Let's break down and **solve Problem 10.15** step by step. --- ### 1. **Restate the Problem** A charge \( q \) moves in a circle of radius \( a \) in the \( xy \)-plane, centered at the origin, with constant angular velocity \( \omega \). At \( t = 0 \), the charge is at \( (a, 0) \). **Find the Liénard-Wiechert potentials for points on the \( z \)-axis.** --- ### 2. **Position of the Charge (as a function of time)** The charge's position as a function of time: \[ \mathbf{r}_q(t) = a \cos(\omega t)\,\hat{\mathbf{x}} + a \sin(\omega t)\,\hat{\mathbf{y}} \] --- ### 3. **Field Point on the \( z \)-axis** Let the field point be at \( \mathbf{r} = (0, 0, z) \). --- ### 4. **Liénard-Wiechert Potentials** The potentials at a field point \( \mathbf{r} \) and time \( t \) are: \[ \boxed{ \begin{align*} V(\mathbf{r}, t) &= \frac{q}{4\pi\epsilon_0} \frac{1}{R - \mathbf{R} \cdot \mathbf{v}/c} \bigg|_{\text{ret}} \\ \mathbf{A}(\mathbf{r}, t) &= \frac{\mu_0 q}{4\pi} \frac{\mathbf{v}}{R - \mathbf{R} \cdot \mathbf{v}/c} \bigg|_{\text{ret}} \end{align*} } \] where: - \( \mathbf{R} = \mathbf{r} - \mathbf{r}_q(t_r) \) - \( R = |\mathbf{R}| \) - \( \mathbf{v} \) is the velocity of the particle at the **retarded time** \( t_r \) - \( t_r \) is given by \( t - t_r = R/c \) --- ### 5. **Components for this Problem** #### **a) Find \( \mathbf{r}_q(t_r) \):** \[ \mathbf{r}_q(t_r) = (a\cos(\omega t_r), a\sin(\omega t_r), 0) \] #### **b) Find \( \mathbf{v}(t_r) \):** \[ \mathbf{v}(t_r) = (-a\omega\sin(\omega t_r), a\omega\cos(\omega t_r), 0) \] #### **c) Find \( \mathbf{R} \):** \[ \mathbf{R} = (0 - a\cos(\omega t_r),\: 0 - a\sin(\omega t_r),\: z) = (-a\cos(\omega t_r),\: -a\sin(\omega t_r),\: z) \] #### **d) Find \( R \):** \[ R = |\mathbf{R}| = \sqrt{a^2 \cos^2(\omega t_r) + a^2 \sin^2(\omega t_r) + z^2} = \sqrt{a^2 + z^2} \] (Notice: \( R \) is independent of \( t_r \), but \( t_r \) still enters through the potentials.) #### **e) Find \( \mathbf{R} \cdot \mathbf{v}/c \):** \[ \mathbf{R} \cdot \mathbf{v} = [-a\cos(\omega t_r)]\cdot [-a\omega\sin(\omega t_r)] + [-a\sin(\omega t_r)]\cdot [a\omega\cos(\omega t_r)] + 0 \] Multiply out: \[ = a^2 \omega [\cos(\omega t_r)\sin(\omega t_r) - \sin(\omega t_r)\cos(\omega t_r)] = 0 \] So, \[ \mathbf{R} \cdot \mathbf{v} = 0 \] --- ### 6. **Retarded Time Condition** \[ t - t_r = \frac{R}{c} = \frac{\sqrt{a^2 + z^2}}{c} \] So, \[ t_r = t - \frac{\sqrt{a^2 + z^2}}{c} \] --- ### 7. **Plug into Liénard-Wiechert Potentials** Since \( \mathbf{R} \cdot \mathbf{v} = 0 \): \[ V(0, 0, z, t) = \frac{q}{4\pi\epsilon_0} \frac{1}{\sqrt{a^2 + z^2}} \] \[ \mathbf{A}(0, 0, z, t) = \frac{\mu_0 q}{4\pi} \frac{\mathbf{v}(t_r)}{\sqrt{a^2 + z^2}} \] Recall: \[ \mathbf{v}(t_r) = (-a\omega\sin(\omega t_r),\; a\omega\cos(\omega t_r),\; 0) \] where \[ t_r = t - \frac{\sqrt{a^2 + z^2}}{c} \] --- ### 8. **Final Answers** \[ \boxed{ \begin{aligned} V(0,0,z, t) &= \frac{q}{4\pi\epsilon_0} \frac{1}{\sqrt{a^2 + z^2}} \\ \mathbf{A}(0,0,z, t) &= \frac{\mu_0 q}{4\pi} \frac{-a\omega\sin(\omega t_r)\,\hat{\mathbf{x}} + a\omega\cos(\omega t_r)\,\hat{\mathbf{y}}}{\sqrt{a^2 + z^2}} \end{aligned} } \] where \[ t_r = t - \frac{\sqrt{a^2 + z^2}}{c} \] --- ### **Explanation Summary** - The scalar potential is static (depends only on distance) because the moving charge is always at the same distance from any point on the \( z \)-axis. - The vector potential oscillates, reflecting the circular motion of the charge, but its magnitude is also determined by the constant distance to the \( z \)-axis. --- **Let me know if you want the explicit electric and magnetic fields as well!**