2}+a^{2}\right)^{3 / 2}} $$ At what point on the axis is the electric field the greatest?

Step-by-Step Solution

Answer

The electric field is greatest at \( x = \frac{a}{\sqrt{2}} \).

Step 1: Set up the Problem

To find where the electric field is greatest, we need to find the maximum of the function \( E(x) = \frac{Q x}{(x^2 + a^2)^{3/2}} \). This means we need to differentiate \( E(x) \) with respect to \( x \) and find the critical points by setting the derivative to zero.

Step 2: Differentiate the Function

Calculate \( \frac{dE}{dx} \). Using the quotient rule where \( u(x) = Qx \) and \( v(x) = (x^2 + a^2)^{3/2} \), we have \[ \frac{dE}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]This leads to\[ \frac{dE}{dx} = \frac{(x^2 + a^2)^{3/2} \cdot Q - Qx \cdot \frac{3}{2}(x^2 + a^2)^{1/2} \cdot 2x}{(x^2 + a^2)^3} \]

Step 3: Simplify the Derivative

Simplify the derivative:\[ \frac{dE}{dx} = \frac{Q(x^2 + a^2)^{3/2} - 3Qx^2(x^2 + a^2)^{1/2}}{(x^2 + a^2)^3} \]Further simplification gives:\[ \frac{dE}{dx} = \frac{Q((x^2 + a^2) - 3x^2)}{(x^2 + a^2)^{5/2}} \]\[ \frac{dE}{dx} = \frac{Q(a^2 - 2x^2)}{(x^2 + a^2)^{5/2}} \]

Step 4: Find Critical Points

Set \( \frac{dE}{dx} = 0 \) to find the critical points:\[ Q(a^2 - 2x^2) = 0 \]Since \( Q eq 0 \), we have\[ a^2 - 2x^2 = 0 \]which gives\[ 2x^2 = a^2 \] thus \[ x^2 = \frac{a^2}{2} \] therefore \[ x = \pm \frac{a}{\sqrt{2}} \] . Since \( x \) represents a distance along the axis, we will use the positive value.

Step 5: Confirm Maximum

We need to confirm that \( x = \frac{a}{\sqrt{2}} \) is indeed a maximum. This can typically be checked by using the second derivative test or observing the behavior of \( E(x) \) around the critical point. For this problem, the analytical behavior suggests a maximum because of the symmetry and nature of the field lines.

Key Concepts

Quotient Rule

The quotient rule is essential when differentiating ratios of two functions. It helps us find the derivative of a function defined as the division of one function by another. When you have a function in the form of \( f(x) = \frac{u(x)}{v(x)} \), the derivative \( \frac{df}{dx} \) is given by:

• \( \frac{df}{dx} = \frac{v(x) \cdot \frac{du}{dx} - u(x) \cdot \frac{dv}{dx}}{[v(x)]^2} \)

Providing clarity, the notation \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) represent the derivatives of \( u(x) \) and \( v(x) \) respectively. In our exercise, \( u(x) = Qx \) and \( v(x) = (x^2 + a^2)^{3/2} \). We apply the quotient rule to differentiate \( E(x) \), which is crucial to identify where the electric field intensity is maximized. This method breaks down complex derivative problems into manageable parts.

Critical Points

Critical points are values of \( x \) where the derivative of a function is zero or undefined. They play a pivotal role in finding maxima, minima, or points of inflection of functions. Once we find \( \frac{dE}{dx} \) using differentiation, we analyze these points by setting \( \frac{dE}{dx} = 0 \). This will help us figure out where the changes happen in the function.
In our exercise, solving \( Q(a^2 - 2x^2) = 0 \) gives potential critical points. Since \( Q eq 0 \), it simplifies to \( a^2 - 2x^2 = 0 \), resulting in \( x = \pm \frac{a}{\sqrt{2}} \). These are the critical points that may lead us to finding the greatest electric field intensity along the ring's axis.

Second Derivative Test

The second derivative test further analyzes the nature of critical points by determining if they are minima, maxima, or points of inflection. After finding a critical point, we calculate the second derivative \( \frac{d^2E}{dx^2} \). If \( \frac{d^2E}{dx^2} \) is positive at a critical point, it indicates a local minimum; if negative, a local maximum; if zero, the test is inconclusive.
In this problem, we seek to confirm the nature of the critical point \( x = \frac{a}{\sqrt{2}} \). While the explicit calculation of the second derivative may be complex, the problem's symmetrical properties and mathematical insights suggest a local maximum at this point, aligning with the electric field's behavior and symmetry around the axis.