Suppose we have a realistic capacitor,connected to a constant voltage source, with plates at some distance $d$ and a varying charge density across plate due to edge effects. Is it correct to assume charge density would be greatest in the middle of the plate and drop as we move towards the edges. Also, Is it correct that the charge density would increase with increasing plates distance ( Electric field increases )?
[Physics] Charge density on a plate of a realistic capacitor
capacitanceelectrostatics
Related Solutions
I think you question can be answered succinctly from this point:
Does a battery always produce equal amount of negative and positive charges or it may produce additional charges ?
A battery never produces additional charge. Very few things can do this (like a Van de Graaff generator), and even those can only do so locally. A battery pumps charge, and that leaves no room for adding to the total charge count (c).
I thought it strange that the problem is starting out saying that the capacitor has on net extra charge, but you have analyzed the situation correctly. Since we're treating it as an infinite parallel plate capacitor, before the battery is connected the charge is concentrated on the outside face of both plates.
When the battery is connected, and the capacitor has fully charged (if you neglect internal resistance this last specifier isn't needed), then you will have additional positive charge on one plate and additional negative charge on the other. Following the eqi-potential principle, these opposite charges will gather on the inner faces of the two plates (a).
Now a deeper question: will connecting the battery affect the charge density on the outer faces of the plates? I will argue "no". The important unwavering assumption is a constant electric potential throughout the interior of a plate. Introducing the battery introduces a field in-between the inner surfaces, which results in potential difference equal to the voltage of the battery. You have 4 points (which are infinite planes) along the number line (since this is 1D symmetry) where charge is located, and the field is $E=\sigma/(2 \epsilon_0)$ pointing away from the plane - that is, proportional to the charge density and constant. That means that the field on the outside of the plates won't change when connecting the battery, since the total charge quantity stays the same, the field doesn't diminish with distance from the charges, and you only moved charges around. The field within a plate is zero and the outside charge density remains the same, thus, in order to transition from a zero field to the unchanged outside field strength, you require the same charge density on the outside face (d). The potential on the surface is different, but the charge is the same - a very important nuance of electronics.
Thus, options "a", "c", and "d" are both correct.
how can we calculate the terminal voltage?
As the problem is stated, the question answer itself:
the capacitor is connected to a power source with constant voltage $V_s$
(emphasis mine). By KVL, the voltage across the capacitor is the voltage across the power source which is the constant $V_s$.
Suppose we move the plates closer to distance d2. Will the terminal voltage be identical to that of d1?
Yes, since the voltage across the capacitor is the constant $V_s$.
However, since changing the distance between the plates changes the capacitance, there will be a current through the capacitor proportional to the rate of change of capacitance:
$Q = CV$
$\dfrac{dQ}{dt}= I_c = C\dfrac{dV_s}{dt} + V_s \dfrac{dC}{dt}$
Since the voltage is stipulated to be constant, the first term on the right hand side is zero.
However, even with a constant voltage across the capacitor, there can be a capacitor current due to time dependent capacitance.
Best Answer
First point is that you would normally get more charge at the edges.
Second point - if you increase the gap then the capacitance would be reduce and the quantity of charge that could be stored for a particular potential difference would be reduced.
Remember electric field is charge / distance so decreasing the gap would increase the electric field
You should also look at this question and the answer. The question addresses realistic capacitors and gap. It is related to part of your question, but not the same.