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Chapter 5: Problem 21

Electric tield of a charge distribution: Suppose we have a distribution of charges and we want to calculate the resulting electric field. One way to do this is to first calculate the electric potential \(\phi\) and then take its gradient. For a point charge \(q\) at the origin, the electric potential at a distance \(r\) from the origin is \(\phi=q / 4 \pi \epsilon_{0} r\) and the electric field is \(\mathbf{E}=-\nabla \phi\). a) You have two charges, of \(\pm 1 \mathrm{C}, 10 \mathrm{~cm}\) apart. Calculate the resulting electric potential on a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) square plane surrounding the charges and passing through them. Calculate the potential at \(1 \mathrm{~cm}\) spaced points in a grid and make a visualization on the screen of the potential using a density plot. b) Now calculate the partial derivatives of the potential with respect to \(x\) and \(y\) and hence find the electric field in the \(x y\) plane. Make a visualization of the field also. This is a little trickier than visualizing the potential, because the electric field has both magnitude and direction. One way to do it might be to make two density plots, one for the magnitude, and one for the direction, the latter using the "hsv"

Short Answer

Expert verified
Calculate the electric potential on a grid by summing the potentials from both charges. Use the negative gradient of the potential to find and visualize the electric field.

Step by step solution

01

Understand the Problem

The problem involves calculating the electric potential and electric field for a distribution of two charges, \(\pm 1 \mathrm{C}\), placed \(10 \mathrm{~cm}\) apart. The solution requires calculating the potential on a grid, visualizing it, and then computing and visualizing the electric field.
02

Electric Potential for a Point Charge

Use the formula \(\phi=\frac{q}{4 \pi \epsilon_{0} r}\) to find the electric potential due to a single point charge. For two charges, the total potential is the sum of the potentials due to each charge.
03

Set Up the Grid

Create a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) grid with points spaced \(1 \mathrm{~cm}\) apart. This corresponds to \(101 \times 101\) points.
04

Calculate Potential for Each Grid Point

For each point \( (x, y) \) on the grid, calculate the distance to each charge and use the potential formula to find the total potential: \[\phi_{\text{total}} = \frac{+1 \mathrm{C}}{4 \pi \epsilon_{0} \sqrt{(x-0.05)^2 + y^2}} - \frac{-1 \mathrm{C}}{4 \pi \epsilon_{0} \sqrt{(x+0.05)^2 + y^2}} \].
05

Visualize the Potential

Create a density plot of the potential values over the grid. Using a plotting library, generate a visualization showing the variation in the potential across the \(1 \times 1\)-meter plane.
06

Calculate the Electric Field

Calculate the electric field by taking the negative gradient of the potential: \[ \mathbf{E} = -abla \phi = -\left(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}\right) \]. Use the central difference method to find the partial derivatives numerically.
07

Visualize the Electric Field

Generate plots for the magnitude \[ |\mathbf{E}| \] and the direction. One way to do this is to create a quiver plot that shows the electric field vectors at grid points. Additionally, create a density plot for the magnitude with an 'hsv' colormap for the direction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Potential
The concept of electric potential \( \phi \) describes the potential energy per unit charge at a given point in space due to an electric field. It’s important for understanding how charges interact.
For a point charge \( q \) at the origin, the electric potential \( \phi \) at a distance \( r \) is given by:
\[ \phi = \frac{q}{4 \pi \epsilon_{0} r} \]
The electric potential due to multiple charges is simply the sum of the potentials from each charge. To calculate the total potential in our exercise, you add the potentials given by the positive and negative charges separately.
Gradient
The gradient ( \( \abla \phi \)) of the electric potential reveals the direction of the steepest increase in potential. It is a vector quantity and essential in calculating the electric field. The electric field \( \mathbf{E} \) is found by taking the negative gradient of the potential:
\[ \mathbf{E} = -\abla \phi = -\left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y} \right) \ \]
In numerical methods, the central difference method is used to approximate these partial derivatives. This involves averaging the differences between adjacent points on the grid.
Point Charge
A point charge is an idealized model of a charged particle with negligible size. The electric potential created by a point charge decreases with distance according to the inverse relationship in the formula \( \phi = \frac{q}{4 \pi \epsilon_{0} r} \ \).
In our exercise, the charges are \(+1 \mathrm{C} \) and \(-1 \mathrm{C} \) placed \(10 \mathrm{cm} \) apart. The potential at any point in the grid is influenced by these two charges. For each grid point \((x, y)\), the contributions from both charges must be computed and summed up.
Visualization
Visualization makes abstract concepts like electric potential and fields easier to understand. By creating density plots and vector fields, we can see how potential and electric fields vary across space.
To visualize the electric potential, we create a density plot where color variations represent different potential values. For electric fields, we can use quiver plots to show the direction and density plots with 'hsv' colormap to indicate the direction and magnitude of the field.
These visual tools help in interpreting complex distributions and interactions of charges.
Numerical Methods
Numerical methods are used to solve problems that may not have simple analytical solutions. For example, calculating derivatives and potentials on a grid:
  • Set up a grid: Create a grid of points spaced at regular intervals (e.g., \(1 \mathrm{cm}\))
  • Calculate values: Compute the potential or any relevant quantity at each grid point
  • Use difference methods: Employ central difference or other numerical techniques to estimate derivatives
These methods facilitate the computation of electric fields and potentials for complex charge distributions and enable detailed visualizations.

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Most popular questions from this chapter

Period of an anharmonic oscillator The simple harmonic oscillator crops up in many places. Its behavior can be studied readily using analytic methods and it has the important property that its period of oscillation is a constant, independent of its amplitude, making it useful, for instance, for keeping time in watches and clocks. Frequently in physics, however, we also come across anharmonic oscillators, whose period varies with amplitude and whose behavior cannot usually be calculated analytically. A general classical oscillator can be thought of as a particle in a concave potential well. When disturbed, the particle will rock back and forth in the well: The harmonic oscillator corresponds to a quadratic potential \(V(x) \propto x^{2}\). Any other form gives an anharmonic oscillator. (Thus there are many different kinds of anharmonic oscillator, depending on the exact form of the potential.) One way to calculate the motion of an oscillator is to write down the equation for the conservation of energy in the system. If the particle has mass \(m\) and position \(x\), then the total energy is equal to the sum of the kinetic and potential energies thus: $$ E=\frac{1}{2} m\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2}+V(x) $$ Since the energy must be constant over time, this equation is effectively a (nonlinear) differential equation linking \(x\) and \(t\). Let us assume that the potential \(V(x)\) is symmetric about \(x=0\) and let us set our anharmonic oscillator going with amplitude \(a\). That is, at \(t=0\) we release it from rest at position \(x=a\) and it swings back towards the origin. Then at \(t=0\) we have \(\mathrm{d} x / \mathrm{d} t=0\) and the equation above reads \(E=V(a)\), which gives us the total energy of the particle in terms of the amplitude. a) When the particle reaches the origin for the first time, it has gone through one quarter of a period of the oscillator. By rearranging the equation above for \(\mathrm{d} x / \mathrm{d} t\) and then integrating with respect to \(t\) from 0 to \(\frac{1}{4} T\), show that the period \(T\) is given by $$ T=\sqrt{8 m} \int_{0}^{a} \frac{\mathrm{d} x}{\sqrt{V(a)-V(x)}} $$ b) Suppose the potential is \(V(x)=x^{4}\) and the mass of the particle is \(m=1 .\) Write a Python function that calculates the period of the oscillator for given amplitude a using Gaussian quadrature with \(N=20\) points, then use your function to make a graph of the period for amplitudes ranging from \(a=0\) to \(a=2\). c) You should find that the oscillator gets faster as the amplitude increases, even though the particle has further to travel for larger amplitude. And you should find that the period diverges as the amplitude goes to zero. How do you explain these results?

Heat capacity of a solid Debye's theory of solids gives the heat capacity of a solid at temperature \(T\) to be $$ C_{V}=9 V \rho k_{B}\left(\frac{T}{\theta_{D}}\right)^{3} \int_{0}^{\otimes_{D} / T} \frac{x^{4} \mathrm{e}^{x}}{\left(e^{x}-1\right)^{2}} \mathrm{~d} x $$ where \(V\) is the volume of the solid, \(\rho\) is the number density of atoms, \(k_{B}\) is Boltzmann's constant, and \(\theta_{D}\) is the so-called Debye temperature, a property of solids that depends on their density and speed of sound. a) Write a Python function \(\mathrm{cv}(\mathrm{T})\) that calculates \(C_{V}\) for a given value of the temperature, for a sample consisting of 1000 cubic centimeters of solid aluminum, which has a number density of \(\rho=6.022 \times 10^{28} \mathrm{~m}^{-3}\) and a Debye temperature of \(\theta_{D}=428 \mathrm{~K}\). Use Gaussian quadrature to evaluate the integral, with \(N=50\) sample points. b) Use your function to make a graph of the heat capacity as a function of temperature from \(T=5 \mathrm{~K}\) to \(T=500 \mathrm{~K}\).

5.17 The gamma function: A commonly occurring function in physics calculations is the gamma function \(\Gamma(a)\), which is defined by the integral $$ \Gamma(a)=\int_{0}^{\infty} x^{n-1} \mathrm{e}^{-x} \mathrm{~d} x . $$ There is no closed-form expression for the gamma function, but one can calculate its value for given \(a\) by performing the integral above numerically. You have to be careful how you do it, however, if you wish to get an accurate answer. a) Write a program to make a graph of the value of the integrand \(x^{n-1} \mathrm{e}^{-x}\) as a function of \(x\) from \(x=0\) to \(x=5\), with three separate curves for \(a=2,3\), and 4 , all on the same axes. You should find that the integrand starts at zero, rises to a maximum, and then decays again for each curve. b) Show analytically that the maximum falls at \(x=a-1\). c) Most of the area under the integrand falls near the maximum, so to get an accurate value of the gamma function we need to do a good job of this part of the integral. We can change the integral from 0 to \(\infty\) to one over a finite range from 0 to 1 using the change of variables in Eq. (5.67), but this tends to squash the peak towards the edge of the \([0,1]\) range and does a poor job of evaluating the integral accurately. We can do a better job by making a different change of variables that puts the peak in the middle of the integration range, around \(\frac{1}{2}\). We will use the change of variables given in Eq. (5.69), which we repeat here for convenience: $$ z=\frac{x}{c+x} . $$ For what value of \(x\) does this change of variables give \(z=\frac{1}{2}\) ? Hence what is the appropriate choice of the parameter \(c\) that puts the peak of the integrand for the gamma function at \(z=\frac{1}{2}\) ? d) Before we can calculate the gamma function, there is another detail we need to attend to. The integrand \(x^{n-1} \mathrm{e}^{-x}\) can be difficult to evaluate because the factor \(x^{2-1}\) can become very large and the factor \(\mathrm{e}^{-x}\) very small, causing numerical overflow or underflow, or both, for some values of \(x\). Write \(x^{n-1}=\mathrm{e}^{(a-1) \ln x}\) to derive an alternative expression for the integrand that does not suffer from these problems (or at least not so much). Explain why your new expression is better than the old one. e) Now, using the change of variables above and the value of \(c\) you have chosen, write a user-defined function gamma (a) to calculate the gamma function for arbitrary argument \(a\). Use whatever integration method you feel is appropriate. Test your function by using it to calculate and print the value of \(\Gamma\left(\frac{3}{2}\right)\), which is known to be equal to \(\frac{1}{2} \sqrt{\pi} \simeq 0.886\). f) For integer values of \(a\) it can be shown that \(\Gamma(a)\) is equal to the factorial of \(a-\) 1. Use your Python function to calculate \(\Gamma(3), \Gamma(6)\), and \(\Gamma(10)\). You should get answers closely equal to \(2 !=2,5 !=120\), and \(9 !=362880\).

Rearranging Eq. (5.19) into a slightly more conventional form, we have: $$ \int_{a}^{b} f(x) \mathrm{d} x=h\left[\frac{1}{2} f(a)+\frac{1}{2} f(b)+\sum_{k=1}^{N-1} f(a+k h)\right]+\frac{1}{12} h^{2}\left[f^{\prime}(a)-f^{\prime}(b)\right]+\mathrm{O}\left(h^{4}\right) . $$ This result gives a value for the integral on the left which has an error of order \(h^{4}-a\) factor of \(h^{2}\) better than the error on the trapezoidal rule and as good as Simpson's rule. We can use this formula as a new rule for evaluating integrals, distinct from any of the others we have seen in this chapter. We might call it the "Euler-Maclaurin rule." a) Write a program to calculate the value of the integral \(\int_{0}^{2}\left(x^{4}-2 x+1\right) \mathrm{d} x\) using this formula. (This is the same integral that we studied in Example 5.1, whose true value is \(4.4\).) The order- \(h\) term in the formula is just the ordinary trapezoidal rule; the \(h^{2}\) term involves the derivatives \(f^{\prime}(a)\) and \(f^{\prime}(b)\), which you should evaluate using central differences, centered on \(a\) and \(b\) respectively. Note that the size of the interval you use for calculating the central differences does not have to equal the value of \(h\) used in the trapezoidal rule part of the calculation. An interval of about \(10^{-5}\) gives good values for the central differences. Use your program to evaluate the integral with \(N=10\) slices and compare the accuracy of the result with that obtained from the trapezoidal rule alone with the same number of slices. b) Good though it is, this integration method is not much used in practice. Suggest a reason why not.

Create a user-defined function \(f(x)\) that returns the value \(1+\frac{1}{2} \tanh 2 x\), then use a central difference to calculate the derivative of the function in the range \(-2 \leq x \leq 2\). Calculate an analytic formula for the derivative and make a graph with your numerical result and the analytic answer on the same plot. It may help to plot the exact answer as lines and the numerical one as dots. (Hint: In Python the tanh function is found in the math package, and it's called simply tanh.)
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