Question

An electric field varies in space according to this equation: \(\vec{E}=E_{0} x e^{-x} \hat{x}\). a) For what value of \(x\) does the electric field have its largest value, \(x_{\max } ?\) b) What is the potential difference between the points at \(x=0\) and \(x=x_{\max } ?\)

Short answer

b) What is the potential difference between the points at \(x=0\) and \(x=x_{\max}\)?

Step-by-step solution

Part a: Find \(x_{\max}\) where electric field has its largest value

To find the maximum electric field, we need to find the critical points of the given electric field function. First, let's differentiate the electric field equation with respect to \(x\): \(E(x) = E_0 x e^{-x}\) \(\frac{dE(x)}{dx} = E_0 (e^{-x} - xe^{-x})\) Now, let's set the first derivative equal to zero to find the critical points: \(E_0 (e^{-x} - xe^{-x}) = 0\) \(e^{-x}(1 - x) = 0\) The exponential function \(e^{-x}\) is never zero, so we need to solve for \(x\) in the following equation: \(1 - x = 0\) \(x = 1\) Now, let's check the second derivative to confirm that this point is a maximum: \(\frac{d^2E(x)}{dx^2} = E_0 (2x - 2)e^{-x}\) Plug in \(x = 1\): \(\frac{d^2E(x)}{dx^2}(1) = E_0 (2 - 2)e^{-1} = 0\) Since the second derivative is zero, we need to study the sign change of the second derivative around \(x=1\). For \(x<1\), \(\frac{d^2E(x)}{dx^2} > 0\) and for \(x>1\), \(\frac{d^2E(x)}{dx^2} < 0\). This means that at \(x=1\), the electric field has a maximum value. Thus, the largest value of the electric field occurs at \(x_{\max}=1\).

Part b: Calculate the potential difference between \(x=0\) and \(x=x_{\max}\)

To find the potential difference between two points in an electric field, we need to calculate the negative integral of the electric field with respect to the distance (\(x\)) between the two points. The electric potential difference between \(x=0\) and \(x=x_{\max}(=1)\) is given by: \(\Delta V = -\int_{0}^{1} E(x) dx = -\int_{0}^{1} E_0 x e^{-x} dx\) Let's use integration by parts: Let \(u = x\) and \(dv = E_0 e^{-x} dx\), then \(du = dx\) and \(v = -E_0 e^{-x}\) \(\Delta V = -\left[-E_0 x e^{-x}\right]_{0}^{1} + \int_{0}^{1} E_0 e^{-x} dx\) \(\Delta V = -(E_0 (1 e^{-1}) - E_0 (0)) + E_0 [ - e^{-x}]_{0}^{1}\) \(\Delta V = E_0 (1 - e^{-1}) + E_0 (1 - e^{-1})\) \(\Delta V = 2E_0 (1 - e^{-1})\) The potential difference between the points at \(x=0\) and \(x=x_{\max}\) is \(\Delta V=2E_0(1 - e^{-1})\).

Key concepts

Electric Field Differentiation

When studying electric fields in physics, it's common to encounter problems where the electric field varies as a function of position. Differentiation plays a vital role in such scenarios, specifically when we seek to find the maximum value of the electric field. For a function describing an electric field, such as \( \vec{E} = E_0 x e^{-x} \hat{x} \), differentiation helps us to determine how the field changes with respect to position.

In particular, determining the point of maximum electric field involves finding the first derivative of the function, \( \frac{dE}{dx} \), and setting it to zero. This identifies the critical points, which are potential candidates for maximum or minimum values. By further examining the second derivative, \( \frac{d^2E}{dx^2} \), we can confirm the nature of these critical points—whether they correspond to a maximum, minimum, or inflection point.

For our example, after differentiation, we find that the maximum electric field occurs at \( x_{max} = 1 \). This process is known as electric field differentiation and is essential for analyzing electric field distributions and their extremities.

Electric Potential Difference

The concept of electric potential difference is fundamental in understanding electric fields and is central to the study of electrodynamics. The potential difference, often denoted as voltage, represents the work done to move a unit positive charge from one point to another within an electric field. It's the electric field’s potential to do work on a charge and is given by the negative line integral of the electric field along the path the charge moves.

In our example, \( \Delta V \) between two points is found by integrating the negative electric field from one point to the other. At points \( x=0 \) and \( x=x_{max} \) in the electric field \( \vec{E} = E_0 x e^{-x} \hat{x} \) the potential difference is computed as \( \Delta V = -\int_{0}^{1} E(x) dx \). This concept is deeply rooted in the conservation of energy and is a measurable quantity in both theoretical physics and practical electronics.

Integration by Parts in Physics

Integration by parts is a powerful mathematical tool often used in physics to simplify the process of integration, especially when dealing with product of functions. It comes from the product rule for differentiation and provides a way to break down more complex integrals into simpler parts. The formula for integration by parts is given by \( \int u dv = uv - \int v du \), where \( u \) and \( dv \) are functions of the variable of integration.

In the context of our electric potential difference calculation, integration by parts is used to evaluate the integral \( \int_{0}^{1} E_0 x e^{-x} dx \). By choosing \( u = x \) and \( dv = E_0 e^{-x} dx \), we find the potential difference without directly integrating the original expression. This technique showcases how physical problems often require a blend of mathematical tactics for their solutions. Integration by parts is particularly helpful when dealing with integrals of polynomial functions multiplied by exponential or trigonometric functions.