1. The problem statement, all variables and given/known data Q .1 The plates of a parallel plate capacitor are given equal positive charges. What will be the potential difference between the plates? What will be the charges on the facing surfaces and on the outer surfaces? Q.2 Each plate of a parallel plate capacitor has a charge q on it. The capacitor is now connected to a battery .Now, a) The facing surfaces of the capacitor have equal and opposite charges. b) The two plates of the capacitor have equal and opposite charges. c) The battery supplies equal and opposite charges to the two plates. d) The outer surfaces of the plates have equal charges. 2. Relevant equations 3. The attempt at a solution The set up of the two problems looks the same, say, +q, charge is given to the two plates . Let us call the two plates be A and B .The +q charge will distribute uniformly on the two surfaces. Both outer and inner surfaces of A will have +q/2 charge .Now applying Gauss Law ,the inner surface of B will have –q/2 charge and the rest +3q/2 on the outer surface .The charges q/2 on inner surface of A and –q/2 on inner surface of B will contribute to the electric field . Ans 1) V=Ed E=q/2εA Thus ,V=qd/2εA Charges on the surfaces of the plates are (q/2 ,q/2)on A and (-q/2,3q/2) on B Is the solution correct ? Ans 2) I am not clear about the role of battery in affecting the charges on the surfaces of the plates. Option a) ,c) looks right . Option b)is surely wrong .But not sure about option d) How will the battery affect the various charges on the inner and outer surfaces of the two plates ?