B-field inside a charging || plate (disc shaped) capacitor

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

I have difficulty in solving part (C).
The surface bounded by the Amperean loop is like an open drum.
The current enclosed by the loop is $I_{en}$ = I through the bottom of the drum + I(s) through that part of the curved surface which touches the plate .
Now, how to find out I(s)?
How to find out the direction of B considering this surface?
If I consider the planar surface through the Amperean loop, then the displacement current is along z-axis and so the B-field is along the $\hat \phi$ direction.

 

[Charles Link]

Homework Statement

View attachment 210261View attachment 210262

Homework Equations

The Attempt at a Solution

I have difficulty in solving part (C).
The surface bounded by the Amperean loop is like an open drum.
The current enclosed by the loop is $I_{en}$ = I through the bottom of the drum + I(s) through that part of the curved surface which touches the plate .
Now, how to find out I(s)?
How to find out the direction of B considering this surface?
If I consider the planar surface through the Amperean loop, then the displacement current is along z-axis and so the B-field is along the $\hat \phi$ direction.

I think for part "c" they are simply trying to show you the more complete Maxwell's law (in differential form) $\nabla \times B=\mu_o J+\mu_o \epsilon_o \dot{E}$. Integrating over $dA$ and using Stokes theorem gives the more complete form of Ampere's law: $\oint B \cdot dl=\mu_o I +\mu_o \epsilon_o \int \dot{E} \, \cdot dA$. Since there is no $E$ from the capacitor outside the capacitor, I think they are just looking for you to evaluate the left side and the first term of the right side to compute $B$. $\\$ (I believe the displacement current for part (b) is $I_D=\epsilon_o \int \dot{E} \cdot \, dA$). $\\$ Editing: One item they usually try to show with this exercise in displacement current is, since we have $\nabla \times E=-\dot{B}$ (which integrates to, along with Stokes theorem) $\oint E \cdot \, dl=-\dot{\Phi}_m$, that there might be a corresponding equation of the form $\nabla \times B=\mu_o \epsilon_o \dot{E}$ which integrates to $\oint B \cdot \, dl=\mu_o I_D$ where $I_D= \epsilon_o \int \dot{E} \cdot \, dA$ . We also can have real currents $I_S$ in which case we have $\oint B \cdot \, dl =\mu_o I_S$, so these two equations are combined into the more complete form of Ampere's law: $\oint B \cdot \, dl=\mu_o(I_S+I_D)$. $\\$ (In differential form, this more complete equation reads $\nabla \times B=\mu_o J +\mu_o \epsilon_o \dot{E}$. This more complete form is used to derive the electromagnetic wave equation).

 

[TSny]

Now, how to find out I(s)?

Consider the net charge q in the shaded region of radius s. Can you express q in terms of the total charge Q on the entire plate?

Consider the rate of change of q and how it is related to I and I(s).