# Tag Info

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the ratio of the change in an electric charge in a system to the corresponding change in its electric potential is called capacitance I.e the ability of system to store charge. You can find more info here. https://en.m.wikipedia.org/wiki/Capacitance

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Because an LED acts like a diode, the negative current from the AC source will be clipped and the capacitor will always be charged and it will act like a second voltage source. Look at my picture below. You can see that the green line represents the voltage going through the capacitor and the blue line (it's a little hard to see since it's covered up by the ...

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Capacitors connected in series $$\frac{1}{C_{\mathrm{total}}} = \frac{1}{C_1} + \frac{1}{C_2}$$ (Where $C$ is the capacitance) Capacitors in parallel $$C_{\mathrm{total}} = C_1 + C_2$$ As you can see, capacitors are total opposite of resistors when connected in series or parallel More capacitance or $C$ means the capacitor can store more charge. As ...

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It's not so much the increase/decrease in charge that is responsible for a change in mutual conductance, as the change in geometry, since by definition the mutual capacitance of charged conductors is independent of their charge. This answer on the meaning of mutual capacitance provides details if needed. The interesting thing though is that the effect ...

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The phenomenon you are talking about is called dielectric absorption. The way it works is this: Let's say you've just discharged a capacitor. An ideal capacitor would remain at zero volts after this. However, in real life, the capacitor will develop a small voltage from time-delayed dipole discharging (also known as dielectric relaxation). Dielectric ...

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Actually charges do accumulate even around a resistor. Consider the following model: two conductors with resistances $R_A$ and $R_B$ and lengths $L_A$ and $L_B$ are connected to each other, and a potential difference of $\Delta U$ is applied to their free ends. By Ohm's law we have: $$\Delta U_A=IR_A,$$ $$\Delta U_B=IR_B.$$ This means that, for equal-sized ...

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I think it is easy to understand. Here I am giving a heuristic picture, current in any wire is generated by charge flow. $I=\frac{dQ}{dt}$ now if you apply alternating current on the sides of capacitor you will find that charge on one plate is constantly increasing and decreasing, which induces the opposite charge on the other plate changing with same rate....

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You are referring to the formula for capacitance of a uniformly charged disk. A metal disk has uniform potential, but NOT uniform charge: the charge density at the periphery of the disk is higher than at the center.

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In principle any acceleration of an electron causes some radiation, and an electron has to accelerate in order to leak from one plate to the other. However: the velocities, and therefore the accelerations, of electrons in electrical circuits are small. Calculating the electron drift velocity is an exercise routinely given to students and the results tend ...

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The force $F$ between capacitor plates is discussed here and shown to be $\frac 12 QE$ where $Q$ is the charge on the capacitor and $E$ the electric field strength. In your example with a constant voltage this is better written as $F=\frac 12 CV \; \frac V d$ where $C$ is the capacitance $=\frac{k\epsilon_o A}{d}$ with $d$ the separation of the plates and ...

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Hint: To find the effective capacitance you probably used the following steps: (i) Replace the $5 \mu F$ and $6 \mu F$ capacitors by a single effective capacitor, say $C_1$; (ii) Replace the $2 \mu F$ and $4 \mu F$ capacitors by another single effective capacitor, say $C_2$; (iii) You now have two capacitors in parallel and you can now replace the ...

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You obviously know the formulae for finding the equivalent capacitance for series and parallel combinations. Step back one pace to remember that capacitors in series have the same charge on them and capacitors in parallel have the same voltage across them. So solve the problem by looking at the pairs of capacitors which are in series with one another not by ...

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$$C = \frac {Q}{V}$$ $$Q = VC$$ You can use the capacitance ratios to find the voltages on them depending on how they're arranged with relation to the battery (it is unclear from your depiction). Then multiply the voltage by its capacitance to get the charge, $Q$.

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Here is another attempt at an answer, from a slightly different angle. It is less sophisticated and may add insight. You say that the capacitors are isolated before they are connected. Literally, that is only possible if they are in different universes. In the real world, they are connected through an infinitesimal capacitance - the capacitance between the ...

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If the charges on the plates of the capacitor are +Q and -Q then the PD between the plates is V = Q/C where C is capacitance. (This is the definition of capacitance.) If we add the same amount of charge Q'>Q to each plate, this has no effect on the PD between the plates, because it will increase the absolute potential of each plate by the same amount. The ...

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Let us say plate A has a charge q1 and plate B, which faces plate A has a charge q2. By making use of the fact that the net field in the bulk of a conductor in static conditions is zero, and that the net field near the outer surface of a conductor equals [local surface charge density/€0], you can prove the following: Charge on the outer surfaces of A and B ...

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Measurements disturb the double slit experiment because of the particle nature of light and matter. In order to measure which slit the electron passes through there must be some sort of interaction to detect the electron. By putting a capacitor in the way, you would drastically affect the particle. Think of it like putting a hose-pipe in front of a ...

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Suppose the surface charge densities on the bottom plate is $\sigma$ and on the top plate $-\sigma$, then the electric field due to the bottom plate is $\frac{\sigma}{2 \epsilon_0}{\bf n}$ and that due to the top plate $-\frac{\sigma}{2 \epsilon_0}{\bf n}$, where ${\bf n}$ is a unit vector pointing from the bottom plate to the top plate. This gives the total ...

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Charges flow until the potential difference across each capacitor is the same. In each capacitor the product $Ed$ will be the same. The electric field in each capacitor will change because the charge density on the capacitor plates changes.

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Start off by considering what the charge distribution would be like without the battery being connected. The charge distribution must be such that there is no electric field inside either of the plates. The consider which of the charges will change when the battery is connected.

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Your diagram shows that you have almost solved the problem. You have obviously correctly decided on how the charges ($4Q$ and $-2Q$) are distributed on the original capacitor. If you double the area of the plates for the first capacitor what happens to the charge distribution on the plates and what happens to the capacitance of the larger area arrangement ...

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The trick to this question is that neither situation can exist in reality. There are no perfect insulators (required to isolate the plates in part 1), and no perfect conductors (required for the wires). Both of these situations are ideal cases that exist at the end of an extrapolation of what can happen in reality. To bring this situation back to reality, ...

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