# Tag Info

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You first need to know that how the current flows through a capacitor. There are two parallel plates with a dielectric between them such as air etc. One plate is connected to the positive terminal of the battery while other is connected to the negative terminal of the battery. Charges from the negative terminal will accumulate on the plate connected to it. ...

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Because potential exactly means that you have its potential at every point and to get from start to finish you need to make a number of steps. That is one leap can be broken into a series of steps, every step is closer to the target. As you move closer to the target, your potential becomes closer to the target potential. That is, you have some potential ...

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There is no limit on the 'ability to store charge' involved. What is different, is the proportional relationship between stored energy per unit charge (voltage) and amount of charge (ampere-seconds) stored. A rechargeable battery keeps its charge chemically bonded (and stores a LOT of charge for a given voltage rise), while a close-spaced-plate capacitor ...

2

A lot of these problems are best done by first redrawing the circuit so that it is in a more accessible form. The correct answer is $\frac 8 3 \;\mu$F.

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Comment and Hint: The following figure may help, as long as you maintain all the connections this is the same circuit (will let you identify the capacitors). I get $8/3 \mu F$

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You can solve this strictly with adding capacitance in parallel and elastance in series. The 2uF is in parallel with the others. The two 1uF on the left are in parallel and the 1uF on the right is in series with the two on the left.

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The component that ensures the current is zero just after the switch is closed is the inductor. Inductors do not like changes in current, since a change in current means the magnetic field linking the inductor is changing and this generates a back emf that opposes the change. If you replace the inductor with a piece of wire you would have an RC circuit and ...

1

Setting up the differential equation $$L \frac {dI}{dt} + RI + \frac {Q}{C} = V$$ will not necessarily answer your question, "What and how can I conclude about the current in this circuit just after switch is closed." If you look at the methods of solving the differential equation somewhere on the way to the solution initial conditions are needed, one of ...

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To question 2: When the electron reaches the end of a conductor, it would have to move into the air, which is an isolator. The entire conductor is at equal potential, which is much much lower than the potential at a point out in the air. So it reaches the end and stops, since it is only driven by the potential difference $$F=\frac{dU}{dx}$$ in it's rush to ...

3

So, we have series LCR circuit. $V$ is a constant voltage source. $L$, $C$, and $R$ represents the inductance, capacitance and resistance in the circuit respectively. A current $I$ flows through the circuit. Now, the current through each component is the same. So, the potential difference between each component added up together gives the emf $V$. ...

3

From Kirchhoff's second law, the sum of all the voltages around a loop is equal to zero. That is, the sum of the voltages across the three elements of your circuit, R, L and C, must be equal to the time varying voltage from the source: $$V_R+V_L+V_C = V(t)$$ As $V_R=RI$, $V_L=L\frac{dI}{dt}$ and $V_C=\frac{Q}{C}$, we get your equation, which is correct: ...

2

But I wanted to know whether we can charge a capacitor while it is in use If, by "while it is in use", you mean while the capacitor is discharging, i.e., energy is flowing out of the capacitor to some load, then the answer is no since, by definition, if a capacitor is charging, energy is flowing into the capacitor. Put another way, a capacitor cannot ...

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The capacitance of a capacitor depends on the dielectric constant (relative permittivity) between the plates. The maximum charge that can be stored on a capacitor depends on the maximum potential difference across the plates of the capacitor which in turn depends on the maximum electric field (breakdown potential gradient - dielectric strength) which the ...

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What you have forgotten is that $u$ is not a potential rather it is a number with no units. I do not know how you derived your expression for $u$ but here is a way without a lot of the intermediate steps. The arrangement you have described is similar to that of two parallel line charges with separation $2a$ and charge per unit length $\pm \lambda$ as shown ...

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$1~\textrm{Ampere-hour}$ equals $3600~\textrm{Coulomb}$ of charge. The amount of charge a battery can hold is determined by the amount of chemicals that are in it and it's fixed by the design of the battery. Usually it only decreases because of loss of electrolyte or changes in the chemical and physical structure of the electrodes. The voltage on a ...

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With the capacitor remaining connected to the battery, then the potential difference between the plates is unchanged. If the potential difference stays the same and the gap between the plates stays the same, then the electric field stays the same. Your initial assumption that the field is reduced by a factor $K$ is only true if the capacitor is disconnected ...

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$Q=CV \Rightarrow \frac {dQ}{dt} = I = C \frac {dV}{dt}$ The voltage across the capacitor is equal to the voltage of the supply. So whatever the voltage of the supply does the voltage across the capacitor exactly follows. At time $t_A$ the capacitor is uncharged and the voltage across the capacitor is zero. However at this time the voltage of the supply ...

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Alternating circuit voltage and current periodically reach their respective maximum values after a certain time interval. Consider the current I=I(m)sin(wt) and Voltage V=V(m)sin(wt+pi/2)= v(m)cos(wt). What this phase difference signifies is that if you start the circuit at t=0, then the voltage at t=0 v=V(m) while i=0, meaning the circuit will gain maximum ...

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I expect that the book question is asking for a qualitative answer, as provided by @Unnikrishnan in his comment, rather than a formula. The capacitance increases because the fringing fields effectively increase the area of the plates. You can see this in diagrams which show the field lines extending onto the back of the capacitor plates. These fields are ...

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I think the answer is clearer if you consider the equipotential as shown in the diagram below. Forgive their straight line nature as they were easier to draw that way. Given that $\vec E$ must be perpendicular to an equipotential surface then in your computation of potential difference $\displaystyle V_{AB}=-\int_B^A\vec{E\,}\cdot\mathrm d\vec{r\,}$ the ...

2

Potential refers to a particular point - or set of points which are "equipotential". So you can talk about the potential of one of the capacitor plates (because each is an equipotential surface) but not the potential of the capacitor (because when charged the $2$ plates are at different potentials). When talking about a capacitor, potential usually means ...

5

As drawn, the circuit, assuming ideal circuit elements, is problematic for the reason you've deduced (KVL gives a contradiction). One interpretation is that there is infinite large current for an infinitesimal time which instantaneously charges the capacitors to their final steady state voltages. To gain some insight, add a resistance $r$ in series with ...

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No, that circuit cannot exist in that regime. You are neglecting the internal resistance of the wires between the voltage source and the capacitors, and if the capacitors are discharged (in which case the voltage over them is zero) that's no longer a good approximation. You therefore need to insert a small resistance on either side of the voltage source, ...

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In the case of a resistance based touch screen, it uses it's the user's pressure to control the flow of current. But the capacitive touch screens work in a different way. Human body is basically a conductor. It can hold charge. The capacitive touch screens are constructed from copper or indium oxide. They could store electric charge in between tiny wires ...

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You have to think in terms of potential . Consider the first plate, say $A$. It can be charged to a maximum value $+Q$ since any further increase in charge will cause a leakage of charge due to the increase in potential. You can imagine that. The potential of a charge distribution decreases slowly than the electric field with distance. So we need no leakage ...

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As I read your statements, I get the impression that the difference between capacity and capacitance is not clear to you. The capacity of a capacitor is defined by its "physical" construction (length, width, area, volume, material, etc. C = kA/d). However, capacitance is a measure of how difficult/easy it is for a capacitor to store charge (C = Q/V , ...

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Schrodinger's Cat explains well why only part of the impendance is taken. Why are they considering a phase difference of $\phi$ ... Also, why are they taking modulus of Z Here's an algebraic explanation: For the RLC circuit, we can write the total impedance in a general form, $$Z_T=R + j Z_r,$$ where $Z_r$ is the total reactive impedance of the $L$'s ...

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Why are they considering a phase difference of $\phi$? Your calculations are not totally correct. The voltages across different impedants $V_C,V_R,V_L$ have a phase relationship between them and hence the different impedances $Z_C,Z_R,Z_L$ are not directly linearly related as you have done. Consider the following phasor diagram: I hope it is clear ...

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I'm assuming that you are asking this question in context of an L-C circuit. The reasonant frequency of an L-C circuit is given by the formula $$f = \frac{1}{2\pi}\sqrt{\frac{L}{C}}$$ where L is the inductance of the inductor and C is the capacitance of the capacitor. Hence if any of these two values are changed the reasonant frequency of the circuit will ...

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Assume that the voltage across an isolated capacitor with air ($\approx$ vacuum) between the plates of separation $d$ is $V_{\text{air}}$ and the charge stored on the capacitor is $Q$. $Q = C_{\text{air}} V_{\text{air}}$ The electric field between the plates is $E_{\text{air}} = \dfrac {V_{\text{air}}} {d}$. Now put a dielectric of relative ...

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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 ...

0

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 ...

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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 ...

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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 ...

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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.

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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 ... 3 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 ... 3 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 ...

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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" ...

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