Question

Given \(\mathbf{E}=-x \mathbf{a}_{x}+y \mathbf{a}_{y},(a)\) find the work involved in moving a unit positive charge on a circular arc, the circle centered at the origin, from \(x=a\) to \(x=y=a / \sqrt{2} ;(b)\) verify that the work done in moving the charge around the full circle from \(x=a\) is zero.

Short answer

Answer: The circle can be parametrized as \(\mathbf{r}(t) = a \cos(t) \mathbf{a}_{x} + a \sin(t) \mathbf{a}_{y}\), where \(t\) is the angle measured from the positive x-axis and varies from \(0\) to \(2\pi\). 2. What is the work done by the electric field when moving a unit positive charge from \(x=a\) to \(x=y=a/\sqrt{2}\)? Answer: The work done by the electric field is \(\frac{a^2}{2}\). 3. What is the total work done by the electric field when moving a unit positive charge around the full circle from \(x=a\)? Answer: The total work done by the electric field is zero.

Step-by-step solution

Setup and parametrize the path

A circle with center at the origin and radius \(a\) can be parametrized as follows: $$ \mathbf{r}(t) = x(t) \mathbf{a}_{x} + y(t) \mathbf{a}_{y} = a \cos(t) \mathbf{a}_{x} + a \sin(t) \mathbf{a}_{y}, \quad 0 \le t \le 2\pi $$ where \(t\) is the angle measured from the positive x-axis.

Compute \(\mathbf{r}'(t)\) and \(\mathbf{E}(\mathbf{r}(t))\)

To compute the line integral along the arc, we need to find \(\mathbf{r}'(t)\) and \(\mathbf{E}(\mathbf{r}(t))\). Differentiating \(\mathbf{r}(t)\) with respect to \(t\), we get: $$ \mathbf{r}'(t) = -a \sin(t) \mathbf{a}_{x} + a \cos(t) \mathbf{a}_{y} $$ Now, we substitute \(\mathbf{r}(t)\) into the given electric field \(\mathbf{E}\): $$ \mathbf{E}(\mathbf{r}(t)) = -x(t) \mathbf{a}_{x} + y(t) \mathbf{a}_{y} = -a\cos(t) \mathbf{a}_{x} + a\sin(t) \mathbf{a}_{y} $$

Calculate the line integral along the path

The work done by the electric field moving a unit positive charge along the path is the line integral: $$ W = \int_C \mathbf{E} \cdot \text{d}\mathbf{r} = \int_C \mathbf{E}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \text{d}t $$ Substitute \(\mathbf{E}(\mathbf{r}(t))\) and \(\mathbf{r}'(t)\) into the line integral: $$ W = \int_C (-a\cos(t) \mathbf{a}_{x} + a\sin(t) \mathbf{a}_{y}) \cdot (-a \sin(t) \mathbf{a}_{x} + a \cos(t) \mathbf{a}_{y}) \text{d}t $$ Evaluating the dot product, we have: $$ W = \int_C (a^2 \sin(t) \cos(t) + a^2 \sin(t) \cos(t)) \text{d}t = \int_C 2a^2 \sin(t) \cos(t) \text{d}t $$

Evaluate the integrals for part (a) and part (b)

Now we can evaluate this integral for both parts of the problem. (a) For part (a), we need to find the work involved in moving the charge from \(x=a\) to \(x=y=a/\sqrt{2}\): $$ W_a = \int_{0}^{\pi/4} 2a^2 \sin(t) \cos(t)\text{d}t = a^2 [\sin^2(t)]_{0}^{\pi/4} = a^2 \left(\frac{1}{2} - 0\right) = \frac{a^2}{2} $$ (b) For part \((b)\), we have to verify that the total work done in moving the charge around the full circle from \(x=a\) is zero: $$ W_b = \int_{0}^{2\pi} 2a^2 \sin(t) \cos(t)\text{d}t = a^2(\sin^2(t) \big|_{0}^{2\pi} = 0 $$ As the total work done in moving the charge around the full circle from \(x=a\) is zero, it verifies part (b) of the problem. So, the work involved in moving a unit positive charge from \(x=a\) to \(x=y=a/\sqrt{2}\) is \(\frac{a^2}{2}\), and the total work done in moving the charge around the full circle from \(x=a\) is zero.

Key concepts

Electromagnetic Theory

Electromagnetic theory is essential in understanding electric fields, especially when considering the movement of charges. Electric fields are vector fields that represent the way charged particles influence each other in space.
This concept is part of a more extensive framework that includes electric force, magnetism, and their interactions.
Electric fields can be visualized as lines pointing in the direction they exert force on positive charges, usually described in equations through vectors.

• The field direction shows where a positive test charge would move.
• The magnitude of the field dictates how much force is involved.

In the exercise, the electric field is given by \(\mathbf{E}=-x \mathbf{a}_{x}+y \mathbf{a}_{y}\). This means, mathematically, it pushes charges depending on their position in an X-Y coordinate system.
This field is static, meaning it does not change over time.

Line Integrals

Line integrals are a tool in calculus used to find the total impact of a vector field along a certain path. They sum up contributions from every point that a particle, or in this case, a charge, traverses.
This type of integral is particularly useful in physics, especially for finding work done by fields like electric fields.

• In our case, we use the path of a circle centered at the origin.
• The electric field's effect is then calculated along this path.

The line integral computes work done or energy transferred, which can be positive, negative, or zero, depending on the vector field and path configuration.
In the exercise, the integral is \(\int_C \mathbf{E} \cdot \text{d}\mathbf{r}\), representing the work needed to move the charge along the specified circular arc.

Vector Calculus

Vector calculus extends normal calculus to account for quantities that have both magnitude and direction, such as electric fields. This branch of mathematics allows us to work with vector fields and compute useful properties.
Derivatives and integrals in vector calculus are essential in analyzing how vectors change over space. In terms of vectors, differentiation can tell us the rate of change, while integration can present the accumulation of effects.

• Parametrizing the circle: This provides a set of coordinates for all points on the circle, useful for calculating distances and resolving vectors into components.
• Finding \(\mathbf{r}'(t)\): This derivative yields the direction and speed of a charge moving along the circle.

Vector calculus is powerful because it simplifies complex physical phenomena into manageable mathematical forms.
By breaking down vectors into components \(x(t)\) and \(y(t)\), our calculations become more straightforward.

Work Done in Electric Fields

The concept of work in electric fields is crucial to understanding energy transfer. Work occurs when a force moves an object across a distance. Similarly, in electric fields, work is done when a charge moves.
This relates to energy expended or required to move the charge against the field's influence. The calculation of work requires not only the strength and direction of the electric field but also the path taken through it.

• Whereas the work along a specific path segment is not always zero, a closed loop can result in zero net work, indicating energy conservation.
• The exercise highlights that work done in full circular paths often sums to zero, aligning with the conservative nature of electric fields.

For our problem, the work done through part \((a)\) is \(\frac{a^2}{2}\), indicating energy expenditure.
However, traversing the complete circle cancels this work, leaving a net zero, confirming the field's conservative property.