A Noob's question on declination The Sun's position at noon is perpendicular to the local x-y plane on the equator during equinox. So the declination of Sun on that day is 0. Today the declination of Sun is close to 23, can this be taken as a virtual equator and hence the tropic of Cancer is in 46.5N (by adding 23.5 to it) and tropic of Capricorn is in 0.5S? Or my assumption is wrong?
I don't think this particularly useful way to think about it. The equator is the equator, Cancer is 23.4 North, and Capricorn is 23.4 South. The equator is a great circle on the sphere of the sky, but the Tropic of Cancer is not a great circle.  These are fixed.  

Now (in Northern Hemisphere summer) the sun is directly above a point on (or near) the tropic of cancer.  This doesn't make the tropic of cancer a "virtual equator". It just makes it the tropic of cancer in summer.

It makes a difference because the path of the sun seen from the equator on midsummer is not the same as the path of the sun seen from the tropic of Capricorn on the equinox.  On the equinox, the sun traces a great circle in the sky (from wherever you view it) On midsummer's day the sun doesn't trace a great circle (anywhere).