Integral vs differential curl theorem implicit condition

[superg33k]

Faraday's law has an integral and a differential version:

[tex]curl \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t} \mbox{ and } \oint_{C} \mathbf{E} \cdot d \mathbf{l}=- \frac{d}{dt} \int_{S} \mathbf{B} \cdot d \mathbf{S}[/tex]

When I use the differential version I always have a constant of integration that doesn't appear in the integral version. Where does it come from? Am I making an implicit assumption using the integral version or am I making an explicit assumption that I have not noticed?

For example given:
[tex]\mathbf{B}=At \mathbf{ \hat{e}_z} \mbox{ and } \mathbf{E}=E(r) \mathbf{ \hat{e}_{\phi}} [/tex]
I solve this as:
[tex]curl \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t} \Rightarrow \frac{1}{r} \frac{\partial}{\partial r}(r E(r)) = - \frac{\partial }{\partial t} At = -A \Rightarrow \frac{\partial}{\partial r} (r E(r)) = -Ar \Rightarrow rE(r) = \int -Ar \partial r = -\frac{Ar^2}{2} +B, r \gt 0[/tex]
[tex]E(r) = -\frac{Ar}{2} +\frac{B}{r}, r \gt 0[/tex]

Now the integral version, the line/surface integral is a circle round the z axis.
[tex]C: r=R, 0 \lt \phi \leq 2\pi, z=0, d \mathbf{l} = R \mathbf{\hat{e}_{\phi}}\partial\phi \mbox{ and } S: 0 \lt r \leq R, 0 \lt \phi \leq 2\pi, z=0, d \mathbf{S}= r \mathbf{\hat{e}_z}\partial \phi \partial r [/tex]
[tex]\oint_{C} \mathbf{E} \cdot d \mathbf{l}=- \frac{d}{dt} \int_{S} \mathbf{B} \cdot d \mathbf{S} \Rightarrow \oint_{0}^{2\pi}(E(R) \mathbf{ \hat{e}_{\phi}}) \cdot (R \mathbf{\hat{e}_{\phi}})\partial\phi = 2\pi R E(R) = - \frac{d}{dt} \int_{0}^{R} \int_{0}^{2 \pi} At \mathbf{ \hat{e}_z} \cdot r \mathbf{\hat{e}_z}\partial \phi \partial r=-\pi AR^2 \Rightarrow E(R)=-\frac{AR}{2}[/tex]
since true for all R>0
[tex]E(r)=-\frac{Ar}{2}, r \gt 0[/tex]

Where have I made any assumption that makes B dissappear? (Sorry for any mistakes in my latex)

Thanks for any help.

 

[superg33k]

Your surface S is encloses the origin, so your integral over S assumes that the B field is contiuious everywhere inside it, including r=0. Therefore you have assumed that B and hence E isn't infinite at r=0. Apply this same assumption to your differential version and your constant B dissappears.

 

[Char. Limit]

Your surface S is encloses the origin, so your integral over S assumes that the B field is contiuious everywhere inside it, including r=0. Therefore you have assumed that B and hence E isn't infinite at r=0. Apply this same assumption to your differential version and your constant B dissappears.

Just to let you know, we don't allow insults here, not even directed at yourself.