Problem 62 - Single Variable Calculus

Question

Electrical Force of a Conductor A ring-shaped conductor of radius \(a\) carrying a total charge \(Q\) induces an electrical force of magnitude $$ F=\frac{Q}{4 \pi \varepsilon_{0}} \cdot \frac{x}{\left(x^{2}+a^{2}\right)^{3 / 2}} $$ where \(\varepsilon_{0}\) is a constant called the permittivity of free space, at a point \(P\), a distance \(x\) from the center, along the line perpendicular to the plane of the ring through its center. Find the value of \(x\) for which \(F\) is greatest.

Step-by-Step Solution

Answer

The value of \(x\) for which the electrical force is greatest is \(x = \frac{a}{\sqrt{2}}\).

Step 1: Take the derivative of the force equation with respect to X

Since we want to find the maximum force concerning the distance x, we first need to find the critical points. We get the critical points by taking the derivative of the force equation with respect to x: \(\frac{dF}{dx}\) Before taking the derivative, let's rewrite the equation in a more convenient form for differentiation: \(F= c \cdot \frac{x}{\left(x^{2}+a^{2}\right)^{3 / 2}}\), where \(c = \frac{Q}{4 \pi \varepsilon_0}\) Now, apply the derivative with respect to x: \(\frac{dF}{dx}= c \cdot \frac{d}{dx}\left[\frac{x}{(x^2 + a^2)^{3/2}}\right]\)

Step 2: Evaluate the derivative using the quotient rule

Using the quotient rule for differentiation \(\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{vu' - uv'}{v^2}\), where \(u = x\) and \(v = (x^2 + a^2)^{3/2}\): \(u = x\), so \(u' = 1\) \(v = (x^2 + a^2)^{3/2}\), so \(v' = \frac{3}{2}(x^2 + a^2)^{1/2} \cdot 2x = 3x(x^2 + a^2)^{1/2}\) Now, plug the values of u, u', v, and v' into the quotient rule: \(\frac{dF}{dx}= c \cdot \frac{((x^2 + a^2)^{3/2} - x(3x(x^2 + a^2)^{1/2}))}{((x^2 + a^2)^{3/2})^2}\)

Step 3: Set derivative to zero and solve for x

To find the maximum value, set the derivative equal to zero and solve for x: \(0 = (x^2 + a^2)^{3/2} - x(3x(x^2 + a^2)^{1/2})\) Now, try to solve for x. First, observe that both terms have a factor of \((x^2 + a^2)^{1/2}\). Factor this out: \(0 = (x^2 + a^2)^{1/2}[(x^2 + a^2) - 3x^2]\) This expression will be equal to zero when either of the factors equals zero. 1st factor: \((x^2 + a^2)^{1/2} = 0\) is not possible because it's a square root, and neither x nor a can be imaginary. 2nd factor: \((x^2 + a^2) - 3x^2 = 0\) Now, let's solve the second factor for x:

Step 4: Solve the equation

From the second factor, we have: \((x^2 + a^2) - 3x^2 = 0\) Now, simplify and solve for x: \(a^2 = 3x^2 - x^2\) \(a^2 = 2x^2\) \(\frac{a^2}{2} = x^2\) \(x = \pm\frac{a}{\sqrt{2}}\) Since distance can't be negative, we'll take the positive value: \(x = \frac{a}{\sqrt{2}}\) So the value of x for which F is greatest is \(x = \frac{a}{\sqrt{2}}\).

Key Concepts

Maximizing Electric Force

When analyzing the behavior of electric forces within a conductor, one problem often encountered is determining the point at which the electric force is at its maximum. In the given exercise, the electric force exerted by a ring-shaped conductor is described by the function
\[ F=\frac{Q}{4 pi epsilon_{0}} \frac{x}{(x^{2}+a^{2} )^{3 / 2}} \] This function depends on the distance from a point to the center of the ring. Maximizing this force involves finding the appropriate value of 'x' that would result in the greatest force.

• We are seeking the point where the first derivative of the force with respect to 'x' is equal to zero, which indicates a possible maximum.
• To find this point, one must derive the force function and apply calculus concepts like the derivative and critical points.
• The solution demonstrates that when the derivative of the force with respect to 'x' is set to zero, the point giving the maximum electric force can be found, yielding \( x = \frac{a}{ \backslash sqrt{2}} \).

This concept helps us understand how to control and predict the behavior of electric fields around conductors, which is crucial in many areas of physics and engineering.

Quotient Rule for Differentiation

Differentiation is a fundamental operation in calculus used to find the rate at which a function is changing at any given point. The quotient rule is a specific technique used when differentiating a function that is the ratio of two other functions. In our example, the electrical force depends on both 'x' and 'x' to a function of a higher power. The quotient rule states that for two differentiable functions, 'u' and 'v', the derivative of their quotient is given by
\[ \frac{d}{dx} \frac{u}{v} = \frac{vu' - uv'}{v^{2}} \]

Application of Quotient Rule

In the exercise, 'u' is 'x' and 'v' is \( (x^{2} + a^{2})^{3/2} \). After calculating the derivatives of 'u' and 'v' individually, they are plugged back into the quotient rule equation.

• Understanding how to properly apply the quotient rule is essential for solving complex derivatives and is used extensively in physics, engineering, and economics.
• Since the force function is a ratio of two other functions, the quotient rule is the most appropriate method to find its derivative, which is a necessary step to find where the force is maximized.

By mastering the quotient rule, students not only learn to solve textbook problems, but also gain a tool for analyzing real-world situations involving rates of change.

Critical Points in Calculus

Critical points are essential concepts in calculus, representing the x-values in a function's domain where the function's derivative is zero or undefined. These points are critical for determining where a function has a local maximum or minimum.

Finding Critical Points

To find a function's critical points, one must first obtain its derivative and then solve for when that derivative equals zero. Going back to our exercise:

• The derivative of the electric force, concerning 'x', will reveal potential maxima or minima for the force experienced at a certain distance.
• By setting this derivative to zero and solving for 'x', we discover the critical points of the force function.
• In our problem, only one critical point is relevant, as negative distances don't make physical sense in this context.

Understanding critical points helps not just with textbook exercises, but in real-world problems where it's important to know the limits or optimal points of a system under examination, such as finding the most efficient shape for a conductor or determining when a physical system reaches its limits. This concept is pervasive in the studies of mathematics, physics, economics, and various engineering disciplines.