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21

The fan motor provides a torque $\tau$ which has to accelerate $\alpha$ the fan blades whose moment of inertia is $I$: $$\tau=I\alpha$$ Given how long it takes for the fan blades to stop the frictional torques must be fairly low and so the torque applied by the motor to keep them going must also be low. With the relatively small torque rating, even if the ...


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" ...


10

A much simpler way of thinking about this is to consider energy. When the fan is spinning it has quite a lot of kinetic energy (try to stop it by putting your finger in the way to confirm this (don't actually do this!)). That kinetic energy goes as the square of the rotation rate, in fact. So as the fan starts, the motor needs to add energy to it. It ...


10

You might be thinking in comparison to a desk or handheld electric fan. As mentioned by @Farcher, $\tau = I\alpha$. $I$, the moment of inertia of a spinning body around a particular axis of rotation, is calculated as follows: $$I = \iiint\rho(x,y,z)||r||^2\ dV$$ Or with uniform density, $$I = \rho\iiint||r||^2\ dV$$ From this formula, you can see that ...


8

It's a bit complicated (Wikipedia). Induction motors work in sync with the AC frequency but have no torque at 0 RPM so they need some arrangement to get them started.


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 $1/2 QE$.


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 = ...


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 ...


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 ...


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 ...


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 ...


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 ...


2

To this EE's eyes, both formulas are incorrect; the reactance of a capacitor is $$X_C = -\frac{1}{\omega C} $$ and so the sign is incorrect in the first formula. Since reactance is the imaginary part of impedance $$Z = R + iX$$ reactance is a real number and so the second formula can't be correct. It is instead the formula for the impedance of a ...


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 ...


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

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 ...


2

When the fan starts spinning, each blade starts from rest. Newton's first law of motion states that unless a force is applied to it, the [velocity][1] of a body does not change. That property is inertia. To speed up, the blades must accelerate. Newton's second law of motion states that the force necessary for an acceleration of a body is proportional to and ...


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.


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 ...


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 ...


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

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 ...


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 ...


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 ...


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 ...


1

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

(a)You need to have the potential difference across both capacitors to be the same so as their capacitances are different so must be the charges on their plates. So think about $V= \frac QC$ to decide how the net charges are distributed. (b) $C_2$ can be thought of a combination of two capacitors in series and that combination then being in parallel with ...


1

(a) Of course after redistribution of charges among plates $A$ and $D$, an excess negative charge of $60\mu C$ will remain on the wire connecting plates $A$ and $D$. But remaining charge of D is not $30\mu C$. You have $V_C-V_D=V_B-V_A$ and $12\mu F (V_C-V_D )+ 18\mu F (V_B-V_A)=60\mu C$. So charge of plate $D$ is $-\frac{18\mu F}{ 12\mu F+ 18\mu F } ...


1

Electrons will flow from the negative plate of the charged capacitor onto one of the plates of the initially uncharged capacitor. At the same time electrons will also flow from the other plate of the initially uncharged capacitor onto the positive plate of the charged capacitor. So the charged capacitor is losing charge to charge the initially uncharged ...


1

Both your questions relate to the special property of the sine wave, whose derivative is another sine wave shifted ahead by a quarter period, or as you write, 90 degrees, in advance. You are right we can express the current as "i=I sin (wt+90)". But this is a different function than "i=I sin (wt-90)", which would have a wrong sign. Therefore it is only ...



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