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.
Best Answer
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.