If I charge a capacitor and connect one lead to ground keeping the other lead floating, will the capacitor discharge ?
G-------||------ open/floating
+q -q
(G for ground)
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The problem is classic. Connect a charged sphere to an other neutral sphere. How does the charge density change ? It depends on the capacity of the spheres. The earth can be modelized as being a very large sphere, so there is a charge variation but it is very small. Physically when electrons try to flow out from the negative electrode to the ground, the positive armature holds them up. |
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(1) For a capacitor to discharge, it is necessary though not sufficient for there to be a means for charge to move from one plate to the other. (2) In the diagram of your question, the plate with -q charge is "open", i.e., there is no means for which charge may move from or to that plate. (1) and (2) together imply that the answer is no, the capacitor will not discharge. EDIT: based on the comments of Anamitra Palit, I think it is important to emphasize that the context of the OPs question, as I understand it, is not a "physicist's" capacitor context but rather an "EE's" capacitor context. By that, I mean that the capacitance associated with the plates dwarfs all others present, i.e., from either plate to a nearby conductor etc., that might be considered and are ignored. If this isn't true, if the "stray" capacitances are significant, then we don't have a capacitor but rather a system of capacitors. For example, $C_{12}, C_{1G}, C_{2G}$ are the plate to plate and plates to ground capacitances respectively. If these are all significant, then connecting the positive plate to ground significantly changes the system. However, for ordinary capacitors as typically used in (low-frequency) electric circuits, $C_{12}$ is the only significant capacitance. |
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Uniqueness Theorem Considerations Initial Situation: The capacitor plates are equally and oppositely charged. Potential on the positive plate:+V Potential on the negative plate:-V PDE:$\nabla^2 \phi=0$ Soln: $\phi=\phi(x,y,z)$ The boundary conditions may be changed without and change in the value of $\vec{E}$ We may write Potential on the positive plate =0(=V-V) Potential on the negative plate=-2V(=-V-V) Soln for potential:$\phi=\phi(x,y,z)+C$ where C=-V is an additive constant The second formulation may be applied when the grounding takes place that is in the final situaion. There is no change in the value of $\vec{E}$. Potential simply changes by an additive constant. Physical Considerations Let's assume that the entire charge on the grounded plate flows out(or gets balanced by electrons from the ground) Charge density on the grounded plate: $\sigma=0$ Therefore, In the vicinity of the plate $E_{n}=0$ (at all points near it,facing the opposite plate) This would be an impossibility if the charge on the floating plate(opposite plate) is still there.The effect of the charge [in the form of non-zero intensity]from the ungrounded plate should reach the grounded one. Thus a full discharge of the grounded plate is impossible For Partial flow of Charge(ie Partial Discharge) from the grounded Plate to the earth: We consider two distinct points on the wire connecting the grounded plate to the earth.It would be impossible for the unequal amounts of opposite charge on the two plates to produce $\vec E=0$ at both the points simultaneously. Presence of non-zero $\vec{E}$ will disturb equilibrium ---that would lead to some peculiar type of electrodynamic situation. The same dynamic situation should prevail if we have equal and opposite amounts of charges[uniformly distributed] on the plates unless the plates are assumed to be of infinite extension which gives $E_{1}=\frac{\sigma}{\epsilon_{0}}$ and $E_{2}=-\frac{\sigma}{\epsilon_{0}}$. But again the connecting wires disrupt the symmetry required for the derivation of the two aforesaid formulas (and the grounding wire itself will upset the values of relative permittivity) This is perhaps consistent with the fact that that the capacity of the earth +grounded plate is infinite only in the physical sense and not in the strict mathematical sense. |
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Since we all see the lightnings from time to time this means that the Earth has charge on its own. From this we may see that earth (ground+atmosphere) is a capacitor itself. It was experimentally checked that the ground has negative charge and so it is the source of electrons. So in your question you plug one capacitor to the half of the other one with huge charge. The answer is - no it will NOT discharge COMPLETELY. What will happen? In your picture the positive charge will be compensated. |
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