# Tag Info

## New answers tagged capacitance

2

In the passage you quoted, McDonald is describing the very edge of the thin conducting disk. Recall that although the disk is thin, it does have a finite thickness. Therefore, its edge looks like this, with the conductor drawn in grey: One way to describe this geometry is to say the internal angle is $\pi/2$. McDonald chooses to instead say that the ...

0

You're confusing charge with voltage/potential. Objects have charge, voltage is measured between objects. If two objects have no charge, the voltage between them is 0. If two objects have the same charge, the voltage between them is still 0. Voltage is always a relative difference between two points. When we say "Terminal A is at 5 volts", what we ...

1

The net charge of any of those internally connected pairs of plates is always zero. That is, when you charge the capacitors, charge doesn't leave the wire between C and D, it only moves along it, and is held in place by the electric field of the adjacent plates. If a circuit is completed that allows charge to flow from D's negative plate to A's positive ...

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The formula $V=\frac{Q}{C}$ gives the amount of charge that is there on one of the plates. The total amount of charge on both the plates taken together is zero! Both of them are oppositely charged.

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For flow of charge, the circuit should be closed. In open circuit, no charge flows. If we connect both the capacitor plates it makes closed circuit, charge flows in the circuit, as a result charges on the plates neutralizes to zero. If only +ve plate of the capacitor is only connected to ground there is no closed circuit. no charges flows from the ground. ...

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since circuit is open no charge is shared. if the circuit is closed ie capacitors are connected by conducting wire then charge will be shared. charge would flow from higher potential to lower until they come to common potential. In the given circuit no charge is shared and potential across both the capacitors remains same.

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Net charge on capacitor is always zero because there is equal and unlike charges on plates. Hence capacitor is not charge storing device. It is electrical energy storing device. In any form of capacitor, stored charge when charged by voltage V is q=cv where +cv is stored in one plate and -cv is stored in another plate. In this question charge stored should ...

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They are different because in a transmission line we have distributed resistance, capacitance, conductance and inductance (meaning that each tiny segment of transmission line has its own tiny resistance, capacitance, conductance and inductance) while in RLC circuits we have lumped resistance, inductance and capacitance. Also RLGC doesn't model a transmission ...

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as $C=E_0\cdot A / d$, the nearness of plates increases the capacitance. if d decrease, capacitance $C$ increases.

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The capacitance $C=\frac{Q}{U}$ is always constant for any types of capacitors. as $Q$ is increased $U$ also increase so that the fraction $C$ remains constant. capacitor is fixed for particular size of capacitor. greater the size of capacitor, greater will be its capacitance. Capacitance is analogous to the capacitance of water tank at our home. larger ...

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Your argument is correct that there if there is no current in the resistor which is in parallel with the $2C$ condenser, then the charge on the $2C$ condenser must be $0$. This, as you probably already know, is because the two elements are in parallel and so they must have the same potential across them. However, you should give a clear argument as to why ...

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One way to compute the electrostatic energy stored into a plate capacitor is simply to calculate it as: $E_{el} \equiv \int d^3r \: \frac{1}{2}\vec{D}\cdot\vec{E}$ which would be simply: $E_{el} \equiv \int d^3r \: \frac{\epsilon_0}{2}\vec{E}^2$ in vacuum. I would just consider the energy difference between before and after insertion of the dielectric ...

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Yes d=0.5d, Basically start thinking in terms of work per unit charge. So work you do on $1$ coulomb is $E$x$0.5d$. as field in conductor is $0$ in electrostatic condition , you don't have to do any work at all in that region . so $\Delta V$ is 0 in that region . Hence , $\Delta V_{net}= E_{outside}$x$0.5d$. = V for the capacitor . And calculate the new ...

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If there is a charge $Q$ and $-Q$ on each plate of the capacitor , when you insert a perfect conductor between the plates (parallel), you simply will have a charge $+Q$ on one side(facing negative plate of capacitor) and $-Q$ on other side of the inserted plate. (Given that the area of plates is large enough to assume constant electric field between them) ...

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There are 3 main reasons for using a capacitor. First it stores the energy, so it can deliver a pulse of energy that is far larger than the battery can. Remember it may take several seconds of battery energy to fully charge the flash capacitor. Then the capacitor releases all that in less than a millisecond ($10^{-3}s$) or even just a few microseconds, so ...

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Because we don’t need lighting, but flash, that means we have to release much energy is a short time interval. Capacitors could accumulated and release energy fast.

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Edge effects. After the electron leaves the capacitor, the electric field winds up slowing it back down. Let's assume the capacitor is infinitely-massive and that the acceleration of the electron is small enough that we can ignore radiation. Then if you were to idealize the electric field of the capacitor, treating it as a uniform field between the plates ...

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When we're calculating the energy stored in a capacitor we normally assume it is isolated i.e. there are no other charges nearby to affect it. This makes the calculation nice and simple: the energy is proportional to $Q^2$ and the energy is stored in the electric field around the capacitor. However in your question you are introducing another charge, your ...

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The energy is of course coming from the electric field of the capacitor. The energy of any capacitor is always stored in it's electric field. If an electron is initially positioned very far away and then moves close to the capacitor, it's being pulled by the field and that means energy is being transferred. The electric field get's a little weaker - loosing ...

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Capacitor is losing energy, potential has changed as field is created even by this charge which is moving under the influence of force between capacitor plates . Take the point charge's potential , and then assume distance between capapcitor plate is d, now as -ve charge approaches +ve plate, it decreases the potential of the +ve capacitor plate more than ...

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The capacitance is the ratio of charge on the plates over the voltage applied. $$C = \frac{Q}{V} \Leftrightarrow Q = C \cdot V$$ The calculation you show determines the capacitance from measured voltage and charge on the plates. You basically know the result you want and determine the size of the capacitor you need. A larger capacitor, with a larger ...

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