Let's analyze the problem step by step. ### **Given Electromagnetic Field** \[ \vec{E}(\vec{r}, t) = -\frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \theta(vt-r) \hat{r} \] \[ \vec{B}(\vec{r}, t) = 0 \] where \(\theta(x)\) is the Heaviside step function: \[ \theta(x) = \begin{cases} 1 & x > 0 \\ 0 & x \leq 0 \end{cases} \] We are to find the corresponding **charge density** \(\rho\) and **current density** \(\vec{J}\). --- ### **Step 1: Use Maxwell's Equations** #### **Gauss's Law (for charge density \(\rho\)):** \[ \nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0} \] Calculate \(\nabla \cdot \vec{E}\): - \(\vec{E}(\vec{r}, t)\) is radial and spherically symmetric except for the step function. - For a spherically symmetric field: \[ \nabla \cdot \left( f(r) \hat{r} \right) = \frac{1}{r^2} \frac{d}{dr}\left( r^2 f(r) \right) \] with \(f(r) = -\frac{q}{4\pi \epsilon_0 r^2} \theta(vt - r)\). So, \[ \nabla \cdot \vec{E} = \frac{1}{r^2} \frac{d}{dr} \left[ r^2 \left( -\frac{q}{4\pi \epsilon_0 r^2} \theta(vt - r) \right) \right] = \frac{1}{r^2} \frac{d}{dr} \left( -\frac{q}{4\pi \epsilon_0} \theta(vt - r) \right) \] \[ = -\frac{q}{4\pi \epsilon_0} \frac{1}{r^2} \frac{d}{dr} \theta(vt - r) \] But, \[ \frac{d}{dr} \theta(vt - r) = -\delta(vt - r) \] So, \[ \nabla \cdot \vec{E} = -\frac{q}{4\pi \epsilon_0} \frac{1}{r^2} \left( -\delta(vt - r) \right) = \frac{q}{4\pi \epsilon_0 r^2} \delta(vt - r) \] Therefore, \[ \frac{\rho}{\epsilon_0} = \frac{q}{4\pi \epsilon_0 r^2} \delta(vt - r) \implies \rho = \frac{q}{4\pi r^2} \delta(vt - r) \] --- ### **Step 2: Use the Continuity Equation (for current density \(\vec{J}\)):** \[ \frac{\partial \rho}{\partial t} + \nabla \cdot \vec{J} = 0 \] Calculate \(\frac{\partial \rho}{\partial t}\): \[ \rho = \frac{q}{4\pi r^2} \delta(vt - r) \] \[ \frac{\partial \rho}{\partial t} = \frac{q}{4\pi r^2} \cdot v \delta'(vt - r) \] Assume \(\vec{J} = J_r \hat{r}\): \[ \nabla \cdot \vec{J} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 J_r) \] So, \[ \frac{qv}{4\pi r^2} \delta'(vt - r) + \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 J_r) = 0 \] \[ \implies \frac{\partial}{\partial r} (r^2 J_r) = -\frac{qv}{4\pi} \delta'(vt - r) \] Integrate with respect to \(r\): Let \(J_r = \alpha \delta(vt - r)\). Try \(J_r = \frac{qv}{4\pi r^2} \delta(vt - r)\): \[ \frac{\partial}{\partial r}(r^2 J_r) = \frac{\partial}{\partial r}\left(qv/4\pi \delta(vt - r)\right) = -\frac{qv}{4\pi} \delta'(vt - r) \] So, \[ \vec{J} = \frac{qv}{4\pi r^2} \delta(vt - r) \hat{r} \] --- ### **Step 3: Match with the Options** - \(\boxed{\rho = \frac{q}{4\pi r^2} \delta(vt - r), \quad \vec{J} = \frac{qv}{4\pi r^2} \delta(vt - r) \hat{r}}\) This matches **option (c)**. --- ## **Final Answer** \[ \boxed{ \text{(c) } \rho = \frac{q}{4\pi r^2} \delta(vt - r), \quad \vec{J} = \frac{qv}{4\pi r^2} \delta(vt - r) \hat{r} } \] **This is the correct answer.**

Certainly! Let's analyze each option carefully, understand their forms, and see why option (c) is the correct one based on our derivation. --- ### **Given Data Recap:** - The electric field: \[ \vec{E}(\vec{r}, t) = -\frac{1}{4\pi \epsilon_0} \frac{q}{r^2} \theta(vt - r) \hat{r} \] - The magnetic field: \[ \vec{B}(\vec{r}, t) = 0 \] - The step function \(\theta(vt - r)\) indicates the field exists only inside a spherical shell expanding outward at speed \(v\). --- ### **Step 1: Find the charge density \(\rho\)** Using Gauss's Law: \[ \nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0} \] For a spherically symmetric field: \[ \nabla \cdot \vec{E} = \frac{1}{r^2} \frac{d}{dr} (r^2 E_r) \] Given: \[ E_r = -\frac{1}{4\pi \epsilon_0} \frac{q}{r^2} \theta(vt - r) \] Compute: \[ r^2 E_r = -\frac{q}{4\pi \epsilon_0} \theta(vt - r) \] Differentiate w.r.t \(r\): \[ \frac{d}{dr} (r^2 E_r) = -\frac{q}{4\pi \epsilon_0} \frac{d}{dr} \theta(vt - r) = -\frac{q}{4\pi \epsilon_0} (-\delta(vt - r)) = \frac{q}{4\pi \epsilon_0} \delta(vt - r) \] Thus: \[ \nabla \cdot \vec{E} = \frac{1}{r^2} \times \frac{q}{4\pi \epsilon_0} \delta(vt - r) = \frac{q}{4\pi \epsilon_0 r^2} \delta(vt - r) \] Applying Gauss's Law: \[ \rho = \epsilon_0 \nabla \cdot \vec{E} = \frac{q}{4\pi r^2} \delta(vt - r) \] **Interpretation:** - The charge density is concentrated on the expanding spherical shell at radius \(r = vt\). - It is proportional to the delta function \(\delta(vt - r)\), indicating a thin shell of charge moving outward at speed \(v\). --- ### **Step 2: Find the current density \(\vec{J}\)** Using the **continuity equation**: \[ \frac{\partial \rho}{\partial t} + \nabla \cdot \vec{J} = 0 \] Calculate \(\partial \rho / \partial t\): \[ \rho = \frac{q}{4\pi r^2} \delta(vt - r) \] \[ \frac{\partial \rho}{\partial t} = \frac{q}{4\pi r^2} v \delta'(vt - r) \] Assuming the current flows radially: \[ \vec{J} = J_r(r, t) \hat{r} \] and the divergence: \[ \nabla \cdot \vec{J} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 J_r) \] From the continuity equation: \[ \frac{q v}{4\pi r^2} \delta'(vt - r) + \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 J_r) = 0 \] This implies: \[ \frac{\partial}{\partial r} (r^2 J_r) = - \frac{q v}{4\pi} \delta'(vt - r) \] Integrating w.r.t \(r\): \[ r^2 J_r = \frac{q v}{4\pi} \delta(vt - r) + C \] Assuming the current vanishes at infinity, \(C = 0\), so: \[ J_r = \frac{q v}{4\pi r^2} \delta(vt - r) \] **Interpretation:** - The current density is localized on the same spherical shell as the charge density. - It points outward (\(\hat{r}\)) with magnitude proportional to \(\delta(vt - r)\). --- ### **Summary of derived \(\rho\) and \(\vec{J}\):** \[ \boxed{ \rho = \frac{q}{4\pi r^2} \delta(vt - r), } \quad \boxed{ \vec{J} = \frac{q v}{4\pi r^2} \delta(vt - r) \hat{r} } \] --- ### **Analysis of Options:** Let's examine each option: --- ### **Option (a):** \[ \rho = -q \delta^{3}(\vec{r}) \quad \text{and} \quad \vec{J} = 0 \] **Interpretation:** - Describes a point charge at origin, static, with no current. - **Contradicts** the given field which is expanding outward; the charge isn't static at the origin. - **Incorrect** for a spherical wavefront expanding at speed \(v\). --- ### **Option (b):** \[ \rho = -\frac{q}{4\pi} \delta^{3}(\vec{r}) \quad \text{and} \quad \vec{J} = 0 \] **Same reasoning as (a):** - Static point charge, no current. - Does **not** match the time-dependent, propagating nature of the given field. --- ### **Option (c):** \[ \rho = \frac{q}{4\pi r^2} \delta(vt - r), \quad \vec{J} = \frac{q v}{4\pi r^2} \delta(vt - r) \hat{r} \] **This matches** our derived expressions: - The charge is localized on an expanding spherical shell. - The current is outward, moving with the shell at speed \(v\). **Correct Choice.** --- ### **Option (d):** \[ \rho = -\frac{q}{4\pi} \delta(r - v t), \quad \vec{J} = -\frac{q v}{4\pi} \delta(r - v t) \hat{r} \] **Differences:** - The negative sign is inconsistent with the positive charge \(q\). - The delta function is written as \(\delta(r - v t)\) instead of \(\delta(v t - r)\). Since \(\delta(x)\) is symmetric, this is a matter of notation, but the sign matters for charge sign. **Incorrect** because of sign discrepancy and inconsistency with the derived positive charge density. --- ## **Final Conclusion:** **The correct physical and mathematical expressions for the charge and current densities corresponding to the given electromagnetic field are those in option (c).** --- ## **Summary:** | Option | \(\rho\) | \(\vec{J}\) | Correct? | Explanation | |---------|-------------|--------------|------------|--------------| | (a) | \(-q \delta^3(\vec{r})\) | 0 | No | Static point charge at origin, no propagation. | | (b) | \(-\frac{q}{4\pi} \delta^3(\vec{r})\) | 0 | No | Same as (a). | | (c) | \(\frac{q}{4\pi r^2} \delta(vt - r)\) | \(\frac{q v}{4\pi r^2} \delta(vt - r) \hat{r}\) | Yes | Represents a spherical shell of charge moving outward at speed \(v\). | | (d) | \(-\frac{q}{4\pi} \delta(r - v t)\) | \(-\frac{q v}{4\pi} \delta(r - v t) \hat{r}\) | No | Sign mismatch and inconsistent notation. | --- ### **Final Answer:** \[ \boxed{ \textbf{Option (c):} \quad \rho = \frac{q}{4\pi r^2} \delta(vt - r), \quad \vec{J} = \frac{qv}{4\pi r^2} \delta(vt - r) \hat{r} } \] This describes a spherical shell of charge moving outward at speed \(v\), perfectly matching the given electromagnetic field.