# Electrostatic Induction and Flow of Charges

I have a couple of questions which come under the above title.

1) Firstly if we bring a negatively charged rod near a sphere (Fig 1 and 2)

The end A would get positively charged and the end B will get negatively charged. Now if I ground the sphere (in that same position ) , the negative charge or electrons will flow down to the ground. Now what would be a correct explanation for the flowing of electrons ? I am asking this because some people say that the negative charges should be held in place by the positive charges. 2) Also what would happen if we connect the end A to the ground (Fig 3)?

3) If I have a battery and I just connect the positive terminal to a wire (the circuit is incomplete). Will any charge develop on end (A) of the wire?

I am asking this because I read here on Stack Exchange that the end A will get charged but the charge would be very less due to the very small capacitance of the wire. How can we calculate the charge developed on A if we know the capacitance of the wire and the voltage of the battery 4) Finally we have a capacitor in place of a battery and we connect the negative plate to a wire ( Fig 4).

Will any charge develop on the end A of the wire. I am asking this because I think that (this should also be in the case of the battery) that electrons shouldn't be able to flow from the negative plate to the end A of the wire because the electrons in the negative plate are held in place due to attraction by the positive charges on the other plate?

It seems to me, since you say twice in your post:

Some people say that the negative charges should be held in place by the positive charges.

and

I think that (this should also be in the case of the battery) that electrons shouldn't be able to flow from the negative plate to the end A of the wire because the electrons in the negative plate are held in place due to attraction by the positive charges on the other plate

that you have some misconceptions on how electrostatic forces work. Let me give an example to highlight what I think is wrong with your understanding. Say there is a charge -q sitting close to a charge +q. You bring a big charge +Q close to this environment. -q will no longer be held in place by +q: it will rush towards +Q

The takeaway is that charges are fickle things--they go where it suits them (regions of higher potential for electrons and lower potentials for protons) It doesn't matter if there is a positive charge next to the electron--see a high potential region that they can get to (there needs to be a conducting medium)? They will go there.

1) Once the charges have been distributed like in your picture, the electrons are at -ve potential. Connect the sphere to ground and they wil go there. ($0$V is higher than -ve volt)

2)It doesn't matter what end you connect to ground. A conductor is an equipotential. The entire sphere will be at $0$V

3)Yes, the wire will now be at a positive potential so it will develop +ve charge by having electrons move to the capacitor plate, but it won't act as a capacitor.

4)The wire is at $0$ volts. It is brought in contact with the -ve plate; this leads it to acquire some negative charge which will spread to its surface untill both the surface of the wire (which can be treated as a conducting cylinder) and the -ve plate are at the same potential.

• You said that if a bigger charge Q is brought near the system of q and -q the charges will move and the electrons in the case of sphere will move to the ground because 0V is higher than -ve . That works because ground is a vast reservoir. But in the case of wire connected to the -ve plate of the capacitor the situation is different. Firstly there is -ve charge on one plate and +ve on another mutually attracting each other. But the wire is not symbolic of the bigger charge Q. How do we know that the potential of the wire is -ve or +ve or 0 wrt to the capacitor plate. – Shashaank Aug 28 '16 at 14:58
• Shouldn't the wire have that much charge ( or potential) to effect the electrons on -ve plate after overcom ing the effect of the charges on the positive plate. And if the wire develops some charge can we detect it using an electroscope after disconnecting the wire . – Shashaank Aug 28 '16 at 15:01
• "a path for electrons to travel between points at different potential ". But what is the point of different potential. It's the end of the wire here isn't it ? – Shashaank Aug 28 '16 at 15:15
• No in the case of the capacitor the circuit is incomplete . One end of the wire is connected to the -ve plate and the other end of the wire is free . There is no material there. The wire itself is the material there. – Shashaank Aug 28 '16 at 15:21
• You are forgetting the fact that there in no conducting medium between the plates. So, even is the positive plate is at 5V, electrons from the -ve plate can't go there. They move to the wire, even though it's at $0$V because they can. – GeeJay Aug 29 '16 at 5:00