# Tag Info

1

The first equation assumes the external electric field (caused by the charges on the plates of the capacitor) doesn't change. When a battery is connected, it can fill and discharge the plates as necessary to maintain the voltage. So, the E_0 value increases as you add the dielectric.

0

Without even doing any circuit calculations, you can conclude the voltage between a and b is zero by symmetry. Proof: Assume there's a voltage between the two points. If you close the switch, a current would flow. If you take the mirror image of the circuit, you'd expect the same current to flow, but in the opposite direction. Except the circuit is left ...

0

First, let's assume the left-most terminal is connected to the positive terminal of the battery and the capacitor voltage reference direction is left-most terminal positive. Now, consider a KVL loop clockwise through the top 2C capacitor, the switch, the bottom C capacitor and the battery: $$10 \mathrm V = V_{2C_{top}} + V_{ab} + V_{C_{bot}}$$ So, the ...

0

The potential difference across the top two capacitors must be the same as the difference across the bottom two. I will number the capacitors $C_{11}$ for top left, $C_{12}$ for top right etc. If we assume the charge on each capacitor is the same, then the voltage difference must be zero. But if we can assume that each capacitor may have a different charge ...

1

I asked a somewhat different, yet similar question.Hope this helps! Why is an $LC$ oscillator lossless, but $C V^2 / 2$ energy is lost to a capacitor connected to an ideal voltage source?

-1

I can't understand your question thoroughly. But if you connect the negative terminal of the battery to any conducting body like a metal can, then there will be a flow of charge till the potential of the negative terminal and the metal are equal because the metal body and the negative terminal are a single conducting body and a conducting body has a ...

0

Everything you say is correct in the steady state. The problem you run into is that when you remove charge from a charged capacitor to an uncharged capacitor, there is a potential difference. And somehow, you have to remove the energy from the electron that moves from one to the other. It turns out, as you calculated, that you in fact remove half of the ...

0

The explanation is simple. Start with a definition of voltage: the work done moving 1 coulomb of charge from point $a$ to point $b$. In this case $a$ to $b$ is one plate of the capacitor to the other, since we are talking about the voltage across the capacitor. And now a definition of the work done: it's $\text{force} \times \text{distance}$. A capacitor ...

1

Perhaps this is what you are looking for: Screen capture: http://www.falstad.com/circuit/ The default circuit, as shown, is an LRC circuit. On the Schematic: Gray is 0V Green is Positive Voltage Red is Negative Voltage The yellow dots are a visualization of current: positive holes. The graphs along the bottom, from left to right, are for the ...

1

The difference is that batteries chemically "pump" electrons from one side to the other. There is a small amount of charge separation in a battery even when it is not connected to a circuit. This charge creates an electric field that opposed the chemical action of the battery to prevent further charge separation. This makes the battery act somewhat like a ...

0

The chemical reaction in a battery can only happen when there's a complete circuit; the chemicals at one electrode can then dump electrons into the circuit, while at the other electrode they take electrons away. The rest of the time, the chemicals are electrically neutral. So while the whole principle of a capacitor is to keep charges separate, batteries ...

0

Using Kirchhoff's First Rule,finding charges in capacitors between D & C (Let it be $Q_m$). $$Q_1-Q_m-Q_2=0\\ Q_3+Q_m-Q_4=0\\$$ Using Kirchhoff's Second Rule,finding voltages loops ADC & DBC. $$\frac{Q_1}{C_1}+\frac{Q_m}{C_m}-\frac{Q_3}{C_3}=0\\ \frac{Q_2}{C_2}-\frac{Q_4}{C_4}-\frac{Q_m}{C_m}=0$$ When the Bridge is balanced,$Q_m=0$,So the set ...

4

The key to understand the issue is that between the upper and lower "corner" of the circuit the voltage is always zero, therefore no current will flow across $\mathrm C_5$ and $\mathrm C_6$. In "corners" I mean the points common to $\mathrm C_3$-$\mathrm C_4$ and $\mathrm C_1$-$\mathrm C_2$. These pairs of capacitors are effectively voltage dividers. ...

0

$$V = {v_r}(t) + {v_c}(t) = i(t)R + {1 \over C}\int\limits_{{t_0}}^t {i(\tau )d} \tau$$ Differentiating wrt t: $$RC{{di(t)} \over {dt}} + i(t) = 0$$ Solving differential equation: $$I(t) = {V \over R}{e^{{t \over {{\tau _0}}}}}$$ From this equation we notice initially at t = 0 $$I = {V \over R}$$ As time is increasing current starts decreasing until at ...

4

By capacitor charge is meant the absolute value of the charge on each capacitor plate: $\mid Q \mid$. If the battery generates the potential difference $V$ and you connect the capacitor to the battery through a conducting wire, as shown in your picture, once the equilibrium is reached each plate of the capacitor will have a charge $Q = CV$, where $C$ is ...

0

Based on equation I just found I was able to solve for the phase angle. phi = arctan((XL-XC)/R) Information from http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rlcser.html

3

Just a simple answer - there's nothing to dissipate energy. If there were a resistor in the circuit, it would dissipate energy as heat. Inductors and capacitors don't dissipate energy. The energy just sloshes back and forth between being stored in the magnetic field, and being stored in the electric field. It's just like a spring-mass system, where energy ...

5

Wouldn't this inductor's emf counteract the discharging capacitor and actually charge it? / stop the capacitor from fully discharging? The inductor doesn't care about what the charge state of the capacitor is. All it cares about is how quickly the current through it is changing, and it generates a back-voltage according to the equation V=L*dI/dt. You ...

2

The inductor never creates a current in the opposite direction. An inductor creates an EMF to counteract the changing B field(Lenz law). The B field is changing because the current in the inductor is changing. So effectively, the inductor resists changes in current. So initially, the capacitor tries to discharge strongly but is slowed down by the ...

2

$E_1 = E_2$ . since $E$ is independent of dielectric as long as potential b/w plates is constant. $$E= = -\frac{dV}{dr}$$ So, it is independent of dielectric b/w it. So, correct statement would be $$E_1d = E_2d$$ $$Ed = Ed$$

1

You seems to assume both capacitors has the same plate separation $d$. So, lets assume that. Assume there is no dielectric material. Therefore, nicely $Ed = Ed$ in both capacitors. Which is nice. :). Now, I think I understand your confusion. Have an isolated capacitor with electric field inside plates of $E$. Insert dielectric $K$. Under this case, the ...

0

Let's accelerate a charged plate capacitor by pulling one of it's plates while carefully adjusting the pulling force so that: A) The distance between plates stays constant in the capacitor frame B) The distance between plates stays constant in the capacitor's original frame What are the energies of the E-fields? (not counting kinetic energy) Case A: ...

-1

It's actually very simple: Energy in motion has kinetic energy. If energy is E in the frame where the energy is not moving, then in the frame where the energy is moving the energy is $\gamma E$ People here have calculated that the energy of a very fast moving E-field without the kinetic energy is very small. OK, but the energy of a very fast moving ...

0

It is not entirely correct to consider the "energy of an electron" in this regard. The energy is not associated with the electron alone. Rather, we call it electric potential energy and it depends on total electric charge at the same spot and is compared with some other area in the circuit where the potential energy is zero as a reference. Because, if there ...

0

$E=\frac{1}{2} CV^2$ is energy stored due to all electrons or to say more correctly due to potential difference created by electrons in plates. As distance between 2 plates is increased electric field between plates decreases and charge also decrease on plates.Since, $C=\frac{{\epsilon}{A}}{D}$ , capacity of capacitor decreases. Hence its energy stored also ...

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