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1

The battery doesn't do any significant work if there's nothing connected to it. You can think of a battery with nothing connected to it as being a battery with an extremely large resistor connected to it. That's even essentially physically accurate, instead of just being an idealization, because although the resistivity of air is extremely high, it's not ...


1

In a perfectly ideal circuit without any resistive losses (even internal to the battery), energy will flow to fill capacitance with charge, and be stored as an electrical field but the transient flow will also create magnetic fields in any inductance. Without loss, the energy will bounce back and forth between the inductance and capacitance as magnetic and ...


0

Since this is a homework question, I will point you in the right direction rather than do your work for you. Reading the question carefully, I believe you are trying to solve the following situation: There is nothing ambiguous here. You have two "red" charges (identical, 1.8 uC), and need to determine the value of the "green" charge such that the ...


1

Dielectric heating is the priciple of a microwave oven. Water $H_2O$ has a strong dipole moment. Since the water molecule is not linear and the oxygen atom has a higher electronegativity than hydrogen atoms, the oxygen atom carries a slight negative charge, whereas the hydrogen atoms are slightly positive. As a result, water is a polar molecule with an ...


-3

I assume that when you say 'microwave' you are talking about a microwave oven. As you say sand unlike water dense foods cannot conduct electricity. But in a microwave the foods are not heated up using electricity but by bombarding it with microwaves. The microwave use electricity to generate microwaves of frequency 2.4 GHz. In your question you have already ...


1

The answer-as requested by HDE. So my foolish misconstruct of the electric field is that since it accelerates charged particles more or less depending on the distance from the source, it should create less motion in electrons a further distance from the source. I'm not sure why I thought resistance had much to do with it-on further analysis resistance ...


0

The key think you need to know is that if you apply a voltage $V$ and a current $I$ flows then the power dissipated is: $$ P = VI \tag{1} $$ You know the power at 220V so you can calculate the current, and if we assume the kettle behaves as a simple resistor then it will obey Ohm's law so you can calculate the resistance. $$ R = \frac{V}{I} \tag{2} $$ ...


4

This table should answer your question which is about intermolecular interactions. The largest distance that the electromagnetic forces can be effective for neutral atoms and molecules is in fractions of a nanometer. When in classical distances then the electric field coming from neutral atoms is effectively zero. Nano structures though, if you make your ...


4

I need to know why that is the answer. (Jump to the summary if you want to skip the details). As you may know, a transformer is essentially two inductors that share most if not all of the magnetic flux linkage. Recall the formula for the voltage across an isolated inductor $$v_L(t) = L \frac{di_L}{dt}$$ Now consider two inductors, $L_1$ and $L_2$ ...


0

I'm going to write a really simple answer... A short summary of what goes on in a step-up transformer: In an AC source, both the magnitude and direction of current changes As the primary windings of the transformer is connected to an AC source, there will be a change in magnetic flux (remember: the magnetic field strength around the coil is dependant on ...


1

The transformer operation is based on Faraday's Law $$ \nabla\times\mathbf{E} = -\dfrac{\partial\mathbf{B}}{\partial t} $$ This law relates the generated electric field (resulting in electromotive force in a circuit) with the variation of the magnetic flux density. Also indicates that a changing electric field generates a density also varying magnetic ...


-1

Your thought is correct if you talk about an isolated system with a DC generator and a capacitor instead of a chemical battery. After the magnetic force moved some electrons through the generator from one side of the generator to the over the process get stopped because the electrostatic forces in the atoms of the capacitor rise exponential. If you place ...


0

The potential difference exists between the two plates of the same battery. You don't know for sure if the potential difference exists between two plates from different batteries. Additionally, you're not allowing any current to flow even if there is some potential difference. In other words, you have not closed the circuit. Current can't flow in a circuit ...


0

In circuit analysis we make some assumptions and we use shorthand notations frequently. For example we assume the potential doesn't vary anywhere in the wire, even it was 1 km long. Hence we don't write the spatial coordinates like we do in electromagnetics, in that they complicate the analysis and don't give much accuracy (the dimensions of the wire and ...


3

The electric potential is always a function of both spatial position and time, in both circuit theory and electrodynamics. $V$ is constant along a wire, so in circuit analysis you can take a short cut by specifying the position by just specifying which wire you're talking about, like $V_{1}(t)$, instead of specifying three spatial coordinates. $V$ doesn't ...


1

When you consider a voltage $V(t)$ in a circuit, you are talking about a voltage at a specific point in the circuit - in other words, implicitly it's $V(t,x,y,z)$ - at a certain time & place. When you have a static situation (things don't change over time) it's possible to talk about a potential as a function of location: $V(x,y,z)$ describing the ...


1

It is a big misunderstanding that we name radio waves and photons electromagnetic radiation and this masked the different nature of this two phenomenons. A single photon or a serie of photons we get by acceleration of a electron (or the nucleus). This we call emission. A single photon travels as a single object through space until he gets absorbed by an ...


4

The radio waves or microwaves that are used for communication don't contain just one photon. They contain a bunch. (Maybe someone will do the math for how many photons a standard radio broadcast antenna is producing each second; it'll blow your knee-high off even if you're wearing sandals over them.) Consider for example a frequency-modulated signal. The ...


1

The other answers are good (especially the I = (V1 - V2) / (R1 + R2) equation that we will use) but I just wanted to give you ballpark estimates of some numbers that you can expect to see. Imagine that you are going to do this to a 9V battery and a 1.2V AA battery, then: V1 - V2 = 7.8V For internal resistances, it's hard to put ballpark numbers in the ...


5

Definitely! Consider Ohm's law, $\vec j=\sigma\vec{E}$, where $\vec j$ is the current density, $\sigma$ is conductivity (the inverse of resistivity) and $\vec E$ is the electric field. Anisotropic conductivity corresponds to turning $\sigma$ into a tensor-valued quantity, a $3\times3$ matrix. Ohm's law is then given by $$j_i=\sigma_{ij}E_j,$$ where ...


4

To a degree this is matter of terminology. Resistance is a scalar quantity, but it is derived from the resistivity, which is a second rank tensor and is anisotropic in many materials. For an isotropic material we have the usual formula: $$ R = \rho\frac{\ell}{A} $$ and the usual: $$ V = IR $$ where $\rho$ is the (isotropic) resistivity, $\ell$ is the ...


1

This is a parallel circuit, not a series one. So the currents not need be the same, but the potencial difference is the same instaed. In a series circuit the current is the same but the potencial is different in the many elements of the circuit. You can think about this like the current is a flow of water, since is parallel, the current (the flow) "divides" ...


1

The two resistors are in parallel. This means that at $A$ the current splits between them relating to their reistance. So the current throw the top resistor is $3\,A$ and throw the bottom resistor is $1 \,A$. If we use Kirchhoff's current law which states, that in any node (like $A$) the current flowing into the node is equal to the current flowing out of ...


3

The $KV$ in $30KV$ refers to the motor velocity constant $K_v$. It does not refer to kilovolt $kV$, cf. comments by John Rennie.


3

It is quite easy to use a step-up converter to generate almost any voltage. The question is "voltage at what current". The power a battery can deliver is finite (power = voltage times current), but you can convert voltages in many different ways. The most obvious is an oscillator (inverter) followed by a transformer and a rectifier, but there are more ...


1

I seriously doubt that the batteries were putting out 30 kV. You surely misread or misheard something. The chemistry of batteries is such that individual cells produce from a few 100 mV to a few volts. A 30 kV battery would require 1000s of cells, which would make no sense at all. In addition, 30 kV is much more difficult to handle and would be much less ...


1

yes, you will have to pay more if your load is inductive. most of energy meters work with the voltage and current to calculate energy. When we have inductive load it takes more current than resistive load to produce same power or output. P=V.I.Cosx where x is the phase angle between voltage(V) and current(I). if the phase angle x goes higher, the power ...


1

Another useful analogy, apart from the gravity one described by David Z, is temperature. You can think temperature as your potential, and the heat flow as your current. Two points of space may be at different temperature, but if they are correctly insulated, they won't exchange heat. The heat will flow only if they are connected somehow. For the current is ...


0

A load is only purely inductive as long as you aren't taking any energy out of it. So a purely inductive load would certainly give us a low (zero!) electricity bill, but it wouldn't be much use as a domestic appliance since it couldn't do anything. Once your appliance starts doing any work it is no longer a purely inductive load. If an appliance does work, ...


3

The analogy of electricity to flowing water may come in handy here. In this analogy, a potential difference is like a difference in height. One lake on top of a mountain and another in a valley, for example, might represent the two terminals of the battery, which are at different potentials. If you think about that situation, it's clear that no water flows ...


0

If you insert a dielectric in a circuit, you will not see any current but obviously there is a potential difference across the dielectric. To have a potential difference, you just need an electric field inside the material. This electric field might drive a current if the charges are mobile.


1

You are increasing the number of electrons in the wire, but only by a very small amount. There's a somewhat clichéd but still excellent analogy for electrical circuits called the hydraulic analogy. In the hydraulic analogy the power supply is a pump, and the pressure is the voltage. The water represents the electrons, so the pressure generated by the pump ...


0

No. The voltage is the chemical potential of photons instead to relating to electrons, in other words, an electromagnetic field is established across the wire which moves the electrons. Back to your question, the number of electrons does not change in the wire at the open-circuit scenario, aka the electrons in the wire can't flow into or from anywhere.


0

Voltage is potential difference. A higher voltage means that there is more energy that can be used from the same amount of current. In effect, increasing the voltage is roughly analogous to adding more potential energy per electron, as opposed to current, which is moving more electrons through the wire.


4

In principle, sure. That's what microphones are, as ACuriousMind points out. But if you want to power anything substantial, an important issue to overcome is the relatively small amount of energy contained in sound waves. According to this website, the front rows of a rock concert have a sound intensity of $10^{-1}~\text{W m}^{-2}$. So even if you had a ...


0

By Gauss' Law, $E\cdot A=\frac {q}{\epsilon_0}$ (assuming that the Electric Field is constant at every $dA$ and that it is always parallel to $dA$, which it is in this case) Let us define the charge contained in the original problem cylinder as being $Q$ whereas the charge in the smaller Guassian cylinder as being $q$. Therefore, the charge in the ...


1

Like Eternal Code said, using a cylinder inside the original problem cylinder is the right approach. If you use Gauss' Law, you should find that the electric field inside the infinitely long, uniformly charged cylinder is $$E=\frac{ρr}{(2ε_0)} $$ Now, to calculate the potential difference between the surface and axis of the cylinder, $${\Delta ...


2

Assuming the ground to be at $0 V$, there is a potential difference of $220V$ between the person's right hand and the ground. The only way to prevent an electric current to flow from his right hand to the ground is if the pathways of current between those two are insulated, blocking the path of current. Holding an insulator in the left hand does not prevent ...


0

Actually, using a cylinder for your Gaussian surface is your best approach. The fact the area is infinite should not matter, if you expression the infinite length of the cylinder as a variable, say $l$. Noting that the Gaussian surface area, $A = 2\pi Rl$, and that $Q = \rho l$, the $l$ term should eventually cancel out in your working out.


0

A spherically symmetric charged body behaves as a point charge. By this I mean that the electric potential outside the sphere is the same as for a point charge at the centre of the sphere. This is a consequance of Gauss's law. So suppose we have a sphere of radius $R$ and charge $Q$, the potential at a distance $r > R$ from the centre of the sphere is ...


0

Classically, electrons do move in a conductor that is passing direct current – but much more slowly than you might think. Let's break this down: Current in a wire is defined as the amount of charge that passes through a cross-section of that wire in a single second. By this definition alone, it is clear that a current relies on the motion of some charged ...


0

When an electric field is applied on a conductor there exists a drift velocity for the electrons in the medium . Electrons in conductors exist in what is called the Fermi level, a band level common to the whole solid. Depending on the field they can accumulate on one side or the other leaving a positive charge from the combined field of the molecules on ...


0

Electrons in an (alternate) current move but do not move (translate) a lot, they oscillate 50 times every second. They flow if the current is direct. Temperature does not affect the oscillation of the electrons in a sesible way: it affects resistivity and consequently resistance slightly increases. But that depends greatly upon the material of the ...


0

SPECULATION: "Does the salt decompose during the process?" I suspect that it does. The salt ionizes easily and the ions would migrate under the influence of the electric field. In so doing, they will further ionize the air they traverse, creating a stream of charge carriers. As the electrodes are pulled further apart, the energy available for ...


1

Well, I have a way of remembering if you live in the UK - you always drive your motor on the left! The diagram is correct - it shows what force will be exerted by the external magnetic field on the current carrying wire. The direction of the force is found from the left hand rule: splay your thumb and first two fingers out so they are mutually at right ...


0

The image you've linked shows us how to find the magnetic field associated with a long current-carrying wire. But we're interested not in the field that the wire creates, but rather the field that the wire experiences. The fields that a charged particle generates don't influence itself (see, perhaps, this question). And so all we need to worry about is: what ...


1

I've found an explanation. As Duncan said, as the sum of the charges stays equal the greatest product of two integers with a certain sum is half of the sum when (for example 5+5=10 and 5x5=25 is the greatest product. NOTE: Here the sum is always even as we take two equal charges 'q' initially). Hence all other series yields a force lesser than the ideal ...


1

As long as your product $|q_{1}||q_{2}|$ remains the same as in the case where you had the equally charged spheres, then yes, you will get the same value for the angle (provided the masses are equal). This is because the electrostatic force acts equally on both charged spheres.


0

For a given amount of resistance (combined resistance of all the circuits in your computer, or home, or city), the amount of current which flows is proportional to the voltage. (I=V/R) When lightning strikes a line, it induces a voltage spike. Traditional circuit breakers are current-sensing devices (whether solid state or electromechanical). So, a ...


1

The Lorentz force on a charge in an electromagnetic field is $$F=q(E+v \times B) \ \ .$$ For an electron between the cylinders, $q$ is negative, and $E$ is defined as pointing outward, so the electron will experience a force radially inward. But due to the unfortunate sign convention used for currents, electrons flowing inward means that the conventional ...



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