Dielectric modification of electrostatic equations? I have learnt that in cases of electrostatic fields inside a dielectric of any source charge, the field is reduced by a factor of K( if K, the dielectric "constant", is taken everywhere to be same). But that means that if there is, say, a fixed (by non-electric forces) charge outside a randomly shaped dielectric block, then the fields experienced by our test charge inside the dielectric will be weaker (by a factor of K) than what it would have experienced had it been outside the dielectric. But should the fields of other charges outside the block  be unaffected??  
If this is so, we take two opposite charges in vacuum separated by a distance and a dielectric block in between them. The lines of force would be passing through the slab and vacuum both. If the fields outside are not affected than there would be a contradiction since the closed line integral of field wouldn't be zero.   
But then how and why would the fields outside the dielectric be changed. The contribution of the induced dielectric charges should be zero at least in the idealised case where infinitely large plates (producing uniform fields) have a dielectric bock (smaller then the separation) between them??  
 A: 
But should the fields of other charges outside the block be unaffected??

Nope. The fields outside are affected. It's not always easy to calculate, though. Basically, inside a (linear) dielectric an induced field is formed in the opposite direction which counteracts the electric field, reducing the net electric field.
Of course, this new field has to be caused by something, in this case it is the charge separation induced on the dielectric by the original field1. This induced charge separation induces a field outside as well (usually), affecting the net field on the outside. 

If the fields outside are not affected than there would be a contradiction since the closed line integral of field wouldn't be zero. [...] The contribution of the induced dielectric charges should be zero at least in the idealised case where infinitely large plates (producing uniform fields) have a dielectric bock (smaller then the separation) between them??

Technically, no, as long as the field inside is equally unaffected, the closed line integral will stay zero. Remember, it is a closed integral. You need to pass through the dielectric twice, once in each direction, and the change of field while going from A to B will be cancelled out by the change of field while coming back. This will happen with a large slab, which, as you correctly noted, has a negligible effect on the outside field. However, since the component of distance travelled along the field is the same in this case, and the change in electric field ($\mathbf{E}_{new}-\mathbf{E}_{old}=(K-1)\mathbf E$) is the same throughout, this change cancels itself out in the integral.
1. In the case of a linear dielectric, $\mathbf P=\mathbf E(K-1)$, and we can calculate the bound charge density (i.e., the induced charge) using $\rho_b=-\nabla\cdot\mathbf P$ 
