When we charge a capacitor using a battery and then remove the battery, the plates of capacitor becomes charged. One holds positive charge and the other one gets equal negative charge. o. k. ?

Now if we attach a wire to the positive plate and connect it to the ground , will the electrons from ground climb on the positive plate and make it neutral ? No. But if we connect positive plate to the negative plate then the capacitor will get discharged.

Now consider a situation when we connect 4 capacitors A,B,C,D of equal capacitance in series and connect them to a 10 Volt battery.

Now the P. D. between positive and negative plate of capacitor A will be (10- 7.5) i.e. 2.5 .

For B it will be (7.5 -5 ) i.e. 2.5 , For C it will be (5- 2.5 ) i.e. 2.5, For D it will be (2.5-0) i. e. 2.5.

So potential at positive plate of A is 10 and potential at negative plate of D is 0 and the Potential Difference is 10 volts, which is the potential difference of the battery.

Now connect the wire joining C and D capacitor to ground and now record the potential difference at A, you will find it 7.5 and at positive plate of D it will be 0, and at negative plate of D it will be -2.5. This happens because negative charge from ground climbs on the positive plate of capacitor D and makes it neutral.

My question is why in this case negative charge climbs on this positive plate of D and makes its potential zero ? But such thing does not happen when we connect positive plate of a charged capacitor to the ground.

  • 2
    $\begingroup$ AFAIK charge doesn't flow (to any significant extent in this context) unless you have a circuit. Connecting one end of a charged capacitor to anything has no significant effect. The explanation about a flow of charge causing D+ to be 0V is spurious. Voltage is relative to a reference point, any point on your collection of capacitors can be considered 0V without needing charges to move. $\endgroup$ May 4, 2013 at 19:09
  • 1
    $\begingroup$ Possible duplicate by OP: physics.stackexchange.com/q/63138/2451 $\endgroup$
    – Qmechanic
    May 4, 2013 at 22:12
  • $\begingroup$ Sort of duplicate of physics.stackexchange.com/q/33598/176 $\endgroup$
    – endolith
    May 10, 2013 at 20:44
  • $\begingroup$ Note that, for the same voltage, the charge on an entire circuit is (usually) many orders of magnitude smaller than the charge on one plate of any capacitor involved. $\endgroup$
    – wbeaty
    Jun 16, 2013 at 6:08

4 Answers 4


The net charge of any of those internally connected pairs of plates is always zero. That is, when you charge the capacitors, charge doesn't leave the wire between C and D, it only moves along it, and is held in place by the electric field of the adjacent plates. If a circuit is completed that allows charge to flow from D's negative plate to A's positive plate, the charges will move back to the right place, but the net charge of the 4 capacitors will always be the same.

Connecting the positive terminal of A will not allow charge to flow back from D, so nothing will happen. Similarly, connecting the wire between C and D won't make charge flow in or out of it, at least not in any way significant to the circuit. It only changes the reference for where we make our measurements from.


You're confusing charge with voltage/potential. Objects have charge, voltage is measured between objects. If two objects have no charge, the voltage between them is 0. If two objects have the same charge, the voltage between them is still 0.

Voltage is always a relative difference between two points. When we say "Terminal A is at 5 volts", what we really mean is "Terminal A is at 5 volts relative to the labeled common reference point". It's still a relative measurement, we've just picked a point in the circuit and labeled it as our reference for all other measurements, for convenience.

That common reference point is called "ground" or "common", and can be any point in the circuit, as long as all other voltages are measured relative to it:

enter image description here

These are all identical circuits. The charge on the pins is identical, the voltages between pins are identical. The only difference is which pin we choose to call ground while measuring. Typically you would choose the bottom or middle, like the first or second pictures, but this is just a convention.

In electronics, "ground" has nothing to do with the ground; "earth" has nothing to do with the Earth. Ground is just a label on a schematic.

When you "charge" a capacitor, have you added charge to the capacitor? No. The total charge of the capacitor is always the same. You've just moved some of the charge from one plate to the other. The word "charge" in this case just means "to fill up with energy", just like you can "charge" an inductor with current or "charge" a scuba tank with air or "charge" a cannon with gunpowder. It's an unfortunate terminology.

The water analogy might help to understand this.

It is possible to add charge to one plate of a capacitor, but you won't be able to add very much. It's like charging a metal ball. In this case, you're connecting a voltage source between the Earth and the ball, and moving charge from the Earth to the ball. You're charging a capacitor made up of the Earth as one plate, and the ball as the other. The capacitance of this capacitor is very small, because the "plates" are so far apart, so to move any noticeable charge, you need to use thousands of volts.


For flow of charge, the circuit should be closed. In open circuit, no charge flows. If we connect both the capacitor plates it makes closed circuit, charge flows in the circuit, as a result charges on the plates neutralizes to zero.

If only +ve plate of the capacitor is only connected to ground there is no closed circuit. no charges flows from the ground.

If the circuit is closed and any one point on the circuit is connected to ground, then potential of that point becomes zero and potential of other points changes accordingly. ground potential is assumed to be zero as it is taken as reference point. Potential itself is not absolute value, it is relative value. the potential of same point may be different if we chose different reference points....


It really helps answer this question if you draw the entire circuit, including Earth and any parasitic capacitances between each conductor/component and the Earth.

For example, if an inductor is in parallel with a resistor, and a decaying current is producing 10V across them ...what if the entire circuit is charged to 5VDC wrt earth? You have a charged resistor and charged inductor involved: an inductor treated as one plate of a capacitor, the other plate being earth.

Charged resistor?!

DON'T DON'T DON'T ask questions about capacitive circuits at this point, since that easily forms a direct road to total confusion. First attack any hidden misconceptions by asking yourself about charged-up resistors, wires, batteries, inductors. Each component becomes one plate of a capacitor, where the other plate is Earth.

Once you've explored the topic of "one-plate capacitors," and understand what's going on, then you probably can approach charged-up capacitors without having everything totally derailed by your prior misconceptions.

Charging up an entire capacitor, treated as a unit, is no different than charging up an entire wire, or a resistor, or a coil. You treat each of these as one plate of an approx. tenth-picofarad capacitor. Think: what's the value of a capacitor where one plate is a half inch across, the dielectric is a yard thick, and the other plate is the ground surface?

Then take a look at: engineer's capacitor, a metal sphere with an extremely narrow gap sliced through it.


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