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.

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.

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 \(\mathbf{E}\): $$ \mathbf{E}(\mathbf{r}(t)) = -x(t) \mathbf{a}_{x} + y(t) \mathbf{a}_{y} = -a\cos(t) \mathbf{a}_{x} + a\sin(t) \mathbf{a}_{y} $$

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: $$ W = \int_C (a^2 \sin(t) \cos(t) + a^2 \sin(t) \cos(t)) \text{d}t = \int_C 2a^2 \sin(t) \cos(t) \text{d}t $$

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.