I'm trying to understand how a very short dipole of length $\ell \ll \lambda$ works for receiving radiation. (It is center-fed, and has two thin perfectly conducting arms each of length $\ell/2$ separated by a negligible distance.)
Say I have such a dipole oriented along the direction of the spatially uniform incident electric field $\vec{E}\left(t\right)$, and at all times in the region where the antenna is places, we have $$\vec{B}\left(t\right) \perp \vec{E}\left(t\right)$$ $$\left|\vec{B}\left(t\right)\right| = \dfrac{\left|\vec{E}\left(t\right)\right|} {c}$$
What will be the open circuit voltage $V\left(t\right)$ seen across the dipole? Do I need to specify $\vec{E}\left(t\right) = \vec{E}_0 \cos \omega t$ for a meaningful answer, or will a general $\vec{E}\left(t\right)$ do?
My first guess would be $\displaystyle V\left(t\right)=\int_{\left(0,-\ell/2\right)}^{\left(0,+\ell/2\right)} \vec{E}\left(t\right)\cdot d\vec{r} = \ell\;\left|\vec{E}\left(t\right)\right|$, but I have two confusions:
The integral above would be the potential difference between the tips of the dipole, but given that the electric and magnetic fields are time varying, I understand that the scalar potential itself becomes meaningless. Is this expression still usable for EMF anyway? Or does the fact that $\ell \ll \lambda$ make the potential difference still usable--and if so, why?
Given that the arms of the dipole are perfect conductors, is the potential difference between the two ends of any arm zero? Should I therefore look at the potential difference between the midpoints of the two arms rather than the tips? I will then get $$\int_{\left(0,-\ell/4\right)}^{\left(0,+\ell/4\right)} \vec{E}\left(t\right)\cdot d\vec{r} = \frac{\ell} 2 \;\left|\vec{E}\left(t\right)\right|$$ Or does the radiation resistance of the arms come into play somehow?
Thanks ...