# What happens here in this capacitor?

What happens when we earth the circuit as shown below, why should it matter? because potential at a point is relative? So if the earlier potential was $V_p$ at that point, it becomes $0$, so every potential becomes $V-V_P$ but the potential difference remains constant, so charge should remain constant and hence no heat should be released. Please help only in this conceptual doubt. • Hi ADG. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. – Qmechanic May 13 '15 at 11:07
• @Qmechanic It says: "any question whose value lies in helping you understand the method by which the question can be solved, rather than getting the answer itself. This includes not just questions from actual homework assignments, but also self-study problems, puzzles, etc.". So maybe you wish to make this a homework question? – RE60K May 13 '15 at 12:21

## 1 Answer

If there were no other absolute voltage references in the problem, then you would be correct. But there is already a connection to ground on the upper trace of the diagram. The voltage of that trace does not change (it starts at $0$ and remains $0$ after closing $S$).

• can't get what you mean. – RE60K May 13 '15 at 9:16
• Do you understand that after closing the switch, the potential of the uppermost trace on the diagram (between the capacitors) does not change? – BowlOfRed May 13 '15 at 9:28
• yes, so how does it matter? – RE60K May 13 '15 at 10:10
• You said in your question that every potential becomes $V - V_p$. But that's incorrect. Some potentials change (like that of the middle trace), while others do not (like the top trace). That means the potential between the top and middle trace does not stay constant. – BowlOfRed May 13 '15 at 15:28
• Oh that's it. Side Question: Isn't it like joining those two points by a conducting wire? – RE60K May 13 '15 at 15:54