How can a conductor be grounded yet there are induced charges on it? A classic example for the method of images is the following, quoted from Griffiths's Introduction to Electrodynamics, page 121:

Suppose a point charge $q$ is held a distance $d$ above an infinite grounded conducting plane. Question: What is the potential in the region above the plane? 

Griffiths continued on solving the example using the method of images setting V=0 on the plane as one of the boundary conditions saying "since the conducting plane is grounded".
Now, of course there will be an induced surface charge density. My question is, how can this be since the plane is grounded? 
Does the word grounded have different meanings? sometimes it means not charged and the others it means the potential there is 0?
 A: Think of the ground plane as being an infinitely big electrical conductor, initially uncharged.
We bring in a (say positive) point charge close to the surface of the ground plane.  An negative image charge is induced on the surface near the point charge, and since the net charge on the ground plane is zero, an opposite positive charge is pushed off "to infinity".
Now suppose we bring a positively charged conductor up to the ground plane, and then connect it electrically.  The positive charge in our conductor gets discharged and cancels the negative induced surface charge, but the positive charge that had been pushed off to infinity is still out there and gets redistributed.  Since it is a finite charge distributed over an infinite body, there is effectively zero charge density.  
A: Here grounded means V=0; V=0 means, there is no work to be done to bring a positive charge near the plane of the grounded conductor. This means that there are no charges on the plane. Had there been any charges on the plane it would either push or pull the test charge and V would not be zero. I think this particular example is poorly worded. later it assumes a different situation of ignoring the plane and introducing a image charge. This is just a substitute to the original problem.Originally, due to +kq(look page 123 griffith, later happens to be +q) at distance d induces -kq((look page 123 griffith, later happens to be -q)) charge at the conductor and introducing the negative image charge essentially induces positive charges on the plane which adds up to 0 and the no charge hence no V.
