# Ampere's loop inside capacitor The figure is a view of one plate of a parallel-plate capacitor from within the capacitor.

In the question, we are required to rank the 4 paths (a, b, c and d) according to the value of $\oint B\cdot ds$.

I thought that all 4 paths are tie because in this case $\oint B\cdot ds=\mu_0 (\epsilon_0 \frac{d\Phi_E}{dt})$, where the value of $\frac{d\Phi_E}{dt}$ is the same for all paths.

But the given answer says that b, c, d tie and a is the smallest one. Why?

• because the electric flux $\phi_E$ is smaller inside the area bounded by path $a$. – theo Dec 8 '14 at 6:01
• @theo I think the flux itself doesn't matter. We only need to consider the rate of change of flux – hklel Dec 8 '14 at 6:09
• For a parallel-plate capacitor with linear, isotropic dielectric and small gap $d$ between plates, the electric field and electric flux is uniformly distributed. – theo Dec 8 '14 at 8:48

The reason for your puzzlement is the way that you are writing Ampere's law. Write it like this: $$\oint \vec{B} \cdot d\vec{s} = \mu_0 \int ( \vec{J} + \epsilon_0 \frac{\partial \vec{E}}{\partial t}) \cdot d\vec{A} ,$$ where this is the vacuum version and the integral on the right is a surface integral that includes the current density and the displacement current density.
Thus within the capacitor plates, $\partial \vec{E}/\partial t$ is uniform and parallel to $d\vec{A}$ $$\oint \vec{B} \cdot d\vec{s} = \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t} \int dA,$$ where $\int dA$ is just the enclosed area between the plates. Outside the plates $$\oint \vec{B} \cdot d\vec{s} = \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t} A = \mu_0 \epsilon_0 \frac{d \Phi_E}{dt}$$ where $A$ is the total area of a plate and $\Phi_E$ is the total flux of electric field.
• That is Ampere's law. I assume you understand that $\vec{J}=0$ between the plates. The second term is equivalent to what you wrote. i.e $\int \epsilon_0 (\partial \vec{E}/\partial t) \cdot d{\vec A} = \epsilon_0 (d\Phi/dt)$. The flux is an integral over a surface bounded by the loop around which you are doing the line integral on the LHS. – Rob Jeffries Dec 8 '14 at 9:03