Chapter 4: Problem 14
Given the electric field \(\mathbf{E}=(y+1) \mathbf{a}_{x}+(x-1) \mathbf{a}_{y}+2 \mathbf{a}_{z}\) find the potential difference between the points \((a)(2,-2,-1)\) and \((0,0,0) ;(b)(3,2,-1)\) and \((-2,-3,4)\).
Short Answer
Expert verified
Answer: The potential difference between points (2, -2, -1) and (0, 0, 0) is 6.5 V, and the potential difference between points (3, 2, -1) and (-2, -3, 4) is 0 V.
Step by step solution
01
Set up the integral for potential difference
The potential difference between two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) is given by:
$$V(x_1, y_1, z_1) - V(x_2, y_2, z_2) = -\int_{(x_1, y_1, z_1)}^{(x_2, y_2, z_2)} \mathbf{E} \cdot d\mathbf{l}$$
To perform the integration, we need to evaluate the line integral of the given electric field along the chosen path between the points.
(task a)
02
Calculate the potential difference between \((2, -2, -1)\) and \((0, 0, 0)\)
For this task, we need to integrate the electric field along the path from \((2, -2, -1)\) to \((0, 0, 0)\). We will choose a straight line between the points and parameterize the path as follows:
$$\mathbf{l}(t) = \langle 2-t, -2 + 2t, -1 + t \rangle$$
with \(0 \le t \le 1\).
Now we need to find the derivative \(\frac{d\mathbf{l}}{dt}\):
$$\frac{d\mathbf{l}}{dt} = \langle -1, 2, 1 \rangle$$
Next, we substitute \(\mathbf{l}(t)\) into the electric field:
$$\mathbf{E}(t) = (t-3)\mathbf{a}_x + (2t+1)\mathbf{a}_y + 2\mathbf{a}_z$$
Now take the dot product of \(\mathbf{E}(t)\) and \(\frac{d\mathbf{l}}{dt}\):
$$\mathbf{E}(t) \cdot \frac{d\mathbf{l}}{dt} = -(t-3) + 4t + 2 = -t + 7$$
Finally, we integrate the dot product from \(t=0\) to \(t=1\) to get the potential difference:
$$V_a = -\int_{0}^{1} (-t+7) dt = \left[-\frac{1}{2}t^2 + 7t\right]{0}^{1} = -\frac{1}{2} + 7 = 6.5\,V$$
(task b)
03
Calculate the potential difference between \((3, 2, -1)\) and \((-2, -3, 4)\)
For this task, we need to integrate the electric field along the path from \((3,2, -1)\) to \((-2, -3, 4)\). We will choose a straight line between the points and parameterize the path as follows:
$$\mathbf{l}(t) = \langle 3 -5t, 2 -5t, -1 + 5t \rangle$$
with \(0 \le t \le 1\).
Now we need to find the derivative \(\frac{d\mathbf{l}}{dt}\):
$$\frac{d\mathbf{l}}{dt} = \langle -5, -5, 5 \rangle$$
Next, we substitute \(\mathbf{l}(t)\) into the electric field:
$$\mathbf{E}(t) = (7 - 10t)\mathbf{a}_x + (3t + 3)\mathbf{a}_y + 2\mathbf{a}_z$$
Now, take the dot product of \(\mathbf{E}(t)\) and \(\frac{d\mathbf{l}}{dt}\):
$$\mathbf{E}(t) \cdot \frac{d\mathbf{l}}{dt} = -35 + 55t - 15t + 10 = 50t - 25$$
Finally, we integrate the dot product from \(t=0\) to \(t=1\) to get the potential difference:
$$V_b = -\int_{0}^{1} (50t - 25) dt = \left[-25t^2 + 25t\right]{0}^{1} = -25 + 25 = 0\,V$$
04
Report the results
The potential difference between the points \((2, -2, -1)\) and \((0, 0, 0)\) is \(6.5\,V\), and the potential difference between the points \((3, 2, -1)\) and \((-2, -3, 4)\) is \(0\,V\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Electromagnetic Field Theory
Electromagnetic field theory is a fundamental concept that describes how electric and magnetic fields interact with physical objects and charged particles. It provides a framework to understand and quantify forces exerted by electromagnetic fields, which in turn play a critical role in various technological applications and natural phenomena. The theory encompasses various laws and equations, including those developed by James Clerk Maxwell, which ultimately link electricity, magnetism, and optics into a unified framework.
In the context of the textbook problem, we focus on the electric field part of this theory. An electric field is a vector field that represents the force per unit charge exerted on a test charge placed in the field. It is described by the electric field vector \( \mathbf{E} \), which varies from point to point in space. Understanding how to calculate the interaction between a charge and an electric field is essential in solving problems related to electric potential and energy.
In the context of the textbook problem, we focus on the electric field part of this theory. An electric field is a vector field that represents the force per unit charge exerted on a test charge placed in the field. It is described by the electric field vector \( \mathbf{E} \), which varies from point to point in space. Understanding how to calculate the interaction between a charge and an electric field is essential in solving problems related to electric potential and energy.
Line Integral of Electric Field
The line integral of an electric field is a powerful mathematical tool used in calculating the potential difference between two points in a field. It directly relates to the work done by the electric field in moving a unit charge between two positions. The integral is expressed as:
\[V(x_1, y_1, z_1) - V(x_2, y_2, z_2) = - \int_{(x_1, y_1, z_1)}^{(x_2, y_2, z_2)} \mathbf{E} \cdot d\mathbf{l}\]
where \(\mathbf{E}\) is the electric field vector, \(d\mathbf{l}\) represents an infinitesimal vector along the chosen path, and \(V(x_1, y_1, z_1) - V(x_2, y_2, z_2)\) is the potential difference between the two points. The line integral takes into account the component of the electric field along the path, ensuring that even in complex fields, the calculation reflects the actual scenario.
\[V(x_1, y_1, z_1) - V(x_2, y_2, z_2) = - \int_{(x_1, y_1, z_1)}^{(x_2, y_2, z_2)} \mathbf{E} \cdot d\mathbf{l}\]
where \(\mathbf{E}\) is the electric field vector, \(d\mathbf{l}\) represents an infinitesimal vector along the chosen path, and \(V(x_1, y_1, z_1) - V(x_2, y_2, z_2)\) is the potential difference between the two points. The line integral takes into account the component of the electric field along the path, ensuring that even in complex fields, the calculation reflects the actual scenario.
Parameterization of Path
Parameterizing a path is a technique used to establish an equation or set of equations that clearly define a trajectory between two points, usually in terms of a variable \(t\), which ranges between 0 and 1. It essentially gives us a mathematical way to describe the path taken through a field, which is crucial for performing line integrals. By defining the path like this, one can calculate how the electric field vector \(\mathbf{E}\) changes as one moves along the path and how to apply the dot product with the path's differential vector \(d\mathbf{l}\).
The parameterized path can be a straight line or a more complex curve, depending on the scenario. For the textbook problem, a straight line path was chosen for simplicity, and it ensures that the integral remains solvable analytically, providing a clear way to extract the potential difference.
The parameterized path can be a straight line or a more complex curve, depending on the scenario. For the textbook problem, a straight line path was chosen for simplicity, and it ensures that the integral remains solvable analytically, providing a clear way to extract the potential difference.
Electric Potential Energy
Electric potential energy is the potential energy stored in a charge due to its position in an electric field. The concept is analogous to gravitational potential energy, with the key difference being that it pertains to electrostatic forces instead of gravitational forces. The amount of this energy is related to the work done by or against the field in moving a charge from one point to another.
Electric potential, often just called potential, is defined as the potential energy per unit charge. It provides a useful scalar value that can describe the energy landscape an electric charge experiences in a field. The difference in electric potential between two points, known as the potential difference or voltage, can tell us how much energy is released or absorbed when a charge moves from one point to another. This concept is central to understanding circuits, electrochemical cells, and several other engineering applications.
Electric potential, often just called potential, is defined as the potential energy per unit charge. It provides a useful scalar value that can describe the energy landscape an electric charge experiences in a field. The difference in electric potential between two points, known as the potential difference or voltage, can tell us how much energy is released or absorbed when a charge moves from one point to another. This concept is central to understanding circuits, electrochemical cells, and several other engineering applications.