AL physics about charge

When a piece of dielectric is inserted in between the two plates, charges are induced on the dielectric. The side of the dielectric near to the +ve plate would have -ve charge induced on it, whereas the side near to the -ve plate would have +ve charge induced on it. These induced +ve and -ve charges on the two sides of the dielectric(which are smaller in magnitude than the charges on the plates) set up an electric field opposing the original electric field from the charges on the plates, thus reducing the original field and a weaker field is resulted.
Hence, although the original field now becomes weaker (i.e. less field lines), there are induced charges on the dielectric which give rise to additional field lines in the region between the dielectric and the plates, making the number of field lines beginning or terminating on the charges on the plates unaltered. Since field lines must begin and terminate at charges, the amount of charges on the plates therefore remains unchanged.
This result is obvious as the +ve plate is isolated, the amount of charges on it is constant. There must be equal and opposite charges on the earthed plate disregard a piece of dielectric is inserted in between as field lines must originate and terminate at equal amount of charges of opposite polarity.
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As explained above, field lines must begin and terminate on equal amount of charges of opposite polarity (e.g. just imagine, say, if 10 +ve charges are on one plate, each of which gives rise to a field line, then there are 10 field lines, these 10 field lines must end at 10 -ve charges on the other plate). Increasing the distance between the plates would not affect the number of field lines, thus the amount of charges must remain unchanged.

because B is earthed, the charge on B will keep constant, even when you insert the dielectric