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1

An electric dipole is some configuration of charge, namely two opposite charges separated by a distance. However, such a description is an effective, quite general description for various physical situations. For example, Carbon monoxide $CO$ can be described as a dipole. Relative to the oxygen-side of the bond there is slightly more negative charge located ...


0

Spark plugs are an open circuit. But there is an exceptional amount of voltage causing a spark to jump the gap, same as lighting.


0

The magnitude of the forces $q_1$ and $q_2$ exert on each other is equal. According to Coulomb's law, the magnitude of the force that a charge $q_1$ will experience due to a charge $q_2$ is $$|\mathbf F_{12}|=k_e{|q_1q_2|\over r^2}\ ,$$ where $k_e$ is Coulomb's constant and $r$ is the distance between the charges. But that equation is symmetrical in $q_1$ ...


-1

We need to create a battery that would instantly store a large amount of electricity at one time. Ex. When a bolt of lighting strikes it gives off a very large amount of power. However a battery needs time to take that energy and change it over to a chemical for storage. Lets say a bolt of lightning is 500 gallons of water and the battery that we ...


3

The answer is simple, the resistor doesn't know what voltage drop to provide, and to some extent it doesn't "care" either (unless it's so high that the current fries it but that's another issue). The "voltage drop" is only governed by the potential difference created by the generator (battery or other). If the circuit is open, the electrons have no path so ...


3

It's not a dumb question. The electrons at the negative side of the battery have energy, and will look for "any path" to get rid of that energy - move in the direction of an electric field. When you create a circuit, you move the potential of the positive terminal closer to the negative terminal. If you assume that "wire" has no resistance (for simplicity) ...


3

The individual resistances are all positive, so the sum $$ \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots \,,$$ is larger than the inverse of any of the individual resistances, and that means that the inverse of the sum is necessarily smaller than any of the resistances. No mucking around with the two-resistor form required.


3

Think about current flow. If we take each individual resistor and determine the current for the applied voltage, we get: $$I_T=\frac {V}{R_1} +\frac {V}{R_2} + ...$$ Dividing everything by the voltage give us: $$\frac {I_T}{V}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Which is the same as: $$\frac {1}{R_{eq}}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Since there ...


2

We can prove it by induction. Let $$ \frac{1}{R^{(n)}_{eq}} = \frac{1}{R_1} + \cdots+ \frac{1}{R_n} $$ Now, when $n=2$, we find $$ \frac{1}{R^{(2)}_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \implies R_{eq}^{(2)} = \frac{R_1 R_2}{R_1+R_2} = \frac{R_1}{1+\frac{R_1}{R_2}} = \frac{R_2}{1+\frac{R_2}{R_1}} $$ Since $\frac{R_1}{R_2} > 1$, we see that $R^{(2)}_{eq} ...


1

It comes from the fact that the voltage in any loop is required to sum to zero. Here is a description of the rule (Kirchhoff's Voltage Law). Your instructor may have told you to follow an electron for the current law, but it is not ideal if the current flows in the opposite direction on one leg of the loop. Determining the voltages on each leg is needed, so ...


2

Short answer: even in non-uniform fields, the speed won't change, but the guiding center can drift with some velocity. In a magnetic field (uniform or otherwise), the force on a charged particle is: $$ F = q\; \vec{v} \times \vec{B(x)} $$ The direction of this force will be tangential to the velocity vector because of the cross product (the result of the ...


0

Does your conductor have any resistivity? In that case the fluctuation-dissipation theorem applies. In the case that your conductor is a perfect superconductor, it would still couple inductively to the electromagnetic field around it, which, per 3rd law of thermodynamics must have a non-zero temperature. To remove those fluctuations, the total field volume ...


1

In the quantum mechanical description of a conductor all energy levels of the conductor are filled up to some specific energy level, called the Fermi level. This is because of the Pauli exclusion principle, which says that electrons with the same spin cannot occupy the same energy level and thus causes higher energy levels to be populated. Therefore, even at ...


0

I don't feel that any of the answers is truly satisfactory. Most merely offer analogies. If I turn on a lamp, the effect does propagate from one end of the connecting conductor to the other almost instantaneously. The effect, however, has not adequately been described as yet by the propagation of electromagnetic fields. Kirchhoff, Sommerfeld, and many others ...


0

If the circuit is closed, one after the other electrons start to move in the chain between the source and the sink approx. with the speed of light. This is because the electrons always feel each over and the signal about the start moves approx. with c. The same happens in a water pipe too. Try it. Of course you can't measure how fast the signal "open wider ...


0

As soon as you complete the circuit an electric field travels though the wire. $$E=\frac{V}{x}=\frac{F}{Q}$$ The electric field travels at the speed of light so for wires in use are too small that the electric field will take noticeable time to travel through the wire. As the electric field is constant around the wire and is (almost) instantly there in ...


0

Under the influence of an applied electric field, electrons in conductors actually do not move very fast, in regards to their bulk flow velocity. For instance, in copper the bulk drift speed of electrons is less than a millimeter per second. However, each electron (specifically, the conduction electrons) has an effective speed of over one million meters ...


-1

GOOD question and thinking But the thickness still effective whatever the value is If thickness value is big and battery has limited charges it will be discharged for example in one second If thickness is very very big for example 1000 times than before the battery will be discharged in one millisecond that's it!


1

I have to cite @CarlWitthoft from the comments. Intensity is not a term used in electric circuits. Intensity of current would not make any sense. However, I can vaguely make out what you are trying to ask: What you are trying to say is: "Two circuits are identical to the one in the figure, except for the fact that one has a nichrome wire and the other has a ...


0

Absolutely not. Part of the reason for this is because the brain's neural pathways and the dendrites of neurons are not compatible with the data transmission through electrical impulses. Electricity moves to the path of least resistance. When you touch another person after receiving a buildup of static electricity, the other person becomes the path of least ...


0

Suppose we have in static electricity two charges negative and positive equal in absolute value and we put electrons along the road between them ,we know that the electrical force or field is different from point to point between them because of distance according to columb law and since the force is different then every electron have different ...


2

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 ...


1

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 ...


2

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, ...



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