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

Answer: The potential difference between the points at \(x = 0\) and \(x = x_{\max}\) is \(\Delta V = E_{0}\left(1-e^{-1}\right)\).

Step-by-step solution

Determine the first derivative of the electric field with respect to x

Using the given equation, the x-component of the electric field can be written as \(E_x=E_{0} xe^{-x}\), where \(E_{0}\) is a constant. To find the maximum value of the electric field, we can take the first derivative of \(E_x\) with respect to x: \(\frac{dE_x}{dx} = E_{0}\left(e^{-x} - xe^{-x}\right)\).

Set the first derivative equal to zero and solve for x

Now, we will find the maximum value of the electric field by setting the first derivative equal to zero: \(0 = e^{-x} - xe^{-x}\). We can rewrite this equation as: \(0 = e^{-x}(1 - x)\). Solving for x, we get: \(x = 1\). So, the electric field has its largest value at \(x_{\max} = 1\).

Calculate the potential difference between the points

The potential difference (\(\Delta V\)) between two points in an electric field can be calculated by integrating the electric field with respect to x: \( \Delta V = V(x_{\max}) - V(0) = - \int_{0}^{x_{\max}} E_x dx\). We have the expression for \(E_x\) as \(E_{0}xe^{-x}\), so we can plug it into the integral: \(\Delta V = -\int_{0}^{1} E_{0}xe^{-x}dx\). To evaluate this integral, we can use integration by parts: Let \(u = x\) and \(dv = E_{0}e^{-x}dx\), then \(du = dx\) and \(v = - E_{0}e^{-x}\). Now we can evaluate the integral: \(\Delta V = -\left(- E_{0}xe^{-x}\Big|_{0}^{1} - E_{0} \int_{0}^{1} e^{-x}dx\right)\) \(= E_{0}\left(1e^{-1} - 0 - \left[-e^{-x}\right]_{0}^{1}\right)\) \(= E_{0}\left(1e^{-1} - 0 - (-e^{-1} + 1)\right)\) \(\Delta V=E_{0}\left(1-e^{-1}\right)\) The potential difference between the points at \(x = 0\) and \(x = x_{\max}\) is \(\Delta V = E_{0}\left(1-e^{-1}\right)\).