Question

An electric field in free space is given by \(\mathbf{E}=x \mathbf{a}_{x}+y \mathbf{a}_{y}+z \mathbf{a}_{z} \mathrm{~V} / \mathrm{m} .\) Find the work done in moving a \(1-\mu \mathrm{C}\) charge through this field \((a)\) from \((1,1,1)\) to \((0,0,0) ;(b)\) from \((\rho=2, \phi=0)\) to \(\left(\rho=2, \phi=90^{\circ}\right) ;(c)\) from \((r=10,\), \(\theta=\theta_{0}\) ) to \(\left(r=10, \theta=\theta_{0}+180^{\circ}\right)\).

Short answer

Answer: In all three scenarios, the work done in moving the \(1 \mu C\) charge is 0 Joules.

Step-by-step solution

Set up the line integral

In this case, we are moving the charge in a straight line from point A to point B. We will represent this line using a parameter \(t\) as \(\mathbf{l}(t) = (1-t)\mathbf{a}_{x} + (1-t)\mathbf{a}_{y} + (1-t)\mathbf{a}_{z}\), with \(t\) varying from 0 to 1. Now let's compute its differential: \(d\mathbf{l} = -dt(\mathbf{a}_{x} + \mathbf{a}_{y} + \mathbf{a}_{z})\).

Compute the dot product of the electric field and the differential

In this step, we will compute the dot product \(\mathbf{E} \cdot d\mathbf{l} = x(-dt) + y(-dt) + z(-dt)\). Since \(\mathbf{l}(t) = (1-t)\mathbf{a}_{x} + (1-t)\mathbf{a}_{y} + (1-t)\mathbf{a}_{z}\), we have \(x = 1-t\), \(y = 1-t\), and \(z = 1-t\). So, the dot product will be \((1-t)(-dt) + (1-t)(-dt) + (1-t)(-dt) = -3(1-t)dt\).

Calculate the line integral

To find the work done on the charge, we need to calculate the line integral: \(W = q\int_{A}^{B} \mathbf{E} \cdot d\mathbf{l} = q\int_{0}^{1} -3(1-t)dt\). Now, let's compute this integral: \(W = q \left[-3t + 3t^2\right]_0^1 = q(-3 + 3) = 0\). Therefore, the work done in moving the charge from \((1,1,1)\) to \((0,0,0)\) is 0 Joules. #b. Path from \((\rho=2, \phi=0)\) to \((\rho=2, \phi=90^\circ)\)#

Convert cylindrical coordinates to Cartesian

We will first convert the given cylindrical coordinates to Cartesian coordinates. Point A: \((\rho=2, \phi=0)\) \(A(x, y, z) = (2\cos(0), 2\sin(0), 0) = (2, 0, 0)\) Point B: \((\rho=2, \phi=90^\circ)\) \(B(x, y, z) = (2\cos(90^\circ), 2\sin(90^\circ), 0) = (0, 2, 0)\)

Set up the line integral

As before, we will now represent the line connecting points A and B using a parameter \(t\): \(\mathbf{l}(t) = (2-2t)\mathbf{a}_{x} + 2t\mathbf{a}_{y}\), with \(t\) varying from 0 to 1. The differential is given by \(d\mathbf{l} = -2dt\,\mathbf{a}_{x} + 2dt\,\mathbf{a}_{y}\).

Compute the dot product of the electric field and the differential

Using the line equation, we have \(x = 2-2t\) and \(y = 2t\). Thus, the dot product of the electric field and the differential is \((2-2t)(-2dt) + 2t(2dt) = -4(1-t)dt\).

Calculate the line integral

As before, we calculate the line integral: \(W = q\int_{A}^{B} \mathbf{E} \cdot d\mathbf{l} = q\int_{0}^{1} -4(1-t)dt\). Computing the integral gives: \(W = q \left[-4t + 4t^2\right]_0^1 = q(-4 + 4) = 0\). Hence, the work done in moving the charge from \((\rho=2, \phi=0)\) to \((\rho=2, \phi=90^\circ)\) is 0 Joules. #c. Path from \((r=10, \theta=\theta_0)\) to \((r=10, \theta=\theta_0+180^\circ)\)#

Convert spherical coordinates to Cartesian

We will first convert the given spherical coordinates to Cartesian coordinates. Point A: \((r=10, \theta=\theta_0)\) \(A(x, y, z) = (10\sin(\theta_0)\cos(\phi_1), 10\sin(\theta_0)\sin(\phi_1), 10\cos(\theta_0))\) Point B: \((r=10, \theta=\theta_0+180^\circ)\) \(B(x, y, z) = (10\sin(\theta_0+180^\circ)\cos(\phi_2), 10\sin(\theta_0+180^\circ)\sin(\phi_2), 10\cos(\theta_0+180^\circ))\) Note that since the problem doesn't specify the values of \(\phi_1\) and \(\phi_2\), we cannot proceed with steps 2, 3 and 4 as above. However, we can deduce the result without computing them. Without loss of generality, we can assume that the charge moves with a constant azimuthal angle, meaning that \(\phi_1 = \phi_2\). Therefore, the radial line connecting A and B lies in a plane where the electric field is perpendicular to the radial distance. In this case, the work done in moving the charge is zero since the electric field and the displacement are orthogonal. In conclusion, the work done in moving the charge from \((r=10, \theta=\theta_0)\) to \((r=10, \theta=\theta_0+180^\circ)\) is 0 Joules.

Key concepts

Line Integral in Electromagnetism

The line integral is a fundamental concept when examining the work done by electric fields. In simple terms, it's a way of adding up the contributions of the electric field along a specific path between two points.

To compute the line integral in electromagnetism, we first need to determine the path a charge takes through the field and then calculate the dot product of the electric field and an infinitesimally small element of that path. This process involves integrating the product of the electric field and the displacement vector over the entire path.

For a charge moving through a uniform electric field, if the path is parallel to the field lines, the work done will be maximal, whereas if the path is perpendicular, no work is done as the field does no 'push' or 'pull' on the charge along the direction of movement. This is why the work done on the charge in all the given paths from the exercise was zero, as the electric field was perpendicular to the displacement in each case.

In the case of complex paths, the integration might require sophisticated calculus techniques and, in some situations, the use of numerical methods to approximate the integral if an exact solution is not attainable.

Electric Field in Free Space

Characteristics of Free Space

Electric fields are represented by vectors describing the force a unit charge would experience at any point in space. In free space, the electric field can vary, but it is not affected by any medium, and hence it diverges, converges, or remains uniform depending on the source configuration.

In the given problem, the electric field depends linearly on the coordinates, which implies that the field strength changes as one moves away from the origin, and the associated work depends on the path taken by the charge. However, due to the specific conditions of the problem, it turns out that no work is done on a charge for any of the three given paths, indicating that the chosen paths are somehow special or symmetrical in the context of the given electric field.

Implications for Work Done

Understanding the nature of the electric field in free space is crucial in determining the work done on a charge. When the electric field is homogenous in free space, the work done on a charge depends only on the endpoint positions, not on the path taken. However, in an inhomogeneous field like the one described in the exercise, typically, the path would influence the work done, but symmetry or specific path choices can result in zero work being done, as in the exercise scenarios.

Coordinate System Conversion

When dealing with electromagnetism problems, the ability to convert between coordinate systems is essential.

Understanding Different Coordinates

Cylindrical and spherical coordinates are often used in electromagnetism because they can simplify calculations involving symmetry around a point (spherical) or an axis (cylindrical). Cylindrical coordinates are defined by a radius \( \rho \), an angle \( \phi \), and a height \( z \), while spherical coordinates use a radius \( r \), an inclination angle \( \theta \), and an azimuthal angle \( \phi \).

Conversion Process

In the exercise, the conversion from cylindrical and spherical to Cartesian coordinates allowed for the evaluation of the electric field at specific points. It essentially involves using the definitions of the coordinate systems and trigonometry to express one set of coordinates in terms of another.

This conversion is crucial for computing work done on charges moving through electric fields in complex geometries where a simple Cartesian coordinate system does not align with the symmetry of the problem. In fact, oftentimes, these conversions can reveal hidden simplifications that make determining the work done much easier, as underlined by the zero-work results found for all paths in the exercise.