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Like this: Source: Wolfram Demonstrations: Lines of Force for Two Point Charges.


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Carlos' answer is fine, but here is a direct and physical way to think about why the voltage increases when the separation between the plates increases. The answer in this video is helpful in my opinion since it doesn't require any discussion of electric field to answer a question that is fundamentally about electric potential. https://youtu.be/NC_j9GQEwU0


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I'm assuming the title of your post is your question. Please note that we can not provide solutions to homework and exercise type questions, only offer guidance. Here is some guidance: In a dc circuit under steady state conditions ideal capacitors look like open circuits unless shorted (as by switch S) and immediately after a switching event the voltages on ...


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If you integrate from the negative plate to the positive plate the potential difference is positive (+V). You have V is the negative integral of E, which is true if you integrate in the direction of E (E dot dS)


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when a surge of electrons moves through the wire to one plate of the capacitor, the electrons on the other plate "see" those electrons across the gap between the plates and are repelled by them- so there is a temporary surge of electrons "fleeing" out of the second plate. As soon as the first plate has become fully occupied with extra ...


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Assuming all capacitors initially discharged, after the battery is connected, and your top-left capacitor becomes fully charged, no current can flow through it. Initially the remaining capacitors will begin storing charge. But, after time passes and the top-left capacitor heads toward battery voltage the remaining capacitors will discharge through the ...


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A good question to ask when bogged down by notation is: "What is happening physically?". In this case, the formula for $V_C$ is actually: $$V_C(t) = \frac{Q(t)}{C},$$ where $Q(t)$ is the net charge on the capacitor at some time $t$. So what is the net charge on the capacitor? Well, you know that there is a "current" in the circuit as the ...


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For an Ohmic resistor, the second equation applies. Similarly for a parallel plate capacitor. $V$ and $I$ or $V$ and $Q$ will have a set relationship; that cannot be varied independently for the given circuit element. However, your first equations apply more generally for any system. i.e. if you take something (doesn't need to be Ohmic) and apply a voltage ...


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Let us begin with a definition, because arguing things without definition may lead to confusion. An electromagnetic wave is defined as an energy wave that has a velocity in vacuum that coincides with the velocity of light in vacuum. According to this definition, gamma rays, X-rays, ultraviolet rays, visible light rays, infra red rays, radio waves, and ...


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The electric potential in not the potential energy, but the potential energy divided by the particle charge. So you need to add the charge of the electron to your equation: $ e{\cdot}ΔV=mv^2/2$


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Here is one way capacitors can self-charge. When dissimilar objects are rubbed together, it is possible to transfer charge (a.k.a. "static electricity") from one to the other. If a capacitor is connected to one of these objects, it can collect the transferred charge and hold it for surprisingly long times. If one of the objects doing the rubbing is ...


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You can add a dielectric b/w the plates in two ways: (1) Keeping the battery connected If you keep the battery connected then, as you've written, a constant potential difference would be maintained. The capacitance is given by EA/d( E is the permittivity of the medium b/w plates, A is the plate area, d is the distance between them). On introducing the ...


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I think you are mixing up assumptions. You can either hold the potential constant or the charge on the plates constant. If you hold the potential constant then of course adding in a dielectric cannot change the potential between the plates. However, the charge on the plates will change so that the potential remains fixed. Similarly, if you hold the charge on ...


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This is an important point which is commonly overlooked. The crux is that the voltage drop on each capacitor is not necessarily the batteries voltage. Let's work out the ideas step by step. Firstly, to find the charge on capacitor system let us take equivalent capacitance. When we replace the two with an equivalent one, all voltage will drop on that, hence ...


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The steady state here does not refer to the capacitor, but to the whole circuit. There is a current flowing through the resistances, which is why it is a steady rather than equilibrium state. (Steady state generally means that nothing changes with time, i.e., there is no charging/discharging, etc.)


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A) ....Since C1 and C2 are two different capacitors, why gain or loss of same amount of charge would cause them to have the same magnitude of potential? They have the same amount of charge but the magnitudes of the voltages (potential differences) of the capacitors cannot be the same if the capacitances are not the same. The relationship between voltage, ...


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The sign we label on both plates of a capacitor represents the sign of charge accumulated on that plate, which is not an indicator of the sign of electric potential. In fact, the electric potential can take any values (the potential across the battery can be from $V_0$ to $0$, or from $0$ to $-V_0$, or even from $100V_0$ to $99V_0$ if you like) as long as ...


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The electric potential is related to the field this way $$\vec{E}=-\vec{\nabla}V.$$ In case you are not familiar with this expression, it means that the electric field points towards regions with lower potential. If you placed the zero at the positive plate, the value of the potential at the negative plate would be negative.


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Could you please elaborate a little more one why the potential at point $a$ is higher than the potential at point $b$? The definition of high and low potential is based on positive electric charge. The potential at $a$ is higher than at $b$ because it takes work by an external agent to move positive charge from the negative plate to the positive plate ...


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The given RLC circuit with an external voltage $V_S(t)$ is described by the following differential equation: $$V_S(t) = LI(t) + RI(t) + \frac{1}{C}Q(t)$$ or using $I(t)=\dot{Q}(t)$ $$V_S(t) = L\ddot{Q}(t) + R\dot{Q}(t) + \frac{1}{C}Q(t).$$ This is very similar to the mechanical oscillator which is described by the differential equation $$F_S(t) = m\ddot{x}(...


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Think from first principles about the movement of electrons near the capacitor plates. On the diagram you posted, electrons on the left capacitor plate would leave and be attracted towards the positive side of the cell. This leaves the left capacitor plate positive. Since that one is positive, electrons would be attracted to the right capacitor plate (from ...


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