Same capacitor with $Q = Q_0+\beta t$ on the positively charged plate. What is the magnitude of the magnetic field $\mathbf{B}$ halfway between the plates, at a radius $r$? <img src="./images/charging_cap_dEdt_smaller.png" align="left" style="width: 400px";/> 1. $\frac{\mu_0 \beta}{2 \pi r}$ 2. $\frac{\mu_0 \beta r}{2 d^2}$ 3. $\frac{\mu_0 \beta d}{2 a^2}$ 4. $\frac{\mu_0 \beta a}{2 \pi r^2}$ 5. None of the above Note: * Correct Answer: E
<img src="./images/cap_plates_boxed.png" align="right" style="width: 400px";/> Consider the surface of an imaginary volume (dashed lines, at right) that partly encloses the left capacitor plate. For this closed surface, is the total flux of the current density, $\iint \mathbf{J} \cdot d\mathbf{A}$ positive, negative or zero? 1. Positive 2. Negative 3. Zero Note: * Correct Answer: B * The charge density is increasing, so the total charge in the region is increasing with time, thus the flux of the current density must be negative (conservation of charge).
<img src="./images/5_locations_charging_cap.png" align="left" style="width: 400px";/> At each location, we will evaluate the sign of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$. **At location 3**, the signs of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$ are: 1. both zero 2. both negative 3. both positive 4. $\partial \rho/\partial t$ is positive and $\nabla \cdot \mathbf{J}$ is negative 5. $\partial \rho/\partial t$ is negative and $\nabla \cdot \mathbf{J}$ is positive *Recall that charge is conserved locally!* Note: * Correct Answer: A * There is no current or charge inthe region
<img src="./images/5_locations_charging_cap.png" align="left" style="width: 400px";/> At each location, we will evaluate the sign of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$. **At location 2**, the signs of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$ are: 1. both zero 2. both negative 3. both positive 4. $\partial \rho/\partial t$ is positive and $\nabla \cdot \mathbf{J}$ is negative 5. $\partial \rho/\partial t$ is negative and $\nabla \cdot \mathbf{J}$ is positive *Recall that charge is conserved locally!* Note: * Correct Answer: D * One way to think about this is that the charge on the plate is increasing (so the density is increasingly more positive - positive charge). So the divergence of the current density is negative (jives with previous CQ also).
<img src="./images/5_locations_charging_cap.png" align="left" style="width: 400px";/> At each location, we will evaluate the sign of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$. **At location 4**, the signs of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$ are: 1. both zero 2. both negative 3. both positive 4. $\partial \rho/\partial t$ is positive and $\nabla \cdot \mathbf{J}$ is negative 5. $\partial \rho/\partial t$ is negative and $\nabla \cdot \mathbf{J}$ is positive *Recall that charge is conserved locally!* Note: * Correct Answer: E * One way to think about this is that the charge on the plate is increasing (so the density is increasingly more negative - negative charge). So the divergence of the current density is positive.
<img src="./images/5_locations_charging_cap.png" align="left" style="width: 400px";/> At each location, we will evaluate the sign of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$. **At location 1**, the signs of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$ are: 1. both zero 2. both negative 3. both positive 4. $\partial \rho/\partial t$ is positive and $\nabla \cdot \mathbf{J}$ is negative 5. $\partial \rho/\partial t$ is negative and $\nabla \cdot \mathbf{J}$ is positive *Recall that charge is conserved locally!* Note: * Correct Answer: A * The divergence of the curl of the magnetic field is zero and there's no electric field out there, so the divergence of J must vanish.
<img src="./images/5_locations_charging_cap.png" align="left" style="width: 400px";/> At each location, we will evaluate the sign of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$. **At location 5**, the signs of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$ are: 1. both zero 2. both negative 3. both positive 4. $\partial \rho/\partial t$ is positive and $\nabla \cdot \mathbf{J}$ is negative 5. $\partial \rho/\partial t$ is negative and $\nabla \cdot \mathbf{J}$ is positive *Recall that charge is conserved locally!* Note: * Correct Answer: A * The divergence of the curl of the magnetic field is zero and there's no electric field out there, so the divergence of J must vanish.
<img src="./images/5_locations_charging_cap.png" align="left" style="width: 400px";/> Suppose the original Ampere's law $\nabla \times \mathbf{B} = \mu_0\mathbf{J}$ were correct without any correction from Maxwell (it’s not, but suppose for a moment that it is). What would this imply about $\nabla \cdot \mathbf{J}$ at points 2 and 4 in the diagram? 1. The remain unchanged 2. They swap signs 3. They become zero 4. ??? Note: * Correct Answer: C
<img src="./images/5_locations_charging_cap.png" align="left" style="width: 400px";/> Let's continue with the (incomplete) definition of Ampere's Law: $\nabla \times \mathbf{B} = \mu_0\mathbf{J}$. What does this form tell you about the signs of $(\nabla \times \mathbf{B})_x$ at locations 1, 3, and 5? 1. All positive 2. All negative 3. Positive at 1 and 5, zero at 3 4. Negative at 1 and 5, zero at 3 5. Something else Note: * Correct Answer: C * All we need here is the direction of Jx, which is zero at 3
<img src="./images/5_locations_charging_cap.png" align="left" style="width: 400px";/> Let's return to the complete definition of Ampere's Law: $\nabla \times \mathbf{B} = \mu_0\mathbf{J} + \varepsilon_0 \mu_0 \frac{d\mathbf{E}}{dt}$. **At location 1**, what are the signs of $J_x$, $dE_x/dt$, and $(\nabla \times \mathbf{B})_x$? 1. $J_x<0$, $dE_x/dt<0$, $(\nabla \times \mathbf{B})_x<0$ 2. $J_x=0$, $dE_x/dt>0$, $(\nabla \times \mathbf{B})_x>0$ 3. $J_x>0$, $dE_x/dt=0$, $(\nabla \times \mathbf{B})_x>0$ 4. $J_x>0$, $dE_x/dt>0$, $(\nabla \times \mathbf{B})_x>0$ 5. Something else Note: * Correct Answer: C * There's no E there, Jx points to the right
<img src="./images/5_locations_charging_cap.png" align="left" style="width: 400px";/> Let's return to the complete definition of Ampere's Law: $\nabla \times \mathbf{B} = \mu_0\mathbf{J} + \varepsilon_0 \mu_0 \frac{d\mathbf{E}}{dt}$. **At location 3**, what are the signs of $J_x$, $dE_x/dt$, and $(\nabla \times \mathbf{B})_x$? 1. $J_x<0$, $dE_x/dt<0$, $(\nabla \times \mathbf{B})_x<0$ 2. $J_x=0$, $dE_x/dt>0$, $(\nabla \times \mathbf{B})_x>0$ 3. $J_x>0$, $dE_x/dt=0$, $(\nabla \times \mathbf{B})_x>0$ 4. $J_x>0$, $dE_x/dt>0$, $(\nabla \times \mathbf{B})_x>0$ 5. Something else Note: * Correct Answer: B * There's no Jx there, Ex points to the right
<img src="./images/capacitor_with_x.png" align="left" style="width: 400px";/> A pair of capacitor plates are charging up due to a current $I$. The plates have an area $A=\pi R^2$. Use the Maxwell-Ampere Law to find the magnetic field at the point "x" in the diagram as distance $r$ from the wire. 1. $B = \frac{\mu_0 I}{4 \pi r}$ 2. $B = \frac{\mu_0 I}{2 \pi r}$ 3. $B = \frac{\mu_0 I}{4 \pi r^2}$ 4. $B = \frac{\mu_0 I}{2 \pi r^2}$ 5. Something much more complicated Note: * Correct Answer: B
<img src="./images/capacitor_with_x.png" align="left" style="width: 400px";/> The plates have an area $A=\pi R^2$. Use the Gauss' Law to find the electric field between the plates, answer in terms of $\sigma$ the charge density on the plates. 1. $E = \sigma/\varepsilon_0$ 2. $E = -\sigma/\varepsilon_0$ 3. $E = \sigma/(\varepsilon_0 \pi R^2)$ 4. $E = \sigma \pi R^2 / \varepsilon_0$ 5. Something much more complicated Note: * Correct Answer: B
<img src="./images/capacitor_with_x.png" align="left" style="width: 400px";/> The plates have an area $A=\pi R^2$. Determine the relationship between the current flowing in the wires and the rate of change of the charge density on the plates. 1. $d\sigma/dt = I$ 2. $\pi R^2 d\sigma/dt = I$ 3. $d\sigma/dt = \pi R^2 I$ 4. Something else Note: * Correct Answer: B
We found the relationship between the current and the change of the charge density was: $\pi R^2 d\sigma/dt = I$. Determine the rate of change of the electric field between the plates, $d\mathbf{E}/dt$. 1. $\sigma/\varepsilon_0 \hat{x}$ 2. $I/(\pi R^2 \varepsilon_0) \hat{x}$ 3. $-I/(\pi R^2 \varepsilon_0) \hat{x}$ 4. $I/(2 \pi R \varepsilon_0) \hat{x}$ 5. $-I/(2 \pi R \varepsilon_0) \hat{x}$ Note: * Correct Answer: B
<img src="./images/capacitor_face_on.png" align="left" style="width: 400px";/> Use the Maxwell-Ampere Law to derive a formula for the manetic at a distance $r<R$ from the center of the plate in terms of the current, $I$. 1. $B=\frac{\mu_0 I}{2\pi r}$ 2. $B=\frac{\mu_0 I r}{2\pi R^2}$ 3. $B=\frac{\mu_0 I}{4\pi r}$ 4. $B=\frac{\mu_0 I r}{4\pi R^2}$ 5. Something else entirely Note: * Correct Answer: B
<img src="./images/capacitor_face_on.png" align="left" style="width: 400px";/> Use the Maxwell-Ampere Law to derive a formula for the magnetic at a distance $r>R$ from the center of the plate in terms of the current, $I$. 1. $B=\frac{\mu_0 I}{2\pi r}$ 2. $B=\frac{\mu_0 I r}{2\pi R^2}$ 3. 0 5. Something else entirely Note: * Correct Answer: A