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.