Let's work through the problem step by step: --- ### **Step 1: Understanding the Problem** We have a **capacitor** (two circular plates of radius \( R \), separated by distance \( d \)) discharging through a **resistor** \( R \). We are to: 1. Form the **Poynting vector** at the "boundary" of the electric field (i.e., at the edge of the capacitor plates). 2. Integrate the **flux** of the Poynting vector through the cylindrical surface surrounding the region between the plates. 3. Show that this flux equals the **rate of change of the energy stored in the capacitor**: \( P_C = \frac{Q}{C} I \). --- ### **Step 2: Electric and Magnetic Fields** #### **Electric Field between the Plates** For a parallel plate capacitor (neglecting fringing fields): \[ E = \frac{Q}{\epsilon_0 A} \] where \( A = \pi R^2 \) is the area of the plates. #### **Magnetic Field at the Edge (using Ampère's Law with Maxwell's correction)** The displacement current between the plates is equal to the actual current in the wires: \[ I(t) = \frac{dQ}{dt} \] By symmetry, the magnetic field at radius \( r = R \) (the edge) inside the capacitor is: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt} \] \[ B(2\pi R) = \mu_0 \epsilon_0 \frac{d}{dt}(E \cdot \pi R^2) \] But \( E = \frac{Q}{\epsilon_0 \pi R^2} \), so: \[ E \cdot \pi R^2 = \frac{Q}{\epsilon_0} \] \[ \frac{d}{dt}(E \cdot \pi R^2) = \frac{1}{\epsilon_0} \frac{dQ}{dt} = \frac{I}{\epsilon_0} \] Thus, \[ B(2\pi R) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi R} \] --- ### **Step 3: Poynting Vector** The **Poynting vector** is: \[ \vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B}) \] - \( \vec{E} \) is vertically between the plates. - \( \vec{B} \) is tangential (azimuthal) around the edge. - \( \vec{E} \times \vec{B} \) points radially inward (through the cylindrical surface at the edge of the plates). Magnitude: \[ S = \frac{1}{\mu_0} E B \] --- ### **Step 4: Surface Integral of the Poynting Flux** The **surface** through which we want the flux is the **cylindrical surface** of radius \( R \) and height \( d \): \[ \text{Flux} = \int \vec{S} \cdot d\vec{A} = S \times (\text{surface area}) \] \[ \text{Area} = 2\pi R d \] Therefore, \[ P = S \cdot 2\pi R d = \frac{1}{\mu_0} E B \cdot 2\pi R d \] Substitute in the values for \( E \) and \( B \): - \( E = \frac{Q}{\epsilon_0 \pi R^2} \) - \( B = \frac{\mu_0 I}{2\pi R} \) So, \[ P = \frac{1}{\mu_0} \left(\frac{Q}{\epsilon_0 \pi R^2}\right) \left(\frac{\mu_0 I}{2\pi R}\right) (2\pi R d) \] Simplify: - \( \mu_0 \) cancels. - \( 2\pi R \) in numerator cancels with denominator. \[ P = \frac{Q}{\epsilon_0 \pi R^2} \cdot \frac{I}{1} \cdot d \] \[ P = \frac{Q I d}{\epsilon_0 \pi R^2} \] But recall for a parallel plate capacitor: \[ C = \epsilon_0 \frac{\pi R^2}{d} \implies \frac{d}{\epsilon_0 \pi R^2} = \frac{1}{C} \] So, \[ P = \frac{Q I}{C} \] --- ### **Step 5: Interpretation** This result is exactly the **rate of change of energy stored in the capacitor**. Recall: \[ U = \frac{1}{2} \frac{Q^2}{C} \] \[ \frac{dU}{dt} = \frac{Q}{C} \frac{dQ}{dt} = \frac{Q}{C} I \] --- ### **Final Answer** \[ \boxed{ \int \vec{S} \cdot d\vec{A} = \frac{Q}{C} I } \] This is the **rate of change of energy stored in the capacitor**, as requested. The Poynting flux through the cylindrical surface surrounding the capacitor equals the power loss from the capacitor as it discharges.