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1

It doesn't. For DC, the current flows in one direction. Charges will flow to the capacitor plate. In the very instant where the capacitor is connected to the battery, the current flows as if there was no capacitor. Only after a short while will a bunch of charges have been accumulated at the plate, starting to repel the incoming charges, lowering the ...


0

Consider a plane of charge. For distances from the plane which are small compared to the size of the plane the electric field will be constant. In the capacitor, with two oppositely-charged plates and relatively small gap, the electric field is a constant, depending only on the overlapping plate area and the amount of charge, $E_{gap}=Q/(k\epsilon_o A)$, ...


1

The working voltage of a capacitor depends on the dielectric strength of the insulator. While electrical breakdown is actually a very complicated process with lots of non-linearities, you can simplify the design of a capacitor by saying "the electric field on the insulator must not exceed X". Once you have said that, and you realize that the electric ...


1

Since the charge Q have to be the same for both capacitors and you need more voltage to to push that charge in the capacitor with less capacitance then you must have more valtage difference in $C_2$ The mechanical analogy is a configuration with 2 springs in parallel that move the same distance from their equilibrium position need more force on the spring ...


2

When you ask questions about things "in the limit", the answer is almost always "It depends". In this case, the answer is "it depends". The equation $Q=CV$ assumes linear behavior of the capacitor - in reality the dielectric of most capacitors has hysteresis as well as a nonlinear component, so as you increase the voltage, the capacitance will change. This ...


0

First, I'm a bit confused by your statement about two batteries. The capacitance of a two-sided capacitor is defined to be the ratio of the magnitude of separated charge (commmonly, the magnitude of charge on each plate) to the resultant potential difference (aka, voltage) between the plates. It's technically a "what if" formula. The actual capacitance ...


0

Batteries are not capacitors. Closing the switch doesn't redistribute some fixed amount of charge in the circuit. Instead, the batteries can create new charge through chemical reactions. The redistribution would slightly lower the potential, allowing the chemical reaction to proceed until it comes back to the resting voltage. While real batteries are ...


1

If there were no other absolute voltage references in the problem, then you would be correct. But there is already a connection to ground on the upper trace of the diagram. The voltage of that trace does not change (it starts at $0$ and remains $0$ after closing $S$).


-1

I think it is possible to produce equipment converting kinetic energy from a bullet. We could use a string attached to a generator and a spring. The string then turns the generator and thereby produces electricity but it is to hard to to something that does not get damaged from the bullet. Here we can use momentum. If it is supplied to create thick but empty ...


1

Actually, you are all way over-thinking the issue. There is a trivial material that can slow a bullet down without being damaged at all. Think water. A couple feet of a water shield will do exactly what you need to stop a bullet enough to make it basically harmless with a bit of additional bullet-proofing. Most of the bullet's kinetic energy will stay in the ...


2

What is your goal - to stop bullets or to generate electricity? The key to stopping bullets is not so much to absorb the energy (although that matters too) but to absorb the momentum. You may know that $$p = F\Delta t$$ In other words, given a certain momentum $mv=p$, you need to apply a force F for a time $\Delta t$ in order to slow it down. The ...


-4

There is no physical challenge (at first glance). This is a technological issue. Man just didn't developed technology to do this (and they are not trying).


0

Your material has to absorb the kinetic energy before converting it into other forms. A thin, lightweight panel would be unable to absorb the bullet's kinetic energy before it punched a hole in the panel and continued on it's way. Plot the tensile strength of the material against the impact energy of the bullet. Unobtainium, impossibrium, phlebotinum and ...


1

If you have an isolated capacitor, so that there is no conducting path for charge to flow from one plate to the other, then the charge on the plates will be conserved as you change the geometry. Since $$Q = CV,$$ a drop in the capacitance $C$ is matched by an increase in the potential $V$. Note that the stored energy $U$ in the capacitor, $$ U = \frac12 ...


0

Model the current through the resistor as $$ i(t) = v(t)/R, $$ the current through the capacitor as you have above, that is $$ i(t) = C \dfrac{dv}{dt}, $$ and the turning on/off of the switch can be modelled using the Heaviside function which would pre-multiply the above expressions. The current has to be continuous throughout so we can equate the two ...


-1

You can substitute the switch with a resistor $R_2$, that has a value of either $0$ or $\infty$.


0

If the switch is open, then no current can pass through the circuit. Your ODE should then take the form $$ \frac{di}{dt}=0\tag{1} $$ When the switch closes, then current can pass through, leading to $$ R\frac{di}{dt}+\frac{i}{C}=0\tag{2} $$ When the switch opens again, we revert the ODE back to Equation (1). Thus, the time range $0\leq t\leq2$ gives you ...


0

In the real world you can never rid yourself of resistance in an electrical circuit. There is resistance in the connecting wires and resistance internal to the capacitor itself. So once you close the circuit and current starts moving, some of that electrical energy will be lost to heat, dissipated by resistance.


3

Comment on the question (v1): If you're confused by long-time behavior of your circuit, its behavior may be sensitive to details which are safe to neglect when describing short-time behavior. If you want a real answer, you should edit your question to include a complete circuit diagram, and perhaps a plot of your data. For example: in one of my labs I have ...


0

The maximum of $$ \left|\cfrac{V_2}{V_1}\right| = \cfrac{\sqrt{1+(\omega RC)^2}}{\sqrt{4 + (\omega RC)^2}} $$ is $1$ at $\omega=\pm \infty$, and you find the half power frequency by solving: $$ \frac{1+(\omega RC)^2}{4+(\omega RC)^2}=\frac{1}{2} $$ which gives $\omega=\pm \sqrt{2}/RC$


1

Suppose you set the zero of potential so both conductors have zero charge at zero potential. If you then set them to potentials $V_A$ and $V_B$, you can prove that they will acquire charges \begin{align} Q_A=C_A V_A+C_{AB}V_B,\\ Q_B=C_{BA}V_A+C_B V_B, \end{align} respectively. This is the real definition of capacitance, and particularly of mutual ...


3

A capacitor stores electric energy in a static electric field between two conductors. The conductors are separated by an insulator called a dielectric, which allows the conductors to be very close to each other without contact. The closer together the conductors, the greater the storage capacity of the capacitor. This storage capacity also is proportional ...


1

I understand the problem now from leongz's excellent explanation and patience in the chat discussion. Thank you very much leongz for helping me out! The capacitor is being discharged through constant current. If it starts with charge Q_0, then from the definition of current we know that the charge decreases by It after time t, where I is the constant ...


2

Electrons flow through the wire from the battery's negative terminal to the battery's positive terminal. If said wire is actually a capacitor, the electrons still flow the same way - from the battery's negative terminal. But since it's a capacitor, the electrons are pushed into the capacitor's negative plate instead of making it all the way to the battery's ...


2

No. Step number two already mentions electrons from the battery flowing into the negative terminal of the capacitor, giving the negative terminal a negative charge. Step number three is talking about electrons flowing out of the positive terminal of the capacitor, giving the positive terminal a net positive charge.


0

as you know that inside a capacitor electric field remains same. If you increase the distance between the two plates electric field does not change just because electric field= surface charge density/ epsilon. so E=V/D gives increment in V as D increses so that electric field remain same.



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