Problem 23 - Calculus with Concepts in Calculus

Question

If the electric charge is uniformly distributed throughout a circular cylinder (such as a telephone wire) of radius \(a\), then at any point whose distance from the axis of the cylinder is \(r\), the electric field intensity is given by $$ E(r)=\left\\{\begin{array}{ll} c r & \text { for } 0 \leq r \leq a \\ c a^{2} / r & \text { for } r>a \end{array}\right. $$ where \(c\) is a positive constant. a. Show that \(E(r)\) is maximum for \(r=a\). b. Is \(E\) differentiable at \(a ?\) Explain your answer.

Step-by-Step Solution

Answer

The maximum value of \(E(r)\) is at \(r=a\), but \(E\) is not differentiable at \(r=a\).

Step 1: Understanding the Problem

For the given problem, we need to evaluate when the electric field intensity \(E(r)\) is at maximum and check if \(E\) is differentiable at \(r=a\). Given the piecewise function for \(E(r)\): \(c r\) for \(0 \leq r \leq a\) and \(c a^{2} / r\) for \(r > a\), we must analyze the behavior of these expressions.

Step 2: Analyze Maximum for the Interval \(0 \leq r \leq a\)

Consider \(E(r) = c r\) for \(0 \leq r \leq a\). To find the maximum value of \(E(r)\) in this interval, we evaluate it at the endpoints, \(r = 0\) and \(r = a\). Since \(E(0) = 0\) and \(E(a) = ca\), the maximum on this interval is \(E(a) = ca\).

Step 3: Analyze Maximum for the Interval \(r > a\)

Now consider \(E(r) = \frac{ca^2}{r}\) for \(r > a\). As \(r\) increases, \(E(r)\) decreases because \(r\) is in the denominator. Thus, \(E(r)\) achieves its highest value just as \(r\) reaches \(a\), giving \(E(a) = ca\).

Step 4: Compare and Confirm Maximum at \(r=a\)

Comparing results from Steps 2 and 3, \(E(r)\) is maximum at \(r = a\) in both intervals since \(E(a) = ca\) is larger than at any other point \(r > a\) and is equal to the maximum at \(0 \leq r \leq a\). Thus, \(E(r)\) is maximum at \(r = a\).

Step 5: Check Differentiability at \(r=a\)

To check differentiability at \(r=a\), we must examine the derivative of \(E(r)\) from the left and right of \(r=a\). For \(0 \leq r < a\), \(E'(r) = c\). For \(r > a\), \(E'(r) = -\frac{ca^2}{r^2}\). Evaluating the right-hand derivative at \(r=a\) gives \(-\frac{ca^2}{a^2} = -c\). Since \(E'(r)\) is not continuous at \(r=a\) (left derivative \(c\) is not equal to right derivative \(-c\)), \(E(r)\) is not differentiable at \(r=a\).

Key Concepts

Piecewise Functions

Piecewise functions are unique mathematical constructs that allow us to define a function differently over multiple intervals. This is particularly useful when a function exhibits different behaviors at various parts of its domain. In the problem at hand, we have the electric field intensity, denoted as \(E(r)\), which is described using a piecewise function. The function changes its definition based on the distance \(r\) from the cylinder's axis:

• For the interval \(0 \leq r \leq a\), the function is given by \(E(r) = cr\).
• For \(r > a\), it changes to \(E(r) = \frac{ca^2}{r}\).

These expressions encapsulate the behavior of the electric field inside and outside the cylinder. The piecewise nature of \(E(r)\) ensures that each segment or interval is addressed with precise calculations, optimizing the function's behavior across its entire domain.

Differentiability of Functions

Differentiability is a vital concept that explores whether a function has a derivative at each point in its domain. Understanding differentiability involves checking the continuity and smoothness of the function's graph. For the function \(E(r)\), evaluating differentiability at \(r = a\) is crucial because it's the transition point between different expressions in the piecewise function.
To analyze differentiability, we compare the left-hand and right-hand derivatives at \(r = a\):

• Left-hand derivative for \(0 \leq r < a\) is \(E'(r) = c\).
• Right-hand derivative for \(r > a\) is \(E'(r) = -\frac{ca^2}{r^2}\). Evaluating this at \(r = a\) results in \(-c\).

Since these derivatives are not equal at \(r = a\) (\(c eq -c\)), \(E(r)\) lacks differentiability at this point. This non-equality indicates a sharp turn or cusp in the function's graph at \(r = a\), where the direction changes abruptly.

Maxima and Minima

Maxima and minima refer to the highest and lowest points that a function can reach within a certain interval. They are critical in understanding the overall behavior and implications of a function, such as efficiency or intensity in physical contexts.
In analyzing \(E(r)\) for maxima, we consider both intervals of the piecewise function separately:

• For \(0 \leq r \leq a\), evaluating \(E(r) = cr\) at the endpoints, we find that the maximum value occurs at \(r = a\), giving us \(E(a) = ca\).
• For \(r > a\), \(E(r) = \frac{ca^2}{r}\) decreases as \(r\) increases, meaning the maximum value right before this decrease is \(E(a) = ca\).

By examining each section, we validate that \(E(r)\) is indeed maximized at \(r = a\). Thus, despite the function's piecewise nature, the maximum electric field intensity from the uniformly distributed charge is located precisely at the cylinder's surface (\(r = a\)). This scenario tells us about the physical properties of the field surrounding the cylinder.