# Tag Info

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As another answer pointed out, your formula is for electric field around an isolated point charge. It doesn't apply to the case of parallel plate capacitor. Normally we use Gauss's Law to find the electric field between the plates of the capacitor. We know that the field between the plates will be uniform from the differential form of Gauss's Law ...

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You reference the equation giving the electric field near a finite point charge. There is no finite point charge in a capacitor (unless we count a single electron, but I think you'll find a single electron won't produce a very large field measurement on a human scale...). The charge is distributed uniformly, and as you (you're a test charge) get closer to ...

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there are lots of questions and explanations, why the field of an infinite plain is homogeneous and does not depend on distance. That is the approximation involved: that the plates are big. So outside the plates the fields add to zero, and inside it's double. No, charge is not brought from one plate to the other. If you have an alternating current, it will ...

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Suppose that you double the separation of the plates of a parallel plate capacitor of initial capacitance $C$ which is connected to a battery of emf $V$. The capacitance of the capacitor become $\frac C 2$ and the energy stored in the capacitor changes from $\frac 12 CV^2$ to $\frac 1 2 \frac C 2 V^2$. There is a decrease in the energy stored in the ...

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I had thought that for part A both light bulbs would begin glowing since the capacitor isn't charged, but i have no idea how to tell which one is brighter. I would reason about this in the following way. (1) This is a series circuit and so, the current through each circuit element (battery, bulbs, capacitor, switch) is identical. (2) The bulbs are ...

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You can solve this problem by using Kirchhoff's two laws. Kirchooff's current law tells you that since this is a series circuit the current in each part of the circuit will be the same all the time. So whatever happens to bulb $A$ will also happen to bulb $B$ as they both have the same current flowing through them. Using Kirchhoff's voltage law you have ...

1

You're on the right track. You're probably familiar with how the current decreases exponentially after closing the switch. $$I(t)=I_0 e^{-t/\tau}$$ Where $\tau$ is the time constant of the circuit given by $\tau = RC$, and $R$ is the total resistance of the bulbs. So when the switch is closed, current will be a maximum, and the bulbs brightest. As time ...

4

a. Immediately after the switch is closed, are either or both bulbs glowing? Explain. They will both glow as some current passes through them as the capacitor is charging. b. If both bulbs are glowing, which is brighter? Or are they equally bright? Explain. They are both equally bright, because an equal and opposite charge is flowing on to ...

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You have to keep this important thing in mind while dealing with problems of these kind. A capacitor behaves as a pure conductor immediately after connecting it with the circuit (i.e., at time t=0). As time passes the capacitor begins to lose its conductance exponentially and finally after a very long time (theoretically when time t tends to infinity) it ...

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If you have covered ac circuit theory you would have found out that for a series circuit the currents in each part of the circuit are in phase but the voltages across the components in the circuit may be out of phase with the current. For a resistor the voltage across it and the current through it are in phase whereas for a capacitor the voltage across it ...

0

This question really boils down to the basics of AC circuits. If you check your theory, you will observe that the voltage (time-varying) applied to a purely capacitive (power source and capacitor) circuit lags the current by $90^0$. But the same voltage applied to a purely resistive (power source and resistor) circuit is in phase with the current. Hence, ...

1

If the inner conducting shell is charged then the charges will reside on the outside of the inner shell as there can be no electric field inside a conducting shell. The charges on the outside of the inner conducting shell will produce a radial electric field and the outer conducting shell will find itself in that electric field. However the outer ...

2

Forget anything about a capacitor and just consider the resistance of the conducting liquid. Think of the liquid as made up of thin $dr$ concentric shells of radius $r$ and find the resistance of a shell in terms of the resistivity, radius and thickness. Then do the integration to find the resistance of all the liquid.

1

For instance, why don't measure the ability to store something by the volume it takes so why not charge per unit volume. There is nothing wrong with you defining a parameter which is the "charge per unit volume" but after defining it then what are you going to do with it? So here you have a capacitor and its charge per unit volume is $3 \;\text{C ... 4$C$is a property of the system. It connects the Voltage across the capacitor$V$and the charge$Q$stored in it. You can do the same thing with a spring using the system equivalence between a capacitor and a spring. The conclusion to draw from your relations is this. 1.) Suppose you have a lot of capacitors with different$C$and you apply the same ... 0 The 2nd and 3rd forms of the equation comes from the following relation $$\boxed{\boldsymbol {C= \frac {Q}{V}}}$$ Where$\boldsymbol {C}$is the capacitance,$\boldsymbol {Q}$is the charge and$\boldsymbol {V}$is the potential. The formula for Energy stored in the the capacitor is $$\boxed{U=\frac{1}{2}QV}$$ Replacing$Q$with$CV$, from the ... 2 Boundary conditions of electromagnetic waves are well looked into because of the nature between polarization and reflection. However in this particular case it is rather simple, it is the evaluation of a contour integral over the boundary. This is done by summing the length elements of the pill-box shape that is typically used in this analysis in the correct ... 2 I understand that capacitance is the ability of a body to store an electrical charge and the formula is$C = {Q \over V}$Perhaps you just need to top thinking of capacitance as that. "Capacitance" sounds like "capacity", which leads to an intuitive trap like this: If I have a basket with a capacity of 2 apples, then a basket with more capacity can ... 2 We Use$C=Q/V$because those were useful things to measure. It's often easy to forget, but many of the equations we use are chosen because the work, and because other equations didn't work. Never underestimate that part of the reality. We don't use "charge per unit volume" because that number is not constant. You can charge a capacitor up without ... 20 You can use a high vertical tube to store water in it (fill it from the bottom by pushing the water in) How much water can you store? It obviously depends on the pressure you apply to push it in. If you push harder, there will be more water stored. The tube is characterized not the amount of water, but by how easy it is to store the water. Its "capacity" ... 2 A capacitor is used to store energy in form of electric fields. This electric field is created by charges on plates of capacitor. So, basically you are storing charge on capacitors. Let someone ask you how much charge you can store in your capacitor.What would you reply? Clearly , you reply " I may store 1mC or 100mC, depending on Potential difference ... 4 Capacitance is "charge over voltage" – and one farad is "coulomb per volt" – because the capacity of capacitors (something that determines their "quality") is the ability to store a maximum charge on the plate ($+Q$on one side,$-Q$on the other side) given a fixed voltage. When you try to separate the charges, you unavoidably create electric fields ... 1 why would increasing voltage, while keeping charge constant, have any effect on the ability of a body to store charge. (1) Capacitors don't store charge, they store electrical energy. For a capacitor, it is understood that one plate has charge$Q$while the other plate has charge$-Q$so there is no net electric charge stored. (2) If you increase ... 0 First off, what is meant by the potential of the conductors? The "absolute" potential is of no importance in a working capacitor. In fact, it's practically impossible to determine. The potential difference between the plates (or between a plate and some ground/earth plane) is the important factor in the capacitance relationship,$C=qV$. Now ... 2 Induction current and displacement current are similar though, but very different terms. Induction current is the normal electric current but displacement current is the term which is only defined by Maxwells' equations. Now generally capacitors have a dielectric material between the +ve and -ve plates. When the capacitor is charged, there exists a strong ... 0 a) Capacitance C increases b) Charge Q remains unchanged c) if charge Q is constant while C increases, that means voltage V decreases (C=Q/V). d) U decreases ( U=Q^2/2C) 0 For a discrete system of$Ncharges, the potential energy associated with their configuration is given by \begin{align}U&= \frac12\,\sum_{j=1}^N \, q_j\sum_{k\ne j}\,\frac1{4\pi\epsilon_0}\,\cdot \frac{q_k}{r_{jk}}\\ &= \frac12\,\sum_{j=1}^N \, q_j\,\varphi(\mathbf r_j)\tag 1 \end{align} where\varphi$is the scalar potential due to all charges ... 0 You have $${U = \frac{1}{2}\frac{Q^2}{\frac{\epsilon_{0}A}{d}}}$$ $${\frac{\epsilon_{0}A}{d}}=C\equiv\textrm{ the capacitance of the capacitor.}$$ Hence $${U=\frac{1}{2}\frac{Q^2}{C}}$$ Now$\displaystyle{Q=CV}$, where V is the p.d. between the capacitor plates. So, $${U=\frac{1}{2}CV^2}$$ or $${U=\frac{1}{2}\frac{\epsilon_{0}A}{d}V^2}$$ ... 0 What 'behaviour' are you asking about? Magnetic fields only affect moving charges. If the charge on the capacitor is not changing, the magnetic field has no effect, regardless of whatever direction it is in. No charges cross the space between the capacitor plates. Perhaps you are thinking of a situation where charged particles are fired into the space ... 0 In your layout, imagine the "antenna/you"-capacitor being parallel to the existing one, the antenna being the upper plate. Parallel capacitors add up their capacity. So how do you become the plate although you are not connected to the wires? The first step is to understand is that this setup (inductor + capacitor) will generate frequencies, as you could ... 0 The player's hand acts as a grounded plate remembering that the player is a reasonable electrical conductor. The capacitor is part of an inductor-capacitor circuit, as you have shown above, which control the frequency of an oscillator. So what is missing is a clear indication that the bottom part of the circuit is connected to the earth/ground. 2 When you apply an AC voltage to a capacitor, current will flow into the capacitor, and back out again. As long as the absolute voltage on the AC generator is higher than on the capacitor, current will flow to increase the charge on the capacitor; and when it's smaller, current will flow to decrease the charge. Note that since there is an equal and opposite ... 2 The electric potential of a point in space is defined by the mechanical energy that it takes to get a (positive) unit charge to that point. We usually define the potential of an infinitely distant point as zero, and then the movement of our test charge is from infinity to the point for which we want to measure the potential. In case of a capacitor the ... 0 The two networks will be equivalent if the e.m.f in the second network is equal to$Q/C$in the first. Then you will get two networks with same initial voltage across their terminals and same impedance too (since$z_{voltage source}=0$), so they are equivalent. As you are asking for an example, consider the simple R-C series circuit, first with a charged ... 0 No. Any two objects which are insulated from each other have a capacitance. The bigger the objects are and the closer together they are, the more capacitance they have. The most efficient way to pack a lot of capacitance into a small device is to use an insulating dielectric film to hold two wide, thin conductors just barely apart from each other over a ... -1 First of all, you should see some properties of a capacitor. A capacitor is a device capable of storing electrical energy. The capacitor can induce an impedance depending on the capacitance of the capacitor and the frequency of input voltage applied. It's called capacitive reactance and is given by:$X_C$=$1/2πfC $where f is the frequency of input ... 0 This is old question; but a nice one. The simple part is simple: Connected means, there are cables - cables do not allow any potential difference. So the voltage over both capacitors will be the same. You have to distribute the net charge (which is the sum or the difference of the single charge, depending on how they are joined) on them in the ratio of their ... 0$C=Q/V$deals with quantities that can be seen from outside: if you connect your device to a voltage, there will be some amount of charge that it will hold till it is "full". In your problem one of the charges is the internal charge, which is something different. Think about what it does: you have some charge on the plates of the capacitor, and the charge ... 0 As an answer to your second question. If you released the dielectric it would reach the equilibrium position and then overshoot as it would have kinetic energy eventually stopping on the other side and then returning. Compare this with the oscillation of a spring-mass system. So the dielectric would undergo oscillatory motion and if damped it would ... 0 The relative permittivity is the ratio of the electric field in a vacuum (due to the charge 7.1$\mu$C) and the electric field with the dielectric present due to the original charge and the induced charge. 4 The existing answers tell you why$F= \frac{1}{2} QE$is right, but I think it's important to say why$F = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{d^2}$is wrong. Coulomb's law is not easily applicable here because the plates are not point charges. In particular, their sizes are not negligible (indeed, much larger) than the distance between them. It would ... 0 In reference to what you are asking, the two similarly charged plates would repel each other. Now if the two plates are kept in an electric field then the two cases can exist: If the outside electric field is opposing the electric field between two plates and is larger than the field between between two plates than the paper bits would be attract ed towards ... 5 The energy of the capacitor is$U= \frac{\epsilon_0}{2} S\,\mathrm d E^2$where$S$is the area of a plate. If we increase of$\Delta d$the distance of, say, the right plate from the left one, keeping fixed the charge$Q$on each plate,$E$does not change and we find a variation of energy $$\Delta U = \frac{\epsilon_0}{2} S E^2 \Delta d = ... 5 The E field due to each plate is E/2 and hence the total field between the plate is E. But a plate won't exert force on itself, so the E field experienced by a plate is E/2 only. Multiplying by charge gives you the force, hence 1/2 QE. 0 You would expect that the force would not depend on whether or not the capacitor was connected to a constant voltage source V. If the capacitor C = \dfrac {\epsilon_o A}{d} is connected to a constant voltage source the the energy stored in the capacitor is U = \frac 1 2 CV^2 = \dfrac 1 2 \dfrac{\epsilon_o A}{d}V^2. The energy stored in the capacitor ... 0 Suppose that initially the capacitor was uncharged. A current would flow in the circuit which would charge the capacitor. The maximum voltage that the capacitor or the resistor can have across them is the voltage of the battery which is 40 V. The sum of the voltages across the capacitor and the resistor must equal the voltage of the battery. No matter how ... 0 Let us write down the equation of motion for the given RC circuit. Using Kirchhoff's voltage law we have U=iR+\frac{q}{C}, where i is the charging current in the circuit at time t, q represents the charge stored in the capacitor at time t. Initially(at t=0), q=0 and i is a maximum.The charging current decreases with time because of a build-up ... 3 While the other two answers are technically correct, they are not actually addressing the engineering aspects of what comprises a "good" capacitor. Ideal capacitors in parallel or series circuits lead to ideal capacitors of different value, i.e. there is no measurable "quality" difference - ideal is ideal. In reality, of course, there are no ideal ... 1 in terms of making a better capacitor that can store more charge would you use in series or in parallel? To be sure, capacitors don't (ordinarily) store charge, capacitors store energy, i.e., a 'charged' capacitor is electrically neutral. If, by better, you mean store more energy for a given voltage, then you want the combination of capacitors to be ... 0 If you want to store more charge (assuming you're not increasing the voltage) you want more capacitance. The way to do that is to put your two capacitors in parallel.$$C_{T}=C_{1}+C_{2}$\$ If you think of each capacitor as pair of parallel you could imagine them in parallel as being side by side and effectively just adding the area of plates together to ...

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