Problem 18 - Calculus

Question

The potential, \(\phi\), of a charge distribution at a point on the \(y\) -axis is given by $$ \phi=\left\\{\begin{array}{ll} 2 \pi \sigma(\sqrt{y^{2}+a^{2}}-y) & \text { for } y \geq 0 \\ 2 \pi \sigma(\sqrt{y^{2}+a^{2}}+y) & \text { for } y<0 \end{array}\right. $$ where \(\sigma\) and \(a\) are positive constants. [Hint: To graph this function, take, for example, \(2 \pi \sigma=1\) and \(a=1.1\) (a) Is \(\phi\) continuous at \(y=0 ?\) (b) Do you think \(\phi\) is differentiable at \(y=0 ?\)

Step-by-Step Solution

Answer

(a) Yes, \(\phi\) is continuous at \(y=0\). (b) No, \(\phi\) is not differentiable at \(y=0\).

Step 1: Evaluate the Potential at y=0

To determine continuity at a point, first evaluate the potential \(\phi\) at \(y = 0\). For \(y \geq 0\), \(\phi(y) = 2 \pi \sigma (\sqrt{y^2 + a^2} - y)\). Substituting \(y = 0\), we have \(\phi(0) = 2 \pi \sigma (\sqrt{0^2 + a^2} - 0) = 2 \pi \sigma a\). For \(y < 0\), \(\phi(y) = 2 \pi \sigma (\sqrt{y^2 + a^2} + y)\). Substituting \(y = 0\) gives the same result: \(\phi(0) = 2 \pi \sigma a\). Thus, the potential is the same from both sides at \(y=0\).

Step 2: Check Continuity at y=0

For continuity, the left-hand limit and right-hand limit as \(y\) approaches 0 from negative and positive sides must equal \(\phi(0)\). As \(y \to 0^+\), \(\phi(y) \to 2 \pi \sigma a\). As \(y \to 0^-\), \(\phi(y) \to 2 \pi \sigma a\). Both limits equal \(\phi(0)\), confirming \(\phi\) is continuous at \(y=0\).

Step 3: Evaluate Differentiability Around y=0

Differentiability requires the derivative to be continuous at \(y=0\). Calculate the derivative from either side of \(y=0\). For \(y \geq 0\), \(\frac{d\phi}{dy} = 2 \pi \sigma \left(\frac{y}{\sqrt{y^2+a^2}} - 1\right)\). For \(y < 0\), \(\frac{d\phi}{dy} = 2 \pi \sigma \left(\frac{y}{\sqrt{y^2+a^2}} + 1\right)\). Evaluate these at \(y = 0\). They yield different values: \(-2 \pi \sigma\) and \(2 \pi \sigma\) respectively, indicating a discontinuity at \(y=0\).

Step 4: Conclusion on Differentiability

Since the derivatives from left and right of \(y=0\) are not equal, \(\phi\) is not differentiable at \(y = 0\). The discontinuity in derivative values implies a sharp change or cusp at this point.

Key Concepts

Continuity

Continuity of a function at a point is fundamental in understanding how smooth a function behaves at that particular point. For the potential function \(\phi\) given in the exercise, continuity needs the function value at \( y = 0\) to be equal to the limits from both the positive and negative sides approaching this point.

• Evaluating \( \phi(0)\) : For both \( y \geq 0\)\ and \( y < 0\),\ the function evaluates to \( 2 \pi \sigma a\). This ensures that \( \phi(0)\) is consistent, regardless of the direction from which \( y\) approaches zero, establishing a potential continuity here.
• Evaluating Limits: Both the left-hand limit, when \( y \to 0^-\),\ and right-hand limit, when \( y \to 0^+\)\, approach \( 2 \pi \sigma a\)\, verifying that these are equal to \( \phi(0)\).\

Since all three quantities \( \phi(0)\),\ the right-hand limit, and the left-hand limit are equal, \( \phi\)\ is indeed continuous at the point \( y=0\).\ This implies no abrupt changes or gaps in its value at that point, and smoothly transitions through \( y=0\).\

Differentiability

Differentiability of a function at a point requires not just that it is continuous, but also that its derivative is continuous and smooth at that point. Let’s explore whether \( \phi\),\ the potential function, is differentiable at \( y = 0\):

• Derivative Evaluation: For \( y \geq 0\),\ the derivative \( \frac{d\phi}{dy} = 2 \pi \sigma \left(\frac{y}{\sqrt{y^2+a^2}} - 1\right)\).\ For \( y < 0\),\ it becomes \( 2 \pi \sigma \left(\frac{y}{\sqrt{y^2+a^2}} + 1\right)\).\
• Checking at \( y = 0\): When evaluated at \( y = 0\),\ the derivatives yield differing outcomes: \( -2 \pi \sigma\)\ from the positive side and \( 2 \pi \sigma\)\ from the negative side.\ These varying derivative values highlight a break in the function's slope at \( y = 0\)\.

Due to this break or discontinuity in slope, the function does not possess a derivative that is smooth and continuous at \( y=0\).\ This discontinuity indicates the presence of a cusp or sharp corner at this point, hence making \( \phi\)\ non-differentiable there. Differentiability and continuity are closely linked, yet a function can be continuous without being differentiable at a point.

Charge Distribution

The charge distribution parameter \( \sigma\)\ in the potential function \( \phi\)\ plays a crucial role in determining how the potential \( \phi\)\ behaves along the \( y\)-axis. Understanding charge distribution is vital in comprehending the electric field's influence produced by a distribution of charges.

• Uniformity: In this problem, \( \sigma\)\ represents a constant, emphasizing a uniform distribution of charge density across the surface or space considered. This uniformity simplifies the mathematical expression of potential, making it predictable and analyzable.
• Effect on Potential: The value of the potential \( \phi\)\ directly scales with \( \sigma\),\ indicating if a charge is more densely packed (higher \( \sigma\)), the potential increases proportionally. <
• \( \sigma\)\ helps quantify the effect of electric charges in terms of voltage potentials, illustrating how concentrated the electric force is in any given region.

Understanding \( \sigma\)'s impact provides insights into how changes in charge distribution lead to variations in \( \phi\)\, influencing electric fields and forces experienced by other charges within the space considered.