electricityelectrochemistrysemiconductor-physics
In chemistry, the reaction at negative electrode produces electrons,which flow to the positive electrode through the circuit. But then, if circuit has a conductor, then wouldn't it also release electrons, thereby increasing the current? Similarly in diodes, the semiconductor would also release electrons.
If a battery only provides a potential difference, then what happens to the electrons produced in the battery due to chemical reactions?
Suppose we take some object, for example a conducting sphere, and start with it at the same electrical potential as its surroundings. Now we add one electron the sphere, and because the sphere now has a higher negative charge that its surroundings the potential of the sphere will be slightly lower (i.e. more negative) than its surroundings. Adding the electron has created a potential difference. Now add a second electron and the potential difference gets bigger. Add more and more electrons and the potential difference keeps getting bigger.
A side note: this is basically how a Van de Graaff generator works. Charge is added to the sphere at the top by mechnical means, and this can create a big enough potential difference to generate some impressive sparks.
Anyhow, if we transfer a charge $Q$ to our metal sphere the resulting voltage change is given by:
$$ V = \frac{Q}{C} \tag{1} $$
where the constant $C$ is called the capacitance of the sphere (strictly speaking it's the self-capacitance). For spheres we can work equation for the out the capacitance fairly easily, and in fact it's given by:
$$ C = 4\pi\varepsilon_0 r $$
where $r$ is the radius of the sphere. For objects with different shapes the equation for the capacitance will be different, but the key point is that any object of any size and shape has some capacitance and we can use equation (1) to work how what voltage difference is created when we add charge to it.
The point of all this is that the ends of your battery also have a capacitance. Suppose the chemical reaction in the battery transfers a charge $Q$:
The if the capacitance of the anode is $C_-$ and the cathode $C_+$, the voltage change due to transfer the charge will be give by equation (1) as shown on the diagram. The total voltage will be:
$$ V = V_+ - V_- = \frac{2Q}{C_b} $$
where I've assumed the capacitances of both ends are the same and I've used $C_b$ for them both. What actually happens in a battery is that the reaction runs and transfers charge until the voltage $V$ builds up to the battery voltage. At that point no more charge can be transferred and the reaction stops.
This probably all seems a bit long winded, but having gone gone through all this we can answer your questions really easily:
$$ Q_T = \tfrac{1}{2} C_T V \tag{2} $$
where $C_T$ is the new bigger total capacitance with the wire attached. So when you increase the capacitance, $C_T$, you increase the charge $Q_T$. That means the reaction in the battery will restart and transfer more electrons until the total charge rises to $C_T$.
$$\begin{align} Q_T &= \tfrac{1}{2} (C_w + C_b) V \\ &= \tfrac{1}{2} C_wV + \tfrac{1}{2} C_bV \\ &= Q_w + Q_b \tag{3} \end{align}$$
where $Q_w = \tfrac{1}{2} C_wV$ is the charge on the wire and $Q = \tfrac{1}{2} C_bV$ is the charge on the battery. So if you disconnect the wire it keeps a charge $Q_w$. The size of the charge is given by $Q_w = \tfrac{1}{2} C_wV$.
In practice the capacitance of a piece of wire is going to be very small, and the charge you'd build up on the wires will be tiny. Nevertheless, the capacitance will be greater than zero so there will be some charge.
There is no effect on the electrons. It is not like a pipe connected to tap and stopped at the other end so that even before unstopping the other end the molecules are being squashed inside. No repulsion as you are thinking of. In the battery case as soon as you connect the + end the current starts to flow. Think of it as not a mechanical force that drives the electrons but the potential difference. so at least two terminals need to be there.
Best Answer
A battery in a circuit normally sources electrons at the negative terminal and sinks electrons at the positive terminal. The chemical reaction at the positive terminal consumes them. The total number of electrons in the battery does not change. Similarly, a diode accepts electrons at the negative terminal and sources electrons at the positive terminal, with no net change to the number of electrons inside.