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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    $\begingroup$ Simple answer: no. There is no circuit here. $\endgroup$ – Siyuan Ren Aug 7 '12 at 5:11
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    $\begingroup$ @KarsusRen Your no should be qualified, since there is some conductivity in the air, particularly dependent on humidity. $\endgroup$ – anna v Aug 7 '12 at 5:37
  • $\begingroup$ @annav, the diagram clearly denotes that the -q plate is open... period. Open means precisely that charge may not flow to or from that plate. $\endgroup$ – Alfred Centauri Aug 7 '12 at 19:34
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    $\begingroup$ The positive charge in the diagram(+q) is simply bound charge which is held in position by the negative charge on the right side plate which is a floating one.In fact this negative charge(-q) has repelled electrons to the ground. This has contributed towards the accumulation of positive charge on the left plate.There was a temporary flow of current which stopped due to the potential on the left plate getting equal to zero.Since the positive plate is connected to the ground ,the ground+plate system has an infinite capacitance. $\endgroup$ – Anamitra Palit Aug 8 '12 at 4:12
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    $\begingroup$ In the above comment I have assumed that you have applied negative charge to the RHS plate and then grounded the LHS plate. But if you charge the two plates simultaneously by charges of opposite nature, the final situation in the last comment gets directly produced.[One should always understand in relation to the previous comment that if capacitance is infinitely large, potential in the equilibrium state should be zero.] $\endgroup$ – Anamitra Palit Aug 8 '12 at 4:16

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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  • $\begingroup$ The above answer is an extremely meaningful one with reference to the issue in the question $\endgroup$ – Anamitra Palit Aug 9 '12 at 6:43

(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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  • $\begingroup$ You first consider a parallel plate capacitor[isolated from its surroundings] where the plates are not grounded.$C=Q/V=\frac{\epsilon_{0}A}{d}$. Now connect the negative plate with a large spherical conductor by a wire. Some charge is expected to flow from the negative plate to the surface of the spherical body.The lines of force between the plates will become distorted modifying the potential difference between the plates.How is the situation going to change with increasing radius of the spherical conductor? The answer by A. Centauri remains incomplete with these issues unaddressed. $\endgroup$ – Anamitra Palit Aug 9 '12 at 6:36
  • $\begingroup$ With reference to my previous comment:You may connect the positive plate to the spherical conductor. The basic considerations/conclusions in the previous comment remain unchanged. $\endgroup$ – Anamitra Palit Aug 9 '12 at 6:40
  • $\begingroup$ In a charged capacitor, let's say the potential of one plate (call it A) is different from that of the ground (relative to an arbitrary point). If I connect the plate to the ground, plate+ground will have identical potential, different than the potential of the plate before it was connected to the ground. So the new potential of the plate will be different. Does it mean the voltage across the plates will change as well? Or will the potential of plate B adjust to the 'updated' potential of plate A somehow? $\endgroup$ – user4205580 Jul 3 '16 at 19:15
  • $\begingroup$ Okay, maybe let's ask a different question: if I connected plate A of a charged capacitor to a different object, such that there's voltage across A and this object, and then disconnect it, will it change the voltage across the plates? I can't figure it out myself. $\endgroup$ – user4205580 Jul 9 '16 at 11:24

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: $\sigma=0$


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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  • $\begingroup$ If you produce a small perturbation to the zero potential on the grounded plate the boundary conditions would change slightly and a solution to Laplace's would change microscopically (and dynamically) $\endgroup$ – Anamitra Palit Aug 9 '12 at 2:11

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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  • $\begingroup$ Asphir - as you said "the positive charge will be compensated", will it really happen? The +q charge is bound by -q (capacitor theory). If +q gets compensated by electrons from ground, then there will be unbalance of charge. What will happen if -q is grounded? If the voltage across the capacitor was 30V after charge, what will be the value of the voltage after one lead is connected to the ground? $\endgroup$ – B Biswas Aug 7 '12 at 9:16
  • $\begingroup$ @Anamitra Palit - Would please answer the above question? $\endgroup$ – B Biswas Aug 9 '12 at 9:34
  • $\begingroup$ For a parallel plate capacitor, the resultant field is zero for points outside the capacitor if the plates are considered to be of infinite extension.There will be no induction effect on a neighboring body.The finite dimensions of the combination will produce a small field in the connecting wire tending to pull up electrons towards the positive plate(grounded one). There will be a $minor$ redistribution of charges on both the plates.That might compensate for the tendency of electrons being pulled towards the grounded +ve plate. The PD between the plates will remain undisturbed. $\endgroup$ – Anamitra Palit Aug 9 '12 at 11:55
  • $\begingroup$ It would be better to view the whole situation in the light of the uniqueness theorem.If you consider the negative charge of the earth the grounded plate should be assigned a small negative potential instead of zero potential.But this will have its own induction effect on the opposite plate.The PD will remain unchanged. $\endgroup$ – Anamitra Palit Aug 9 '12 at 11:55
  • $\begingroup$ case 1.You charge one plate of the capacitor and ground the other plate. An equal and opposite amount of charge will accumulate on the grounded one.Case2. Both the plates are initially charged and then one is earthed.Effective intensity outside the capacitor system is zero.There will be no effect on some uncharged body external to the system. A charged external body may redistribute the charges on the plates and the plates again will produce a secondary effect on the said external body. The extent of disruption will depend amount of charge and its distribution on the ext. body. $\endgroup$ – Anamitra Palit Aug 9 '12 at 12:01

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